Download CMAT 2022 Probability Questions pdf by Cracku. Very Important Probability Questions for CMAT 2022 based on asked questions in previous exam papers. These questions will help your CMAT preparation. So kindly download the PDF for reference and do more practice.<\/p>\n
Download Probability Questions for CMAT<\/a><\/p>\n Get 5 CMAT mocks at just Rs.299<\/a><\/p>\n Download CMAT previous papers PDF<\/a><\/p>\n Question 1:\u00a0<\/b>A bag contains 16 eggs out of which 5 are rotten. The remaining eggs are in good condition. If two eggs are drawn randomly, what is the probability that exactly one of the eggs drawn is rotten ?<\/p>\n a)\u00a0$\\frac{11}{24}$<\/p>\n b)\u00a0$\\frac{13}{24}$<\/p>\n c)\u00a0$\\frac{65}{12}$<\/p>\n d)\u00a0$\\frac{17}{24}$<\/p>\n e)\u00a0$\\frac{7}{12}$<\/p>\n Question 2:\u00a0<\/b>Find the probability that a number from 1 to 300 is either divisible by 3 or 7?<\/p>\n a)\u00a0$\\frac{37}{75}$<\/p>\n b)\u00a0$\\frac{32}{75}$<\/p>\n c)\u00a0$\\frac{36}{75}$<\/p>\n d)\u00a0$\\frac{28}{75}$<\/p>\n e)\u00a0$\\frac{26}{75}$<\/p>\n Question 3:\u00a0<\/b>In a sample , if a person is picked up randomly, the probability that the person is a smoker is $\\frac{3}{5}$, and that of the person being male is $\\frac{1}{2}$ .What is the probability that the person is both male as well as a smoker ?<\/p>\n a)\u00a0$\\frac{10}{11}$<\/p>\n b)\u00a0$\\frac{1}{5}$<\/p>\n c)\u00a0$\\frac{3}{5}$<\/p>\n d)\u00a0Cannot be determined<\/p>\n e)\u00a0None of these<\/p>\n Question 4:\u00a0<\/b>Uma has three children, what is the probability that none of the children is a girl ?<\/p>\n a)\u00a0$\\frac{1}{2}$<\/p>\n b)\u00a0$\\frac{1}{16}$<\/p>\n c)\u00a0$\\frac{1}{3}$<\/p>\n d)\u00a0$\\frac{3}{4}$<\/p>\n e)\u00a0None of these<\/p>\n Question 5:\u00a0<\/b>A committee of 3 members is to be selected out of 3 men and 2 women What is the probability that the committee has at least one woman ?<\/p>\n a)\u00a0$\\frac{1}{10}$<\/p>\n b)\u00a0$\\frac{9}{20}$<\/p>\n c)\u00a0$\\frac{9}{10}$<\/p>\n d)\u00a0$\\frac{1}{20}$<\/p>\n e)\u00a0None of these<\/p>\n Download CMAT Previous Papers PDF<\/a><\/p>\n Take CMAT Mock Tests<\/a><\/p>\n Question 6:\u00a0<\/b>Out of 5 girls and 3 boys, a total of 4 children are to be randomly selected for a quiz contest. What is the probability that all the four children are girls?<\/p>\n a)\u00a0$\\frac{1}{14}$<\/p>\n b)\u00a0$\\frac{1}{7}$<\/p>\n c)\u00a0$\\frac{5}{17}$<\/p>\n d)\u00a0$\\frac{2}{17}$<\/p>\n e)\u00a0None of these<\/p>\n Question 7:\u00a0<\/b>A die is thrown twice. What is the probability of getting a sum 7 from both the throws?<\/p>\n a)\u00a0$\\frac{5}{18}$<\/p>\n b)\u00a0$\\frac{1}{18}$<\/p>\n c)\u00a0$\\frac{1}{9}$<\/p>\n d)\u00a0$\\frac{1}{6}$<\/p>\n e)\u00a0$\\frac{5}{36}$<\/p>\n Question 8:\u00a0<\/b>A bag contains 2 red, 3 green and 2 blue balls. 2 balls are to be drawn randomly. What is the probability that the balls drawn contain no blue ball ?