Here you can download a free Probability questions PDF with answers for MAH MBA CET 2022 by Cracku. These are some tricky questions in the MAH MBA CET 2022 exam that you need to find the Probability of answers for the given questions. These questions will help you to make practice and solve the Probability<\/strong>\u00a0questions in the MAH MBA CET exam. Utilize this best PDF practice set<\/strong> which is included answers in detail. Click on the below link to download the Probability MCQ<\/strong> PDF for MBA-CET 2022 for free.<\/p>\n Download Probability Questions for MAH-CET<\/a><\/p>\n Enroll to MAH-CET 2022 Crash Course<\/a><\/p>\n Question 1:\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 1)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/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 Instructions<\/b><\/p>\n Study the information carefully to answer the following questions:<\/p>\n A bucket contains 8 red, 3 blue and 5 green marbles.<\/p>\n Question 2:\u00a0<\/b>If 3 marbles are drawn at random, what is the probability that none is red ?<\/p>\n a)\u00a0${3 \\over 8}$<\/p>\n b)\u00a0${1 \\over {16}}$<\/p>\n c)\u00a0${1 \\over {10}}$<\/p>\n d)\u00a0${3 \\over {16}}$<\/p>\n e)\u00a0None of these<\/p>\n 2)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Number of ways of drawing 3 marbles out of 16<\/p>\n $n(S) = C^{16}_3 = \\frac{16 \\times 15 \\times 14}{1 \\times 2 \\times 3}$<\/p>\n = $560$<\/p>\n Out of the three drawn marbles, none is red, i.e., they will be either blue or green.<\/p>\n => $n(E) = C^8_3 = \\frac{8 \\times 7 \\times 6}{1 \\times 2 \\times 3}$<\/p>\n = $56$<\/p>\n $\\therefore$ Required probability = $\\frac{n(E)}{n(S)}$<\/p>\n = $\\frac{56}{560} = \\frac{1}{10}$<\/p>\n Question 3:\u00a0<\/b>If 2 marbles are drawn at random, what is the probability that both are green?<\/p>\n a)\u00a0${1 \\over 8}$<\/p>\n b)\u00a0${5 \\over {16}}$<\/p>\n c)\u00a0${2 \\over 7}$<\/p>\n d)\u00a0${3 \\over 8}$<\/p>\n e)\u00a0None of these<\/p>\n 3)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n Number of ways of drawing 2 marbles out of 16<\/p>\n $n(S) = C^{16}_2 = \\frac{16 \\times 15}{1 \\times 2}$<\/p>\n = $120$<\/p>\n Out of the two drawn marbles, both are green<\/p>\n => $n(E) = C^5_2 = \\frac{5 \\times 4}{1 \\times 2}$<\/p>\n = $10$<\/p>\n $\\therefore$ Required probability = $\\frac{n(E)}{n(S)}$<\/p>\n = $\\frac{10}{120} = \\frac{1}{12}$<\/p>\n Question 4:\u00a0<\/b>If 4 marbles are drawn at random, what is the probability that 2 are red and 2 are blue ?<\/p>\n a)\u00a0${{11} \\over {16}}$<\/p>\n b)\u00a0${3 \\over {16}}$<\/p>\n c)\u00a0${11 \\over {72}}$<\/p>\n d)\u00a0${3 \\over {65}}$<\/p>\n e)\u00a0None of these<\/p>\n 4)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Number of ways of drawing 4 marbles out of 16<\/p>\n => $n(S) = C^{16}_4 = \\frac{16 \\times 15 \\times 14 \\times 13}{1 \\times 2 \\times 3 \\times 4}$<\/p>\n = $1820$<\/p>\n Out of the four drawn marbles, 2 are red and 2 are blue.<\/p>\n => $n(E) = C^8_2 \\times C^3_2 = \\frac{8 \\times 7}{1 \\times 2} \\times \\frac{3 \\times 2}{1 \\times 2}$<\/p>\n = $28 \\times 3 = 84$<\/p>\n $\\therefore$ Required probability = $\\frac{n(E)}{n(S)}$<\/p>\n = $\\frac{84}{1820} = \\frac{3}{65}$<\/p>\n Instructions<\/b><\/p>\n Study the given information carefully and answer the questions that follow: Question 5:\u00a0<\/b>If four marbles are picked at random, what is the probability that one is green, two are blue and one is red ?<\/p>\n a)\u00a04\/15<\/p>\n b)\u00a017\/280<\/p>\n c)\u00a06\/91<\/p>\n d)\u00a011\/15<\/p>\n e)\u00a0None of these<\/p>\n 5)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Total number of marbles in the urn = 15<\/p>\n P(S) = Total possible outcomes<\/p>\n = Selecting 4 marbles at random out of 15<\/p>\n => $P(S) = C^{15}_4 = \\frac{15 \\times 14 \\times 13 \\times 12}{1 \\times 2 \\times 3 \\times 4}$<\/p>\n = $1365$<\/p>\n P(E) = Favorable outcomes<\/p>\n = Selecting 1 green, 2 blue and 1 red marble.