# CBSE Class 10th Math 15 – Probability MCQs

#### If a pair of dice is thrown simultaneously, then the possible outcomes are ____.

When a pair of dice is thrown simultaneously, the possible outcomes are (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5) and (6, 6). Total number of possible outcomes = 36

#### If two dice are thrown simultaneously, then the probability that the sum of the two numbers appearing on them is 6, is ____.

Total number of outcomes, when two dice are thrown simultaneously = 36 ⇒ Number of possible outcomes = 36 Let E be the event of getting the sum 6. The outcomes favorable to E are (1, 5), (5, 1), (2, 4), (4, 2) and (3, 3). ⇒ Number of outcomes favorable to E = 5 P(E) = Number of outcomes favorable to E/Number of all possible outcomes of the experiment = 5/36 Hence, the required probability is 5/36.

#### If two dice are thrown simultaneously, then the probability that the sum of the two numbers appearing on them is 12, is ____.

Total number of outcomes, when two dice are thrown simultaneously = 36 ⇒ Number of possible outcomes = 36 Let E be the event of getting the sum 12. The outcome favorable to E is (6, 6). ⇒ Number of outcomes favorable to E = 1 P(E) = Number of outcomes favorable to E/Number of all possible outcomes of the experiment = 1/36 Hence, the required probability is 1/36.

#### A box contains 100 toys. Out of which 84 are good, 12 have minor defects and 4 have major defects. Pavel, a shop keeper accepts only good toys and Valentina, another shop keeper rejects only the toys with major defects. If one toy is drawn at random from the box, then the probability that it is acceptable to Pavel is ____.

Total number of toys = 100 ⇒ Number of possible outcomes = 100 Let E be the event that a toy is acceptable to Pavel. Number of good toys = 84 ⇒ Number of outcomes favorable to E = 84 P(E) = Number of outcomes favorable to E/Number of all possible outcomes of the experiment = 84/100 = 0.84 Hence, the required probability is 0.84.

#### A box contains 12 balls, out of which 5 are black. If one ball is drawn at random, then the probability of getting a black ball is ____.

Total number of balls = 12 ⇒ Number of possible outcomes = 12 Let E be the event of getting a black ball. Number of black balls = 5 ⇒ Number of outcomes favorable to E = 5 P(E) = Number of outcomes favorable to E/Number of all possible outcomes of the experiment = 5/12 Hence, the required probability is 5/12.

#### John and Ray are friends. The probability that both will have the birthday on same day in a non-leap year is ____.

Number of days in a non-leap year = 365 ⇒ Total number of possible outcomes = 365 Let E be the event that John and Ray will have the birthday on same day. The number of favorable outcomes that John and Ray will have the same birthday is 1. P(E) = Number of outcomes favorable to E/Number of all possible outcomes of the experiment = 1/365 Hence, the probability that both will have different birthdays in a non-leap year is 1/365.

#### Two coins are tossed simultaneously for 1000 times. If the frequencies of two tails, one tail and no tail are 210, 550 and 240 respectively, then the probability of obtaining one tail is ____.

Number of possible outcomes = 1000 Let E be the event of getting one tail. Frequency of one tail in the experiment = 550 ⇒ Number of outcomes favorable to E = 550 P(E) = Number of outcomes favorable to E/Number of all possible outcomes of the experiment = 550/1000 = 0.55 Hence, the required probability is 0.55.

#### If one card is drawn from a well shuffled deck of 52 cards, then the probability of getting a red card is ____.

Total number of cards = 52 ⇒ Number of possible outcomes = 52 Let E be the event of getting a red card. Number of red cards = 26 ⇒ Number of outcomes favorable to E = 26 P(E) = Number of outcomes favorable to E/Number of all possible outcomes of the experiment =26/52 = 1/2 Hence, the required probability is 1/2.

#### If two dice are thrown simultaneously, then the probability of getting a doublet is ____.

Total number of outcomes, when two dice are thrown simultaneously = 36 ⇒ Number of possible outcomes = 36 Let E be the event of getting a doublet. The outcomes favorable to E are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6). ⇒ Number of outcomes favorable to E = 6 P(E) = Number of outcomes favorable to E/Number of all possible outcomes of the experiment = 6/36 = 1/6 Hence, the required probability is 1/6.