{"id":6112,"date":"2023-10-09T04:53:41","date_gmt":"2023-10-09T04:53:41","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=6112"},"modified":"2023-10-09T04:53:41","modified_gmt":"2023-10-09T04:53:41","slug":"ncert-solutions-class-10-maths-chapter-15-probability-ex-15-1","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-15-probability-ex-15-1\/","title":{"rendered":"NCERT Solutions Class 10 Maths Chapter 15 Probability Ex 15.1"},"content":{"rendered":"<h2 style=\"text-align: center;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 10 Maths\u00a0<\/span><\/strong><br \/>\n<strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Chapter &#8211; 15 (Probability)\u00a0<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">The NCERT Solutions in English Language for Class 10 Mathematics <strong>Chapter &#8211; 15 Probability <\/strong>Exercise 15.1 has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n<p><strong><span style=\"color: #000000;\">Chapter : 15 Probability<\/span><\/strong><\/p>\n<ul>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-15-probability-ex-15-2\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 15.2<\/a><\/span><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 15.1<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>1. Complete the following statements:<br \/>\n<\/strong><\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(i) <\/strong>Probability of an event E + Probability of the event \u2018not E\u2019 = ___________.<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(ii) <\/strong>The probability of an event that cannot happen is __________. Such an event is called ________.<strong><br \/>\n<\/strong><\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(iii) <\/strong>The probability of an event that is certain to happen is _________. Such an event is called _________.<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(iv) <\/strong>The sum of the probabilities of all the elementary events of an experiment is __________.<strong><br \/>\n<\/strong><\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(v) <\/strong>The probability of an event is greater than or equal to ___ and less than or equal to __________.<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>Solution &#8211;<br \/>\n<\/strong><\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(i)<\/strong> Probability of an event E + Probability of the event \u2018not E\u2019 =\u00a0<span style=\"text-decoration: underline;\"><strong>1<\/strong><\/span>.<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(ii)<\/strong> The probability of an event that cannot happen is\u00a0<span style=\"text-decoration: underline;\"><strong>0<\/strong><\/span>. Such an event is called\u00a0<span style=\"text-decoration: underline;\"><strong>an impossible event<\/strong><\/span>.<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(iii)<\/strong> The probability of an event that is certain to happen is\u00a0<span style=\"text-decoration: underline;\"><strong>1<\/strong><\/span>. Such an event is called\u00a0<span style=\"text-decoration: underline;\"><strong>a sure or certain event<\/strong><\/span>.<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(iv)<\/strong> The sum of the probabilities of all the elementary events of an experiment is\u00a0<span style=\"text-decoration: underline;\"><strong>1<\/strong><\/span>.<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(v)<\/strong> The probability of an event is greater than or equal to\u00a0<span style=\"text-decoration: underline;\"><strong>0<\/strong><\/span>\u00a0and less than or equal to\u00a0<span style=\"text-decoration: underline;\"><strong>1<\/strong><\/span>.<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>2. Which of the following experiments have equally likely outcomes? Explain.<br \/>\n<\/strong><\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(i) <\/strong>A driver attempts to start a car. The car starts or does not start.<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(ii) <\/strong>A player attempts to shoot a basketball. She\/he shoots or misses the shot.<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(iii) <\/strong>A trial is made to a Solution &#8211; a true-false question. The solution is right or wrong.<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(iv) <\/strong>A baby is born. It is a boy or a girl.<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>Solution &#8211;<br \/>\n<\/strong><\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(i)<\/strong> It is not an equally likely outcome because the probability of it depends on many factors.<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(ii)<\/strong> It is not an equally likely outcome because the probability depends on the ability of the player.<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(iii)<\/strong>\u00a0It is an equally likely outcome because the probability is equal in both cases, it can be either true or false.<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(iv)<\/strong> It is an equally likely outcome because the probability is equal in both cases, it can either be a boy or a girl.<br \/>\n<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>3. Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?