<\/p>\n a)\u00a05\/7<\/p>\n b)\u00a010\/21<\/p>\n c)\u00a02\/7<\/p>\n d)\u00a011\/21<\/p>\n e)\u00a0None of these<\/p>\n Question 9:\u00a0<\/b>A bag A contains 4 green and 6 red balls. Another bag B contains 3 green and 4 red balls. If one a)\u00a013\/70<\/p>\n b)\u00a01\/4<\/p>\n c)\u00a06\/35<\/p>\n d)\u00a08\/35<\/p>\n e)\u00a0None of these<\/p>\n Question 10:\u00a0<\/b>In a box carrying one dozen of oranges one third have become bad.If 3 oranges taken out from the box random ,what is the probability that at least one orange out of the 3 oranges picked up is good ?<\/p>\n a)\u00a01\/55<\/p>\n b)\u00a054\/55<\/p>\n c)\u00a045\/55<\/p>\n d)\u00a03\/55<\/p>\n e)\u00a0None of these<\/p>\n Take CMAT Mock Tests<\/a><\/p>\n Question 11:\u00a0<\/b>There are 8 brown balls, 4 orange balls and 5 black balls in a bag. Five balls are chosen at random. What is the probability of their being 2 brown balls, 1 orange ball and 2 black balls ?<\/p>\n a)\u00a0$\\frac{191}{1547}$<\/p>\n b)\u00a0$\\frac{180}{1547}$<\/p>\n c)\u00a0$\\frac{280}{1547}$<\/p>\n d)\u00a0$\\frac{189}{1547}$<\/p>\n e)\u00a0None of these<\/p>\n Question 12:\u00a0<\/b>In a bag there are 4 white, 4 red and 2 green balls. Two balls are drawn at random. What is the probability that at least one ball is of green colour ?<\/p>\n a)\u00a0$\\frac{4}{5}$<\/p>\n b)\u00a0$\\frac{3}{5}$<\/p>\n c)\u00a0$\\frac{1}{5}$<\/p>\n d)\u00a0$\\frac{2}{5}$<\/p>\n e)\u00a0None of these<\/p>\n Question 13:\u00a0<\/b>A bag contains 3 white balls and 2 black balls. Another bag contains 2 white and 4 black balls. A bag and a ball are picked at random. What is the probability that the ball drawn is white ?<\/p>\n a)\u00a0$\\frac{7}{11}$<\/p>\n b)\u00a0$\\frac{7}{30}$<\/p>\n c)\u00a0$\\frac{5}{11}$<\/p>\n d)\u00a0$\\frac{7}{15}$<\/p>\n e)\u00a0$\\frac{8}{15}$<\/p>\n Question 14:\u00a0<\/b>A bag contains 4 red balls, 6 green balls and 5 blue balls. If three balls are picked at random, what is the probability that two of them are green and one of them is blue in colour ?<\/p>\n a)\u00a0$\\frac{20}{91}$<\/p>\n b)\u00a0$\\frac{10}{91}$<\/p>\n c)\u00a0$\\frac{15}{91}$<\/p>\n d)\u00a0$\\frac{5}{91}$<\/p>\n e)\u00a0$\\frac{25}{91}$<\/p>\n Question 15:\u00a0<\/b>In a bag, there are 6 red balls and 9 green balls. Two balls are drawn at random, what is the probability that at least one of the balls drawn is red ?<\/p>\n a)\u00a029\/35<\/p>\n b)\u00a07\/15<\/p>\n c)\u00a023\/35<\/p>\n d)\u00a02\/5<\/p>\n e)\u00a019\/35<\/p>\n Enroll to CAT 2022 Complete Course<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Out of the 16 eggs, 5 eggs are rotten and 11 eggs are in good condition.<\/p>\n According to the question, out of the two eggs drawn one is rotten and the other is in good condition.<\/p>\n Hence, required probability = $ \\frac{^5C_{1} * ^{11}C_{1}}{^{16}C_{2}} = \\frac{5*11}{16*15\/2} = \\frac{11}{24}$<\/p>\n Hence, option A is the right choice.<\/p>\n 2)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Numbers divisible by 3 = {3, 6, 9, …… 300} 3)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let’s assume the sample size is 100. Let the number of male smokers be x.