<\/p>\n => $P(E) = C^2_1 \\times C^6_2 \\times C^3_1$<\/p>\n = $2 \\times \\frac{6 \\times 5}{1 \\times 2} \\times 3$<\/p>\n = $90$<\/p>\n $\\therefore$ Required probability = $\\frac{P(E)}{P(S)}$<\/p>\n = $\\frac{90}{1365} = \\frac{6}{91}$<\/p>\n Take Free MAH-CET mock tests here<\/a><\/strong><\/span><\/p>\n Enroll to 5 MAH CET Latest Mocks For Just Rs. 299<\/a><\/strong><\/span><\/p>\n Question 6:\u00a0<\/b>If two marbles are picked at random, what is the probability that either both are red or both are green ?<\/p>\n a)\u00a03\/5<\/p>\n b)\u00a04\/105<\/p>\n c)\u00a02\/7<\/p>\n d)\u00a05\/91<\/p>\n e)\u00a0None of these<\/p>\n 6)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Total number of marbles in the urn = 15<\/p>\n P(S) = Total possible outcomes<\/p>\n = Selecting 2 marbles at random out of 15<\/p>\n => $P(S) = C^{15}_2 = \\frac{15 \\times 14}{1 \\times 2}$<\/p>\n = $105$<\/p>\n P(E) = Favorable outcomes<\/p>\n = Selecting 2 green or 2 red marbles.<\/p>\n => $P(E) = C^2_2 + C^3_2$<\/p>\n = $1 + \\frac{3 \\times 2}{1 \\times 2}$<\/p>\n = $4$<\/p>\n $\\therefore$ Required probability = $\\frac{P(E)}{P(S)}$<\/p>\n = $\\frac{4}{105}$<\/p>\n Question 7:\u00a0<\/b>If four marbles are picked at random, what is the probability that at least one is yellow ?<\/p>\n a)\u00a091\/123<\/p>\n b)\u00a069\/91<\/p>\n c)\u00a0125\/143<\/p>\n d)\u00a01\/3<\/p>\n e)\u00a0None of these<\/p>\n 7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Total number of marbles in the urn = 15<\/p>\n P(S) = Total possible outcomes<\/p>\n = Selecting 4 marbles at random out of 15<\/p>\n => $P(S) = ^{15}C_4 = \\frac{15 \\times 14 \\times 13 \\times 12}{1 \\times 2 \\times 3 \\times 4}$<\/p>\n = $1365$<\/p>\n Let no yellow marble is selected.<\/p>\n P(E) = Favorable outcomes<\/p>\n = Selecting 4 out of 11 marbles.<\/p>\n => $P(E) = ^{11}C_4$<\/p>\n = $\\frac{11 \\times 10 \\times 9 \\times 8}{1 \\times 2 \\times 3 \\times 4}$<\/p>\n = $330$<\/p>\n $\\therefore$ Required probability = $1 – \\frac{P(E)}{P(S)}$<\/p>\n = $1 – \\frac{330}{1365} = 1 – \\frac{22}{91}$<\/p>\n = $\\frac{91 – 22}{91} = \\frac{69}{91}$<\/p>\n Question 8:\u00a0<\/b>If three marbles are picked at random, what is the probability that two are blue and one is yellow ?<\/p>\n a)\u00a02\/15<\/p>\n b)\u00a06\/91<\/p>\n c)\u00a012\/91<\/p>\n d)\u00a03\/15<\/p>\n e)\u00a0None of these<\/p>\n 8)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Total number of marbles in the urn = 15<\/p>\n P(S) = Total possible outcomes<\/p>\n = Selecting 3 marbles at random out of 15<\/p>\n => $P(S) = ^{15} C_3 = \\frac{15 \\times 14 \\times 13}{1 \\times 2 \\times 3}$<\/p>\n = $455$<\/p>\n P(E) = Favorable outcomes<\/p>\n = Selecting 2 blue and 1 yellow marble.<\/p>\n => $P(E) =C^6_2 \\times C^4_1$<\/p>\n = $\\frac{6 \\times 5}{1 \\times 2} \\times 4$<\/p>\n = $60$<\/p>\n $\\therefore$ Required probability = $\\frac{P(E)}{P(S)}$<\/p>\n = $\\frac{60}{455} = \\frac{12}{91}$<\/p>\n Question 9:\u00a0<\/b>If two marbles are picked at random, what is the probability that both are green ?<\/p>\n a)\u00a02\/15<\/p>\n b)\u00a01\/15<\/p>\n c)\u00a02\/7<\/p>\n d)\u00a01<\/p>\n e)\u00a0None of these<\/p>\n 9)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n Total number of marbles in the urn = 15<\/p>\n P(S) = Total possible outcomes<\/p>\n = Selecting 2 marbles at random out of 15<\/p>\n => $P(S) = C^{15}_2 = \\frac{15 \\times 14}{1 \\times 2}$<\/p>\n = $105$<\/p>\n P(E) = Favorable outcomes<\/p>\n = Selecting 2 green marbles.