<br \/>\n<\/strong><\/span><span style=\"font-family: Georgia, Palatino;\"><strong>Solution &#8211;\u00a0<\/strong>Tossing a coin is considered to be a fair way of deciding which team should get the ball at the beginning of a football game because it is an equally likely event, that is, the probability of getting heads is 1\/2 which is the same as the probability of getting tails 1\/2; hence, the probability for both the teams are equal.<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>4. Which of the following cannot be the probability of an event?<br \/>\n<\/strong><\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(A) <\/strong>2\/3,<strong> (B)<\/strong> -1.5, <strong>(C) <\/strong>15%, <strong>(D) <\/strong>0.7<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\"><strong>Solution &#8211;\u00a0 <\/strong><\/span><span style=\"font-family: Georgia, Palatino;\">The probability of any event (E) always lies between 0 and 1, i.e. 0 \u2264 P(E) \u2264 1. So, from the above options, option (B) -1.5 cannot be the probability of an event.<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>5. If P(E) = 0.05, what is the probability of \u2018not E\u2019?<br \/>\n<\/strong><\/span><span style=\"font-family: Georgia, Palatino;\"><strong>Solution &#8211;\u00a0 <\/strong><\/span><span style=\"font-family: Georgia, Palatino;\">We know that,<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P(E) + P(not E) = 1<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">It is given that, P(E) = 0.05<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">So, P(not E) = 1 &#8211; P(E)<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P(not E) = 1 &#8211; 0.05<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">\u2234 P(not E) = 0.95<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>6. A bag contains lemon-flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she will take out<br \/>\n<\/strong><\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(i) <\/strong>an orange-flavoured candy?<strong><br \/>\n<\/strong><\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(ii) <\/strong>a lemon-flavoured candy?<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>Solution &#8211;<br \/>\n<\/strong><\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(i)<\/strong> We know that the bag only contains lemon-flavoured candies.<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">So, The number of orange-flavoured candies = 0<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">\u2234 The probability of taking out orange-flavoured candies = 0\/1 = 0<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(ii)<\/strong> As there are only lemon-flavoured candies, P(lemon-flavoured candies) = 1 (or 100%)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>7. It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>Solution &#8211;\u00a0 <\/strong><\/span><span style=\"font-family: Georgia, Palatino;\">Let the event wherein 2 students having the same birthday be E<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">Given, P(E) = 0.992<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P(E) + P(not E) = 1<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P(not E) = 1 \u2013 0.992 = 0.008<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">\u2234 The probability that the 2 students have the same birthday is 0.008<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>8. A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is<br \/>\n<\/strong><\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(i) <\/strong>red?<strong><br \/>\n<\/strong><\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(ii) <\/strong>not red?<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\"><strong>Solution &#8211;\u00a0 <\/strong><\/span><span style=\"font-family: Georgia, Palatino;\">The total number of balls = No. of red balls + No. of black balls<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">So, the total number of balls = 5 + 3 = 8<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">We know that the probability of an event is the ratio between the number of favourable outcomes and the total number of outcomes.<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P(E) = (Number of favourable outcomes\/Total number of outcomes)<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(i)<\/strong> Probability of drawing red balls = P (red balls) = (no. of red balls\/total no. of balls) = 3\/8<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(ii)<\/strong> Probability of drawing black balls = P (black balls) = (no. of black balls\/total no. of balls) = 5\/8<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>9. A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be<br \/>\n<\/strong><\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(i) <\/strong>red?<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(ii) <\/strong>white?<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(iii) <\/strong>not green?<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>Solution &#8211;\u00a0 <\/strong><\/span><span style=\"font-family: Georgia, Palatino;\">The Total no. of balls = 5 + 8 + 4 = 17<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P(E) = (Number of favourable outcomes\/ Total number of outcomes)<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>(i)<\/strong> Total number of red balls = 5<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P (red ball) = 5\/17 = 0.29<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>(ii)<\/strong> Total number of white balls = 8<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P (white ball) = 8\/17 = 0.47<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>(iii)<\/strong> Total number of green balls = 4<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P (green ball) = 4\/17 = 0.23<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">\u2234 P (not green) = 1-P(green ball) = 1 &#8211; (4\/7) = 0.77<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>10. A piggy bank contains hundred 50p coins, fifty \u20b91 coins, twenty \u20b92 coins and ten \u20b95 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin<br \/>\n<\/strong><\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(i) <\/strong>will be a 50 p coin?