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3 Solved Probability Problems To Prepare For Accenture Placement Tests Dear Reader,Below are three problems based on probability, you have to find the minimum or maximum probabilityof the random event. Note : Consider there are two boxes. Separately they contain P number of item1 and Q number of item2. Arrangement is to maximize the probability of choosing item1 : Put 1number of item1 in one box and move remaining (P - 1) items to other box. Question 1 There are two boxes namely A and B. A contains 20 green and 15 blue balls and B contains 7green and 8 blue balls. You can move the balls between the two boxes. If you are allowed tochoose a box at random then what will be the maximum probability of getting a green ballfrom the chosen box? a) 55/58 b) 40/58 c) 41/58 d) 39/58 Solution : Total number of green balls = 20 + 7 = 27 and total number of blue balls = 15 + 8 = 23.Given that, we can move the balls between the boxes.(As discussed in above note) To get a maximum probability of choosing a green ball, we have to keep 1green ball in one box (box A) and the remaining 23 blue and 26 green balls in other box (box B).From the above arrangement, Probability to choose a green ball =Probability of choosing box A x Probability of choosing green ball from box A +Probability of choosing box B x Probability of choosing green ball from box B ...(1)Since there are only two boxes, the probability of choosing box A = probability of choosing box B = ½Probability of choosing a green ball from box A = number of green balls in box A / total number of ballsin box A = 1/1 = 1Probability of choosing a green ball from box B = number of green balls in box B / total number of ballsin box B = 26/(26+23) = 26/49 = 26/49.Required probability (from eqn(1)) = 1/2 x 1 + 1/2 x 26/29 = 1/2 (1 + 26/29) = (1/2) x (55/29) = 55/58.Hence the required maximum probability = 55/58. Question 2 There are three boxes namely A, B and C. A contains 20 – one rupee coins, B contains 12 – fiverupee coins and C contains 31 – two rupee coin. You can move the coins between the threeboxes. If you are allowed to choose a box at random then what will be the maximumprobability of getting a two rupee coin from the chosen box? a) 111/183 b) 169/183 c) 151/183 d) 171/183 Solution : Given that, we can move the coins between the boxes.(As discussed in above note) To get a maximum probability of choosing a two rupee coin, we have tokeep 1 – two rupee coin in 1st box (say A), another 1- two rupees coin in 2nd box (say B) and move theremaining 29 – two rupees coins with 20 – one rupee coins and 12 five rupee coins to the third box.Now, the probability of choosing a two rupee coin from the above arrangement =(Probability of choosing box A x probability of choosing a two rupee coin from A) +(Probability of choosing box B x probability of choosing a two rupee coin from B) + (Probability of choosing box C x probability of choosing a two rupee coin from C) …(1) Since there are three boxes, the probability of choosing box A = probability of choosing box B =probability of choosing box C = 1/3Probability of choosing a two rupee coin from A = number of two rupee coins in box A / total number of coins in box A = 1/1 = 1Probability of choosing a two rupee coin from B = number of two rupee coins in box B / total number of coins in box B = 1/1 = 1Probability of choosing a two rupee coin from C = number of two rupee coins in box C / total number of coins in box C = 29/(29+20+12) =29/61.Put the values in eqn(1),1/3 x 1 + 1/3 x 1 + 1/3 x 29/61 = 1/3 ( 1 + 1+ 29/61) = 1/3 ( 151/61) = 151/183 Question 3 There are three boxes, namely A, B and C. A contains 12 white and 6 green balls, B contains19 green and 6 orange balls and C contains 24 orange balls. You can move the balls between Band C. If you are allowed to choose a box at random then what will be the maximumprobability of getting a green ball from the chosen box? a) 71/72 b) 69/72 c) 41/72 d) 37/72 Solution : Total number of balls in box A = 12 + 6 = 18.Given that, we can move the balls between B and C.(As discussed in above note) We have to keep 1 green ball in box B and move the remaining 18 greenballs with 6 orange to box C to get a maximum probability of choosing a green ball.Now, total number of green balls in box B = 1Total number of green balls in box C = 18Total number of orange balls in box C = 6 + 24 = 30.Now, the probability of choosing a green ball from the above arrangement =Probability of choosing box A x Probability of choosing green ball from box A +Probability of choosing box B x Probability of choosing green ball from box B +Probability of choosing box C x Probability of choosing green ball from box C...(1)Since there are three boxes, the probability of choosing box A = probability of choosing box B =probability of choosing box C = 1/3Probability of choosing a green ball from box A = number of green balls in box A / total number of ballsin box A = 6/18 = 1/3.Probability of choosing a green ball from box B = number of green balls in box B / total number of ballsin box B = 1/1 = 1Probability of choosing a green ball from box C = number of green balls in box C / total number of ballsin box C = 18/(18+30) = 18/48 = 3/8.Put the values in eqn (1), 1/3 x 1/3 + 1/3 x 1 + 1/3 x 3/8 = 1/9 + 1/3 + 1/8 = 41/72. Wipro Sample Probability Questions Dear Reader, Below are three different types of probability questions (including one from conditionalprobability). Question 1 The exciting super final IPL match between Chennai Superkings and Delhi Daredevils was about to start.The thrill and excitement of the fans were shadowed by the forecast that it may rain. Before match itwas predicted that the probability of rain was 40%. If it rained, the Chennai Superkings had a 30%chance of winning. If it did not rain, the Chennai Superkings had a 55% chance of winning. If you areasked to calculate the overall probability of Chennai winning the match what will be your answer ?(a)58% (b)45% (c)55% (d)48%Solution:Probability that Chennai will win = Probability that it rains X Probability of Chennai winning during rain +Probability that it does not rain X Probability of Chennai winning when it is not raining= 40/100 X 30/100 + 60/100 X 55/100 = 1200/10000 + 3300/10000 = 4500/10000 = 45/100 = 45% Question 2 In a college, semester marks are given to I year students.23% of students passed in both I and IIsemesters and 45% of students passed in I semester. Find what percent of student who have passed in Isemester also passed in the II semester.a) 35% b)42% c)51% d)56%Solution :To solve such problems, we have to use the conditional probability formula,P(B|A) = P(A and B)/P(A).In the above formula,P(A and B) denotes percentage of students passed in both semesterP(A) denotes percentage of students passed in I semester.P(B|A) denotes percentage of student those who have passed in II semester after having passed IsemesterP(A and B) = 23% or 0.23 , P(A) = 45% or 0.45