<\/p>\n => $P(E) = C^2_2 = 1$<\/p>\n $\\therefore$ Required probability = $\\frac{P(E)}{P(S)}$<\/p>\n = $\\frac{1}{105}$<\/p>\n Question 10:\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 10)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/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 Question 11:\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 11)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/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 Question 12:\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 12)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Probability of choosing bag 1 = (1\/2) Question 13:\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 13)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Probability that at least 1 ball is red = 1 – probability that none of them is red. 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 14)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/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> Question 15:\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 15)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/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 Question 16:\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 16)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/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 Instructions<\/b><\/p>\n Answer the following questions based on the information given below. Question 17:\u00a0<\/b>If three marbles are drawn at random, what is the probability that at least one is red?<\/p>\n a)\u00a0$\\frac{2}{7}$<\/p>\n b)\u00a0$\\frac{4}{91}$<\/p>\n c)\u00a0$\\frac{61}{91}$<\/p>\n d)\u00a0$\\frac{2}{13}$<\/p>\n e)\u00a0None of these<\/p>\n 17)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Probability that at least one marble is red = 1 – Probability that none of the balls are red<\/p>\n = 1 – $^{10}C_3 \/ ^{14}C_3$<\/p>\n = 1 – 720 \/ 2184 = 1 – 30\/91 = 61 \/ 91<\/p>\n Question 18:\u00a0<\/b>If three marbles are drawn at random, what is the probability that none of them are black ?<\/p>\n a)\u00a0$\\frac{6}{13}$<\/p>\n b)\u00a0$\\frac{3}{91}$<\/p>\n c)\u00a0$\\frac{5}{91}$<\/p>\n d)\u00a0$\\frac{3}{13}$<\/p>\n e)\u00a0None of these<\/p>\n 18)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n The total number of marbles present is $5+4+3+2 = 14$ If no marble is black, the three marbles should be picked from $4+3+2 = 9$ marbles. Hence, the required probability is $\\frac{84}{364} = \\frac{21}{91} = \\frac{3}{13}$<\/p>\n Question 19:\u00a0<\/b>If four marbles are drawn at random, what is the probability that two are red and two are blue ?<\/p>\n a)\u00a0$\\frac{6}{1001}$<\/p>\n b)\u00a0$\\frac{1}{143}$<\/p>\n c)\u00a0$\\frac{1}{13}$<\/p>\n d)\u00a0$\\frac{6}{91}$<\/p>\n e)\u00a0None of these<\/p>\n 19)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Probability of drawing two red and two blue balls = $^4C_2 * ^2C_2 \/ ^{14}C_4$ = 6 \/ 1001<\/p>\n Option a) is the correct answer.<\/p>\n Question 20:\u00a0<\/b>If two marbles are drawn random, what is the probability that they are green ?<\/p>\n a)\u00a0$\\frac{4}{91}$<\/p>\n b)\u00a0$\\frac{51}{91}$<\/p>\n c)\u00a0$\\frac{4}{13}$<\/p>\n d)\u00a0$\\frac{1}{13}$<\/p>\n e)\u00a0None of these<\/p>\n 20)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n Probability of drawing 2 green = $^3C_2 \/ ^{14}C_2$ = 3 \/ (7*13) = 3 \/ 91<\/p>\n
\nAn urn contains 3 red, 6 blue, 2 green and 4 yellow marbles.<\/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 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
\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
\nA bowl contains 4 red, 3 green, 2 blue and 5 black marbles.<\/p>\n
\nThe number of ways of picking three marbles from them is $^{14}C_3 = 364$<\/p>\n
\nThe number of ways in which this can be done is $^9C_3 = 84$<\/p>\n