<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(ii) <\/strong>will not be a \u20b95 coin?<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>Solution &#8211;\u00a0 <\/strong><\/span><span style=\"font-family: Georgia, Palatino;\">Total no. of coins = 100 + 50 + 20 + 10 = 180<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P(E) = (Number of favourable outcomes\/ Total number of outcomes)<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>(i)<\/strong> Total number of 50 p coin = 100<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P (50 p coin) = 100\/180 = 5\/9 = 0.55<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>(ii)<\/strong> Total number of \u20b95 coin = 10<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P (\u20b95 coin) = 10\/180 = 1\/18 = 0.055<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">\u2234 P (not \u20b95 coin) = 1-P (\u20b95 coin) = 1 &#8211; 0.055 = 0.945<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>11. Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish (see Fig. 15.4). What is the probability that the fish taken out is a male fish?<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6752\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex15.1-Que-1.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"200\" height=\"155\" \/><br \/>\n<\/strong><\/span><span style=\"font-family: Georgia, Palatino;\"><strong>Solution &#8211;\u00a0 <\/strong><\/span><span style=\"font-family: Georgia, Palatino;\">The total number of fish in the tank = 5 + 8 = 13<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">Total number of male fish = 5<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P(E) = (Number of favourable outcomes\/Total number of outcomes)<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P (male fish) = 5\/13 = 0.38<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>12. A game of chance consists of spinning an arrow which comes to rest, pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (see Fig. 15.5), and these are equally likely outcomes. What is the probability that it will point at<br \/>\n<\/strong><\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(i) <\/strong>8?<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(ii) <\/strong>an odd number?<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(iii) <\/strong>a number greater than 2?<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(iv) <\/strong>a number less than 9?<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6753\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex15.1-Que-12.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"129\" height=\"128\" \/><br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\"><strong>Solution &#8211;\u00a0 <\/strong><\/span><span style=\"font-family: Georgia, Palatino;\">Total number of possible outcomes = 8<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P(E) = (Number of favourable outcomes\/ Total number of outcomes)<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>(i)<\/strong> Total number of favourable events (i.e. 8) = 1<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">\u2234 P (pointing at 8) = \u215b = 0.125<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>(ii)<\/strong> Total number of odd numbers = 4 (1, 3, 5 and 7)<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P (pointing at an odd number) = 4\/8 = \u00bd = 0.5<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>(iii)<\/strong> Total numbers greater than 2 = 6 (3, 4, 5, 6, 7 and 8)<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P (pointing at a number greater than 4) = 6\/8 = \u00be = 0.75<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>(iv)<\/strong> Total numbers less than 9 = 8 (1, 2, 3, 4, 5, 6, 7, and 8)<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P (pointing at a number less than 9) = 8\/8 = 1<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>13. A die is thrown once. Find the probability of getting<br \/>\n<\/strong><\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(i) <\/strong>a prime number<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(ii) <\/strong>a number lying between 2 and 6<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(iii) <\/strong>an odd number<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>Solution &#8211;\u00a0 <\/strong><\/span><span style=\"font-family: Georgia, Palatino;\">Total possible events when a dice is thrown = 6 (1, 2, 3, 4, 5, and 6)<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P(E) = (Number of favourable outcomes\/ Total number of outcomes)<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>(i)<\/strong> Total number of prime numbers = 3 (2, 3 and 5)<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P (getting a prime number) = 3\/6 = \u00bd = 0.5<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>(ii)<\/strong> Total numbers lying between 2 and 6 = 3 (3, 4 and 5)<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P (getting a number between 2 and 6) = 3\/6 = \u00bd = 0.5<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>(iii)<\/strong> Total number of odd numbers = 3 (1, 3 and 5)<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P (getting an odd number) = 3\/6 = \u00bd = 0.5<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>14. