Perfect square numbers from 5 to 74 are 9,16, 25, 36, 49, 64. No. It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. Let total number of balls = x 1. Let E be the event of having the same birthday. ∴ Favourable number of elementary events = 2 The number divisible by 3 and 5 in given number is 15. Question 4. = \(\frac{25}{50}\) = \(\frac{1}{2}\), (ii) A perfect square number Three different coins are tossed together. What is the probability of getting a card labelled 'S' when the card is chosen at random? (v) at least two tails The sample space is {HH, HT, TH, TT} Solution: Prime numbers less than 15 are 3, 5, 7,11,13. A box contains 70 cards numbered from 1 to 70. The function \(\frac{a}{b}\) is formed, where ‘a’ is the number of sector on which arrow stops on the first spin and ‘b’ is the number of the sector in which the arrow stops on second spin. If 2 more red balls are put in the bag, the probability of drawing a red ball will be 9/8 times the probability of drawing a red ball in the first case. P(a perfect square) =6/70=3/35, 2010 After removing all the red face cards from a pack of playing cards, total number of cards= 52-6 = 46, Question 21. 8 cases. P (a prime number on each die) = \(\frac{9}{36}\) or \(\frac{1}{4}\) ∴ Probability for Apoorv getting the number 36 ∴ Total number of balls in the bag = (5 + x) Favourable outcome is TT; (iii) Probability … A coin is tossed two times. Elementary events associated to random experiment of tossing three coins are No. Students are advised to solve the Probability Multiple Choice Questions of Class 9 Maths to know different concepts. Find the probability of getting such numbers on the two dice, whose product is 12. The outcomes associated with this experiment are given by Number of favourable outcomes when card is neither red nor queen = 28. Total number of cases of Ay are 16. Total outcomes are 4, i.e. What is the probability that it will point at: (i) 8? Solution: Out of 52 cards, one card can be drawn in 52 ways. ∴ Probability that arrow points at a number less than 9 = \(\frac{8}{8}\) = 1. Find the probability that the card selected will be: (i) an even number (ii) a multiple of 3 (iii) an even number and a multiple of 3 Write down all the possible outcomes. If one ball is drawn at random from the bag, find the probability that it is not red. What is the probability that, Question 7. Number of outcomes when product is even = 27 [(1, 2), (1, 4) … (6,6)], Question 1. Total no. Find the probability that the drawn card is. Solution: (ii) Probability that 5 will come atleast once = 1 – P (5 will not come up either time) of possible outcomes = 60 – 11 + 1 = 50. False, because the outcome 3 is more likely than the other numbers. Also, there are 4 kings, two red and two black. ∴ P(E) = \(\frac{3}{6}\) = \(\frac{1}{2}\) Face cards are taken as (4 jacks, 4 queens and 4 kings): Solution: Find the probability of getting at least one tail. Favourable cases of getting not more than one head are 3, i.e. A ball is drawn from the bag at random. ∴ Favourable number of elementary events = 13 Solution: A bag contains 5 red balls and some blue balls. [(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3,6), (4, 1), (4,2), (4, 3), (4, 4), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 6)] Number of cards with a prime number are 11 (i.e. Favourable cases when both numbers are prime are (2, 2), (2, 3), (2, 5), (3, 2), (3, 3),(3, 5), (5, 2), (5, 3), (5,5), i.e. Question 23. Total number of cases while tossing a coin two times are 4, i.e. ∴ Probability of getting red face card = \(\frac{6}{52}\) = \(\frac{3}{26}\), (iv) Since, there is only one jack of hearts. Cards numbered from 11 to 60 are kept in a box. Probability Class 10 Extra Questions Very Short Answer Type Question 1. From a well-shuffled pack of cards, a card is drawn at random. As we know that, Find the probability that product of x and y is less than 16. P (a total of 9 or 11) = \(\frac{6}{36}\) or \(\frac{1}{6}\), Question 14. B : Number on the cards is divisible by 5. (2011D) Solution: S = {HH, HT, TH, TT) = 4 P (both heads or both tails) = P (both heads) + P (both tails) = \(\frac{1}{4}+\frac{1}{4}=\frac{2}{4}=\frac{1}{2}\) Question 11. Number of extremely honest = 6 Total number of balls = 5 + 8 + 7 = 20 (i) Face Card Number of favourable outcomes Number of favourable