<\/p>\n Total number of smokers = 3\/5 * 100 = 60<\/p>\n Number of men = 1\/2 * 100 = 50<\/p>\n Hence, number of male non-smokers is 50-x. Number of female smokers is 60-x and number of female non-smokers is x-10.<\/p>\n Hence, probability of a person picked at random being a smoker and a male = x\/100<\/p>\n As we do not know the value of x, we cannot determine the probability. Hence, option D.<\/p>\n 4)\u00a0Answer\u00a0(E)<\/strong><\/p>\n The number of possible combinations is 2 * 2 * 2 = 8<\/p>\n The probability that none of the children is a girl is 1\/8<\/p>\n Option e) is the correct answer.<\/p>\n 5)\u00a0Answer\u00a0(C)<\/strong><\/p>\n The number of ways = Two men and one woman + One man and two women = $\\frac{6+3}{20}$<\/p>\n = 9\/10<\/p>\n Hence, the correct option is C.<\/p>\n 6)\u00a0Answer\u00a0(A)<\/strong><\/p>\n The total number of ways in which 4 children can be selected is $^8C_4 = 70$<\/p>\n The number of favorable ways in which this can be done is $^5C_4 = 5$<\/p>\n Hence, the required number of ways equals $\\frac{5}{70} = \\frac{1}{14}$<\/p>\n 7)\u00a0Answer\u00a0(D)<\/strong><\/p>\n The number 7 can be obtained in 6 ways. (6,1), (1,6), (5,2), (2,5), (4,3), (3,4).<\/p>\n Required probability = 6\/36 = 1\/6<\/p>\n 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Total number of balls = 2 + 3 + 2 = 7<\/p>\n Total number of outcomes = Drawing 2 balls out of 7<\/p>\n = $C^7_2 = \\frac{7 \\times 6}{1 \\times 2} = 21$<\/p>\n Favourable outcomes = Drawing 2 balls out of 5 (so that none is blue)<\/p>\n =\u00a0$C^5_2 = \\frac{5 \\times 4}{1 \\times 2} = 10$<\/p>\n => Required probability = $\\frac{10}{21}$<\/p>\n 9)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Total balls in bag A = 4 + 6 = 10<\/p>\n Probability that ball is green = $\\frac{4}{10}$<\/p>\n Total balls in bag B = 3 + 4 = 7<\/p>\n Probability that ball is green = $\\frac{3}{7}$<\/p>\n => Required probability = $\\frac{4}{10} \\times \\frac{3}{7}$<\/p>\n = $\\frac{6}{35}$<\/p>\n 10)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Total number of oranges in the box = 12<\/p>\n Number of ways of selecting 3 oranges out of 12 oranges, n(S) = $C^{12}_3$<\/p>\n = $\\frac{12 \\times 11 \\times 10}{1 \\times 2 \\times 3} = 220$<\/p>\n Number of oranges which became bad = $\\frac{12}{3}=4$<\/p>\n Number of ways of selecting 3 oranges out of 4 bad oranges = $C^4_3 = C^4_1 = 4$<\/p>\n Number of desired selection of oranges, n(E) = 220 – 4 = 216<\/p>\n $\\therefore$ $P(E) = \\frac{n(E)}{n(S)}$<\/p>\n = $\\frac{216}{220}= \\frac{54}{55}$<\/p>\n => Ans – (B)<\/p>\n 11)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Total number of balls in the bag = 8 + 4 + 5 = 17<\/p>\n P(S) = Total possible outcomes<\/p>\n = Selecting 5 balls at random out of 17<\/p>\n => $P(S) = C^{17}_5 = \\frac{17 \\times 16 \\times 15 \\times 14 \\times 13}{1 \\times 2 \\times 3 \\times 4 \\times 5}$<\/p>\n = $6188$<\/p>\n P(E) = Favorable outcomes<\/p>\n = Selecting 2 brown, 1 orange and 2 black balls.