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting<br \/>\n<\/strong><\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(i) <\/strong>a king of red colour<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(ii) <\/strong>a face card<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(iii) <\/strong>a red face card<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(iv) <\/strong>the jack of hearts<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(v) <\/strong>a spade<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(vi) <\/strong>the queen of diamonds<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>Solution &#8211;\u00a0 <\/strong><\/span><span style=\"font-family: Georgia, Palatino;\">Total number of possible outcomes = 52<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P(E) = (Number of favourable outcomes\/ Total number of outcomes)<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>(i)<\/strong>\u00a0Total number of the king of red colour = 2<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P (getting a king of red colour) = 2\/52 = 1\/26 = 0.038<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>(ii)<\/strong>\u00a0Total number of face cards = 12<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P (getting a face card) = 12\/52 = 3\/13 = 0.23<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>(iii)<\/strong>\u00a0Total number of red face cards = 6<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P (getting a king of red colour) = 6\/52 = 3\/26 = 0.11<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>(iv)<\/strong> Total number of jack of hearts = 1<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P (getting a king of red colour) = 1\/52 = 0.019<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>(v)<\/strong>\u00a0Total numbers of the king of spade = 13<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P (getting a king of red colour) = 13\/52 = \u00bc = 0.25<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>(vi)<\/strong>\u00a0Total number of the queen of diamonds = 1<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P (getting a king of red colour) = 1\/52 = 0.019<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>15. Five cards, the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.<br \/>\n<\/strong><\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(i) <\/strong>What is the probability that the card is the queen?<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(ii) <\/strong>If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>Solution &#8211;\u00a0 <\/strong><\/span><span style=\"font-family: Georgia, Palatino;\">Total number of cards = 5<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P(E) = (Number of favourable outcomes\/Total number of outcomes)<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>(i)<\/strong>\u00a0Number of queens = 1<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P (picking a queen) = \u2155 = 0.2<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>(ii)<\/strong>\u00a0If the queen is drawn and put aside, the total number of cards left is (5-4) = 4<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(a)<\/strong> Total numbers of ace = 1<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P (picking an ace) = \u00bc = 0.25<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>(b)<\/strong> Total number of queens = 0<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P (picking a queen) = 0\/4 = 0<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>16. 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>Solution &#8211;\u00a0 <\/strong><\/span><span style=\"font-family: Georgia, Palatino;\">Numbers of pens = Numbers of defective pens + Numbers of good pens<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">\u2234 Total number of pens = 132 + 12 = 144 pens<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P(E) = (Number of favourable outcomes\/Total number of outcomes)<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P(picking a good pen) = 132\/144 = 11\/12 = 0.916<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>17.<br \/>\n(i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?<br \/>\n<\/strong><\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?