outcomes = 3 ∴ Probability for Peehu getting the number 36 Required probability=3/4, Question 46. Solution: Probability Class 9 MCQs Questions with Answers. Find the probability of getting the following: Solution: Find the probability that the arrow will point at any factor of 8. All the black face cards are removed from a pack of 52 playing cards. Total outcomes = 36 In figure is shown a disc on which a player spins an arrow twice. of cards in the box = 18 What is the likelihood that both cards are clubs? Solution: Now, P(R) + P(B) + P(W) = 1 (b) a queen? Harpreet tosses two different coins simultaneously (say, one is of 1 and other of 2). ∴ Favourable number of outcomes = 8 1,4,9,16, 25, 36,49), Number of face cards in remaining cards = 6 (i.e. Therefore, favourable number of outcomes = 6 So, its probability is 0. Apoorv throws two dice once. A box contains 80 discs which are numbered from 1 to 80. Favourable outcomes of drawing a queen = 0 (as there is no card of queen). Total possible outcomes = 36, Question 10. Solution: We know that, P(E)+P(not E) = 1. (5.1),(5, 2),(5, 3),(5,4),(5, 5),(5, 6) P(at least one head)=3/4, Question 67. (i) The outcomes favourable to the event the sum of the two numbers is 8′ denoted by E, are : Total number of bulbs in the box = 400 Total number of cases = 36 Solution: A ball is drawn at random from the bag. Solution: Solution: Let S and R denote the events that Sangeeta and Reshma wins the match, respectively. (iii) a perfect square. P(Hanif will lose the game)=6/8=3/4, Question 72. Favourable cases to win the game are HHH or TTT, i.e. . A card is drawn at random from the remaining cards, after reshuffling them. Required probability=23/49, Question 49. Favourable outcomes for sum of the numbers appearing on two dice is 10 = 3, i.e. Find the probability that the drawn card is, Solution: Find the probability that product of JC and y is less than 16. (1, 4), (4,1), (2,3), (3,2)], Number of favourable cases of event B = 9 [i.e. Therefore, P(E) = \(\frac{3}{4}\) In the above four types, we will have 13 cards in each type. Solution: P(value of xy more than 16) =6/16=3/8, Question 14. Favourable cases when number is a perfect square and is divisible by 9 are 9, 36 and 81. (1, 1) (1, 2) (1,3) (1, 4) (1,5) (1,6) (2, 1) (2, 2) (2,3) Total number of outcomes = 36[(1, 1), (1, 2) … (6,6)] Solution: Solution: (2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6,4), (6, 6)]. (ii) Favourable outcomes are (3,6) (4, 5) (5, 4) (6, 3) (5, 6) (6, 5) i.e., 6 outcomes Now, Ramesh will lose the game if he gets A coin is tossed two times. A box contains 75 cards which are numbered from 1 to 75. Find the probability that in a leap year there will be 53 Tuesdays. .’. Number lying between 2 and 5 are 2 (i.e. Three coins are tossed simultaneously. Required Probability=8/20=2/25, Question 51. number of blue balls in the jar = y All the black face cards are removed from a pack of 52 playing cards. Example 1: Two cards are drawn randomly from a deck of cards. Find the probability that the number on the drawn card is: Question 27. Find the probability that the number on the drawn ticket is a multiple of 3 or 7. Let us see more. Find the probability that. a = 5, b can take 4 value, i.e. Find the probability that the number on the drawn card is: 1 and 2 Cards remaining after removing black face cards = red cards + black cards excluding face cards Solution: ∴ P(E) = \(\frac{3}{6}\) = \(\frac{1}{2}\), Question 5. Assuming that each person is equally likely to be selected, find the probability of selecting a person who is, Solution: One card is selected at random from a pack of 52 cards. (i) an odd number. Find the probability of getting at least one If the probability of getting a white ball is 3/10 and that of a black ball is 2/5,then find the probability of getting a red ball. Question 12. HHH, HHT, HTH, THH, TTH, THT, HTT, TTT Numbers from 1 to 80 which are perfect square are 1, 4, 9,16, 25, 36, 49, 64. A game consists of spinning an arrow which comes to rest pointing at one of the regions (1, 2 or 3) (Fig. Number of ways to draw a black card = 23 Therefore, number of ways of selecting a prize ticket = 5 . ∴ Favourable number of elementary events = 3 (i.e., 1, 3, 5) Find the probability of: Question 6. ∴ Probability of getting red marble = \(\frac{5}{17}\), (ii) Since, there are 8 white marbles in the box. 4 cards). If I toss a coin 3 times and get head each tir ne, then I should expect a tail to have a higher chance in the 4th toss. Hence, required probability = \(\frac{12}{52}\) = \(\frac{3}{13}\), (vi) There are 6 red face cards 3 each from diamonds and hearts. Find the probability that the drawn bulb is not defective. 