Latest TCS Aptitude Question SOLUTION: there are two boxex, one containing 39 red balls and other 29 green balls. yao are allowed to move the balls betn the booxes so that you choose a box randomly and a ball at random

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(#M40000963) TCS question Keep an EYE

there are two boxex, one containing 39 red balls and other 29 green balls. yao are allowed to move the balls betn the booxes so that you choose a box randomly and a ball at random from the choosen box, the probability of getting a red ball is maximized. the maximum probability is
a. .60
b. .50
c. .80
d. .30

Asked In TCS oindrila majumder (13 years ago)

Unsolved Read Solution (5)

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suchandra raavi  (13 years ago)

To maximise the probability of red balls, you need to keep 1 red ball in 1 box and transfer remaining to the other. So box1= 1red ball, box2 = 29 green+38 red. So first to select box1, probability is 1/2.Now only 1 red ball is there, so prbability for red is 1. Total probability is (1/2)*1.Similarly probability of red ball in box2= (1/2)(38/67). total probability = (1/2)*1 + (1/2)(38/67)~ 0.8

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Pargat Pal Singh  (13 years ago)

Keep one red ball in one box and put 38 red balls and 29 green balls in other box.

Then probability of getting red ball = 1/2 *1 + 1/2 * 38/67 = 105/134 = 0.7835

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surajit saha  (13 years ago)

the ball may b redor blue.
the max probability is 1/2+1/2*(38/67)=.60

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tk  (13 years ago)

(1/2)*(1)+(1/2)*(38/67)=.79=.8
C) .8

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teja  (11 years ago)

atleast one red ball should be picked.
so probability to take a red ball from a box is (1/2)*1
red balls in second box is(1/2)*(38/67)
sol:(1/2)+(1/2)*(38/67)
so probability is 0.7835=0.80

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