<\/strong><\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>Solution &#8211;<br \/>\n<\/strong><\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(i)<\/strong> Number of defective bulbs = 4<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">The total number of bulbs = 20<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P(E) = (Number of favourable outcomes\/ Total number of outcomes)<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">\u2234 Probability of getting a defective bulb = P (defective bulb) = 4\/20 = \u2155 = 0.2<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>(ii)<\/strong>\u00a0Since 1 non-defective bulb is drawn, then the total number of bulbs left is 19<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">So, the total number of events (or outcomes) = 19<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">Number of non-defective bulbs = 19 &#8211; 4 = 15<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">So, the probability that the bulb is not defective = 15\/19 = 0.789<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>18. A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears<br \/>\n<\/strong><\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(i) <\/strong>a two-digit number<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(ii) <\/strong>a perfect square number<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(iii) <\/strong>a number divisible by 5 <\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>Solution &#8211;\u00a0 <\/strong><\/span><span style=\"font-family: Georgia, Palatino;\">The total number of discs = 90<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P(E) = (Number of favourable outcomes\/ Total number of outcomes)<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>(i)<\/strong> Total number of discs having two digit numbers = 81<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">(Since 1 to 9 are single-digit numbers and so, total 2-digit numbers are 90-9 = 81)<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P (bearing a two-digit number) = 81\/90 = 9\/10 = 0.9<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>(ii)<\/strong> Total number of perfect square numbers = 9 (1, 4, 9, 16, 25, 36, 49, 64 and 81)<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P (getting a perfect square number) = 9\/90 = 1\/10 = 0.1<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>(iii)<\/strong> Total numbers which are divisible by 5 = 18 (5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85 and 90)<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P (getting a number divisible by 5) = 18\/90 = \u2155 = 0.2<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>19. A child has a die whose six faces show the letters as given below:<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6754\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex15.1-Que-19.jpeg\" alt=\"NCERT Class 10 Maths Solution\" width=\"251\" height=\"39\" \/><br \/>\n<\/strong><\/span><span style=\"font-family: Georgia, Palatino;\"><strong>The die is thrown once. What is the probability of getting<br \/>\n<\/strong><\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(i) <\/strong>A?<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(ii) <\/strong>D?<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>Solution &#8211;\u00a0 <\/strong><\/span><span style=\"font-family: Georgia, Palatino;\">The total number of possible outcomes (or events) = 6<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P(E) = (Number of favourable outcomes\/Total number of outcomes)<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>(i)<\/strong> The total number of faces having A on it = 2<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P (getting A) = 2\/6 = \u2153 = 0.33<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>(ii)<\/strong> The total number of faces having D on it = 1<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P (getting D) = \u2159 = 0.166<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>20. Suppose you drop a die at random on the rectangular region shown in Fig. 15.6. What is the probability that it will land inside the circle with a diameter 1m?<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6755\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex15.1-Que-20.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"289\" height=\"156\" \/><br \/>\n<\/strong><\/span><span style=\"font-family: Georgia, Palatino;\"><strong>Solution &#8211;\u00a0 <\/strong><\/span><span style=\"font-family: Georgia, Palatino;\">First, calculate the area of the rectangle and the area of the circle. Here, the area of the rectangle is the possible outcome, and the area of the circle will be the favourable outcome.<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">So, the area of the rectangle = (3 \u00d7 2) m<sup>2<\/sup>\u00a0= 6 m<sup>2<br \/>\n<\/sup><\/span><span style=\"font-family: Georgia, Palatino;\">and,<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">The area of the circle = \u03c0r<sup>2<\/sup>\u00a0= \u03c0(\u00bd)<sup>2<\/sup>\u00a0m<sup>2<\/sup>\u00a0= \u03c0\/4 m<sup>2<\/sup>\u00a0= 0.78<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">\u2234 The probability that die will land inside the circle = [(\u03c0\/4)\/6] = \u03c0\/24 or, 0.78\/6 = 0.13<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>21. A lot consists of 144 ball pens, of which 20 are defective, and the others are good. Nuri will buy a pen if it is good, but will not buy it if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that<br \/>\n<\/strong><\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(i) <\/strong>She will buy it?<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(ii) <\/strong>She will not buy it?