<\/p>\n => $P(E) = C^8_2 \\times C^4_1 \\times C^5_2$<\/p>\n = $\\frac{8 \\times 7}{1 \\times 2} \\times 4 \\times \\frac{5 \\times 4}{1 \\times 2}$<\/p>\n = $28 \\times 4 \\times 10 = 1120$<\/p>\n $\\therefore$ Required probability = $\\frac{P(E)}{P(S)}$<\/p>\n = $\\frac{1120}{6188} = \\frac{280}{1547}$<\/p>\n 12)\u00a0Answer\u00a0(D)<\/strong><\/p>\n There are 4 white, 4 red and 2 green balls and two balls are drawn at random.<\/p>\n Total possible outcomes = Selection of 2 balls out of 10 balls<\/p>\n = $C^{10}_2 = \\frac{10 * 9}{1 * 2} = 45$<\/p>\n Favourable outcomes = 1 green ball and 1 ball of other colour + 2 green balls<\/p>\n = $C^2_1 \\times C^8_1 + C^2_2$<\/p>\n = 2*8 + 2 = 18<\/p>\n $\\therefore$ Required probability = $\\frac{18}{45} = \\frac{2}{5}$<\/p>\n 13)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Probability of choosing bag 1 = (1\/2) 14)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Probability of drawing blue ball in first attempt = 5\/15<\/p>\n Probability of drawing two green balls in the next two attempts = (6\/14)(5\/13) Probability of drawing green ball in first attempt = 6\/15 Probability of drawing two green balls in first two attempts = (6\/15)(5\/14) Probability of drawing 2 red balls and 1 green ball= 150\/2730 + 150\/2730 + 150\/2730 = 3(150\/2730) = 150\/910 = 15\/91<\/span><\/span> 15)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Probability that at least 1 ball is red = 1 – probability that none of them is red.
\nball is drawn from each bag, and the probability that both are green.<\/p>\n
\nNumbers divisible by 7 = {7,14, …….. 294}
\nNumbers divisible by both 3 and 7 (i.e. multiples of 21) = {21, 42,…. 294}
\nRequired probability = $\\frac{100 + 42 – 14}{300} = \\frac{128}{300} = \\frac{32}{75}$<\/p>\n
\n= $\\frac{3C2 * 2C1}{5C3} + \\frac{3C1 * 2C2}{5C3}$<\/p>\n
\nProbability of choosing bag 2 = (1\/2)
\nProbability of choosing white ball from bag 1 = 3\/5
\nProbability of choosing white ball from bag 2 = 2\/6
\nProbability of choosing bag 1 and white ball from it = (1\/2)(3\/5) = 3\/10
\nProbability of choosing bag 2 and white ball from it = (1\/2)(2\/6) = 2\/12
\nProbability of choosing a bag and drawing a white ball = (3\/10) + (2\/12) = (28\/60) = (7\/15)
\nOption D is the correct answer.<\/span><\/span><\/p>\n
\nProbability of drawing 2 green and 1 blue ball = (5\/15)(6\/14)(5\/13) = 150\/2730<\/p>\n
\nProbability of drawing blue ball in the next attempt = (5\/14)
\nProbability of drawing green ball in the next attempt = (5\/13)
\nProbability of drawing 2 red and 1 green ball = (6\/15)(5\/14)(5\/13) = 150\/2730<\/p>\n
\nProbability of drawing blue ball in the next attempt =(5\/13)
\nProbability of drawing 2 red and 1 green ball = (6\/15)(5\/14)(5\/13) = 150\/2730<\/p>\n
\nOption C is the correct answer<\/span><\/p>\n
\nProbability that none if the two balls is red = (9\/15)(8\/14)
\nProbability that at least 1 ball is red = 1 – probability that none of them is red. = 1- [(9\/15)(8\/14)] = (210-72)\/210
\n= 138\/210
\n=23\/35
\nOption C is the correct answer.<\/p>\n