Latest TCS Aptitude Question SOLUTION: There are two boxes, one containing 39 red balls & the other containing 26 green balls. You are allowed to move the balls b/w the boxes so that when you choose a box random & a bal

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There are two boxes, one containing 39 red balls & the other containing 26 green balls. You are allowed to move the balls b/w the boxes so that when you choose a box random & a ball at random from the chosen box, the probability of getting a red ball is maximized. This maximum probability is
a)60 b )50 c)80 d)30

Asked In TCS (10 years ago)

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sruthi  (10 years ago)

Initially probability is never greater than 1. it always lies b/w 0 and 1. If the problem is solved the answer would be , first move 38 red balls from 1st box (leaving one red ball) to the 2nd box. then probability would be 1/2(1)+1/2(38/64)
=0.80 apprx. !!! where 64 is sum of red and green balls in the second box.

Is this solution Helpfull? Yes (48) | No (2)  

kadali naga sindhuja  (10 years ago)

choosing a box at random from 2 boxes is=> 2c1
since already required box is choosen from the 2 boxes so req is box containing red balls....
so for choosing red ball from the choosen box is=> 39c1
so, total = 2c1*39c1 => 2*39 => 78 (i.e.,approx 80)
so, ANS IS 80

Is this solution Helpfull? Yes (7) | No (9)  

Tanya Maheshwari  (10 years ago)

for maximizing the number of red balls, we will leave 1 red ball in the first box and shift the remaining 38 balls into the 2nd box...
now we have 2 boxes,one with 1 red ball and another with 38 red and 26 green balls..
now for the calculating the maximum probability,

the probability of choosing the first box is 1/2 as we have to chose 1 box out of 2... and then choosing the red ball in it is 1/1 because there is only 1 ball left.... (1/2)*(1/1)=(1/2)..... eq(i)

similarly for the second box.
probability of choosing the second box is (1/2) and red balls is (38/64)
where 38 are the no of red balls in the second box..and 64 are the total no of balls..ie red + green...
hence probability = (1/2)*(38/64) ....eq(ii)

on adding both the probabilities total probability = .64%

Is this solution Helpfull? Yes (5) | No (6)  

Tanya Maheshwari  (10 years ago)

sorry .79=.80

Is this solution Helpfull? Yes (4) | No (1)  

RAKESH BHARDWAJ  (10 years ago)

For probability of choosing red ball is to be maximum, we have to
-move all red balls to other box.
probability of choosing box= 1/2

ans will be-- 1/2*1+1/2*38/64
= .64 is ans,,, options are wrongly written.

Is this solution Helpfull? Yes (2) | No (11)  

Rohit Bohra  (10 years ago)

answer is:0.80;
((1/2)*1)+((1/2)*38/64)=0.80

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Rincy juliet  (10 years ago)

Ans. Well Keep a single red ball in 1 box and move 38 red balls to the other

1st box = 1R
2nd box = 38R + 26G

p(red) = 1/2 * 1 + 1/2 * 38/64 ≈ 0.8

Is this solution Helpfull? Yes (1) | No   

Ayushman Tiwari  (7 years ago)

The options are wrong, how can Probability be ever greater than '1' !!
The correct answer is 51/64

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utkarsh bisht  (10 years ago)

ans will be .8

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monica  (10 years ago)

box n ball we cannt choose ...it's selected at random so i guess .80 is nt d correct ans.

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RAVI GANGWAR  (10 years ago)

correct ans=80

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