<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>Solution &#8211;\u00a0 <\/strong><\/span><span style=\"font-family: Georgia, Palatino;\">The total number of outcomes, i.e. pens = 144<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">Given, the number of defective pens = 20<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">\u2234 The numbers of non defective pens = 144-20 = 124<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P(E) = (Number of favourable outcomes\/Total number of outcomes)<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>(i)<\/strong> Total number of events in which she will buy them = 124<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">So, P (buying) = 124\/144 = 31\/36 = 0.86<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>(ii)<\/strong> Total number of events in which she will not buy them = 20<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">So, P (not buying) = 20\/144 = 5\/36 = 0.138<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>22. Refer to Example 13. (i) Complete the following table:<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6756 size-full\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex15.1-Que-22.jpeg\" alt=\"NCERT Class 10 Maths Solution\" width=\"586\" height=\"113\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex15.1-Que-22.jpeg 586w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex15.1-Que-22-300x58.jpeg 300w\" sizes=\"auto, (max-width: 586px) 100vw, 586px\" \/><br \/>\n<\/strong><\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(ii) A student argues that \u2018there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability of 1\/11. Do you agree with this argument? Justify your Solution &#8211;\u00a0 <\/strong><\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>Solution &#8211;\u00a0 <\/strong><\/span><span style=\"font-family: Georgia, Palatino;\">If 2 dice are thrown, the possible events are:<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">So, the total number of events: 6 \u00d7 6 = 36<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>(i)<\/strong> It is given that to get the sum as 2, the probability is 1\/36 as the only possible outcomes = (1, 1)<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">For getting the sum as 3, the possible events (or outcomes) = E (sum 3) = (1, 2) and (2, 1)<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">So, P(sum 3) = 2\/36<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">Similarly,<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">E (sum 4) = (1,3), (3,1), and (2,2)<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">So, P (sum 4) = 3\/36<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">E (sum 5) = (1,4), (4,1), (2,3), and (3,2)<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">So, P (sum 5) = 4\/36<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">E (sum 6) = (1,5), (5,1), (2,4), (4,2), and (3,3)<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">So, P (sum 6) = 5\/36<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">E (sum 7) = (1,6), (6,1), (5,2), (2,5), (4,3), and (3,4)<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">So, P (sum 7) = 6\/36<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">E (sum 8) = (2,6), (6,2), (3,5), (5,3), and (4,4)<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">So, P (sum 8) = 5\/36<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">E (sum 9) = (3,6), (6,3), (4,5), and (5,4)<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">So, P (sum 9) = 4\/36<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">E (sum 10) = (4,6), (6,4), and (5,5)<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">So, P (sum 10) = 3\/36<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">E (sum 11) = (5,6), and (6,5)<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">So, P (sum 11) = 2\/36<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">E (sum 12) = (6,6)<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">So, P (sum 12) = 1\/36<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">So, the table will be as:<\/span><\/span><\/p>\n<table class=\"table table-bordered\" border=\"1\">\n<tbody>\n<tr>\n<td><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>Event:<\/strong> <\/span>Sum on 2 dice<\/span><\/td>\n<td><span style=\"color: #000000; font-family: Georgia, Palatino;\">2<\/span><\/td>\n<td><span style=\"color: #000000; font-family: Georgia, Palatino;\">3<\/span><\/td>\n<td><span style=\"color: #000000; font-family: Georgia, Palatino;\">4<\/span><\/td>\n<td><span style=\"color: #000000; font-family: Georgia, Palatino;\">5<\/span><\/td>\n<td><span style=\"color: #000000; font-family: Georgia, Palatino;\">6<\/span><\/td>\n<td><span style=\"color: #000000; font-family: Georgia, Palatino;\">7<\/span><\/td>\n<td><span style=\"color: #000000; font-family: Georgia, Palatino;\">8<\/span><\/td>\n<td><span style=\"color: #000000; font-family: Georgia, Palatino;\">9<\/span><\/td>\n<td><span style=\"color: #000000; font-family: Georgia, Palatino;\">10<\/span><\/td>\n<td><span style=\"color: #000000; font-family: Georgia, Palatino;\">11<\/span><\/td>\n<td><span style=\"color: #000000; font-family: Georgia, Palatino;\">12<\/span><\/td>\n<\/tr>\n<tr>\n<td><span style=\"color: #000000; font-family: Georgia, Palatino;\">Probability<\/span><\/td>\n<td><span style=\"color: #000000; font-family: Georgia, Palatino;\">1\/36<\/span><\/td>\n<td><span style=\"color: #000000; font-family: Georgia, Palatino;\">2\/36<\/span><\/td>\n<td><span style=\"color: #000000; font-family: Georgia, Palatino;\">3\/36<\/span><\/td>\n<td><span style=\"color: #000000; font-family: Georgia, Palatino;\">4\/36<\/span><\/td>\n<td><span style=\"color: #000000; font-family: Georgia, Palatino;\">5\/36<\/span><\/td>\n<td><span style=\"color: #000000; font-family: Georgia, Palatino;\">6\/36<\/span><\/td>\n<td><span style=\"color: #000000; font-family: Georgia, Palatino;\">5\/36<\/span><\/td>\n<td><span style=\"color: #000000; font-family: Georgia, Palatino;\">4\/36<\/span><\/td>\n<td><span style=\"color: #000000; font-family: Georgia, Palatino;\">3\/36<\/span><\/td>\n<td><span style=\"color: #000000; font-family: Georgia, Palatino;\">2\/36<\/span><\/td>\n<td><span style=\"color: #000000; font-family: Georgia, Palatino;\">1\/36<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(ii)<\/strong>\u00a0The argument is not correct as it is already justified in (i) that the number of all possible outcomes is 36 and not 11.