20 tickets, on which numbers 1 to 20 are written, are mixed thoroughly and then a ticket is drawn at random out of them. MCQ Questions For Class 10 Maths Probability Question 2. Solution: a = 5, b can take 4 values, Favourable number of elementary events = 3 Find the probability that the drawn card is neither a king nor a queen. Cards marked with number 3, 4, 5, …., 50 are placed in a box and mixed thoroughly. Find the probability that the drawn card is, Solution: (ii) a number lying between 2 and 6. Favourable cases are (1,1), (1,4), (2,2), (3, 3), (4,1), (4,4), (5,5), (6, 6), Question 73. If the queen is drawn and put aside, and a second card is drawn, find the probability that the second card is, Favourable outcomes for drawing a queen = 1, When queen is kept aside, then remaining cards = 4, favourable outcomes of drawing an ace is 1. When two coins are tossed simultaneously, the possible outcomes are (H, H), (H, T), (T, H), (T, T) which are all equally likely. (ii) a perfect square number. Favourable outcomes when the required number is less than 9 are 1,2,3,4,5,6,7,8 i.e 8 outcomes, Number of favourable cases of event A = 4 [i.e. Question 26. Possible outcomes of drawing one card from 48 cards is 48. So, P(F) = \(\frac{0}{36}\) = 0. Number of total possible outcomes = 4 Determine the probability that the pen taken out is a good one. Solution: (i) red. (iii) P (neither white nor black) = P (Red balls) = \(\frac{5}{20}\) = \(\frac{1}{4}\). Number of favourable cases whose sum is 7 are 6, i.e. ∴ Total number of possible outcomes = 17. 10th Grade Worksheet on Probability |Probability Questions and Answers. Hence, required probability = \(\frac{1}{8}\), (ii) The event “getting two heads” will occur, if one of the elementary events HHT, THH, HTH occurs. The probability that the card selected is red card and black card is (1, 1) (3, 3) (5, 5) 1.e, 3 Extra Questions for Class 10 Maths Chapter 15 Probability. Another number y is selected at random from the numbers 1, 4, 9 and 16. Leap year = 366 days = (52 × 7 + 2) days = 52 weeks and 2 days. (ii) exactly two heads. What is the probability as getting at least one '4'? Favourable number of dies =8 If one disc is drawn at random from the box, find the probability that it bears a perfect square number. Ramesh, a shopkeeper will buy only those shirts which are good but ‘Kewal’ another shopkeeper will not buy shirts with major defects. 2 + 2 = 4 cards are removed from a pack of 52 playing cards. A game of chance consists of spinning an arrow on a circular board, divided into 8 equal parts, which comes to rest pointing at one of the numbers 1,2,3,…, 8 which are equally likely outcomes. Find the probability that the card drawn is an ace and black. (i) Since, there are 5 red marbles in the box. Total possible outcomes = 20 (vi) a red face card. Find the probability of getting at least one head. A box contains 100 red cards, 200 yellow cards and 50 blue cards. NCERT Solutions for Class 10 Maths Probability - Exercise 15.1 Q1: The first question contains five fill in the blanks, and you need to have a proper understanding of the text to answer the same. (i) an ace. The number of possible outcomes = 52 .’. Find the probability that the drawn card is of black colour. Find the probability of getting a black queen. Question 17. Total number of cards in the bag = 51. A shirt is taken out of the box at random. If a marble is drawn at random from the box, find the probability that the drawn marble is, Solution: Probability Exercise 25(B) – Selina Concise Mathematics Class 10 ICSE Solutions. State true or false and give the reason. So, favourable number of outcomes = 3 (i.e., 2, 3, 5) of favourable outcomes = 10 Suppose we throw a die once. Favourable number of elementary events = 12 2 Total possible outcomes = 36 Question 24. Two dice are rolled once. Two dice are thrown. Club - 13 cards 2. Two players, Sangeeta and Reshma, play a tennis match.
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