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>23. A game consists of tossing a one-rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result, i.e. three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>Solution &#8211;\u00a0<\/strong><\/span><span style=\"font-family: Georgia, Palatino;\">The total number of outcomes = 8 (HHH, HHT, HTH, THH, TTH, HTT, THT, TTT)<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">Total outcomes in which Hanif will lose the game = 6 (HHT, HTH, THH, TTH, HTT, THT)<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P (losing the game) = 6\/8 = \u00be = 0.75<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>24. A die is thrown twice. What is the probability that<br \/>\n<\/strong><\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(i) <\/strong>5 will not come up either time?<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(ii) <\/strong>5 will come up at least once?<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\"><strong>[Hint: Throwing a die twice and throwing two dice simultaneously are treated as the same experiment] <\/strong><\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>Solution &#8211;\u00a0 <\/strong><\/span><span style=\"font-family: Georgia, Palatino;\">Outcomes are:<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">So, the total number of outcomes = 6 \u00d7 6 = 36<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>(i)\u00a0<\/strong><\/span><span style=\"font-family: Georgia, Palatino;\">Consider the following events.<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">A = 5 comes in the first throw,<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">B = 5 comes in the second throw<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P(A) = 6\/36,<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P(B) = 6\/36 and<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">P(not B) = 5\/6<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">So, P(not A) = 1 &#8211; (6\/36) = 30\/36 = 5\/6<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">\u2234 The required probability = (5\/6) \u00d7 (5\/6) = 25\/36<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>(ii)<\/strong> Number of events when 5 comes at least once = 11(5 + 6)<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">\u2234 The required probability = 11\/36<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>25. Which of the following arguments are correct and which are not correct? Give reasons for your Solution &#8211;<br \/>\n<\/strong><\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(i) If two coins are tossed simultaneously, there are three possible outcomes \u2013 two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is 1\/3<br \/>\n<\/strong><\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(ii) If a die is thrown, there are two possible outcomes \u2013 an odd number or an even number. Therefore, the probability of getting an odd number is 1\/2<\/strong><\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>Solution &#8211;<br \/>\n<\/strong><\/span><span style=\"font-family: Georgia, Palatino;\"><strong>(i)<\/strong> All the possible events are (H,H); (H,T); (T,H) and (T,T)<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">So, P (getting two heads) = \u00bc<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">and, P (getting one of each) = 2\/4 = \u00bd<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">\u2234 This statement is incorrect.<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(ii)<\/strong>\u00a0Since the two outcomes are equally likely, this statement is correct.<\/span><\/p>\n<p><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths\/\" target=\"_blank\" rel=\"noopener\"><b><i>Go Back To Chapters<\/i><\/b><\/a><\/span><\/p>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 10 Maths\u00a0 Chapter &#8211; 15 (Probability)\u00a0 The NCERT Solutions in English Language for Class 10 Mathematics Chapter &#8211; 15 Probability Exercise 15.1 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter : 15 Probability NCERT Class 10 Maths Solution Ex &#8211; 15.2 Exercise &#8211; 15.1 [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1043],"tags":[1045,1120,1121,1044,1049,1048],"class_list":["post-6112","post","type-post","status-publish","format-standard","hentry","category-class-10-maths","tag-class-10-ncert-mathematics-solutions","tag-ncert-class-10-mathematics-chapter-15-probability-solutions","tag-ncert-class-10-mathematics-exercise-15-1-solutions","tag-ncert-class-10-mathematics-solutions-class-10-ncert-solutions","tag-ncert-solutions-class-10-mathematics","tag-ncert-solutions-class-10-maths"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.9 (Yoast SEO v27.3) - 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