Latest Accenture Aptitude Question SOLUTION: There are two boxes, one containing 32 black balls and other containing 31 red balls you are allowed to move the boxes at random so that when you choose the box at random and a bal

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(#M40155467) ACCENTURE Numerical Ability question Keep an EYE

There are two boxes, one containing 32 black balls and other containing 31 red balls you are allowed to move the boxes at random so that when you choose the box at random and a ball at random from the chosen box, the probability of getting the black ball is maximized. This maximum probability is
(a).25
(b).51
(c).75
(d).50

Asked In Accenture Shubha Dewangan (8 years ago)

Unsolved Read Solution (15)

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Read Solution (15) :

ditty k joy  (8 years ago)

keeping one black ball in a box and move rest to the box containing red balls. So now selecting a black ball from the two balls is as :-
selecting black from first box =1/2*1C1=0.5
selecting black from 2nd box=1/2*(31C1)/(62C1)=0.25
therfore probability =0.5 +0.25=0.75

Is this solution Helpfull? Yes (5) | No   

Dhanesh  (8 years ago)

Move 31 balls from black ball box to other box having 31 red balls.

now one box has one black ball
and
other box is having 31 black balls and 31 red balls.

then prob of getting black ball = 1/2 (1+31/62) = 1/2 *3/2 =3/4=0.75

Is this solution Helpfull? Yes (5) | No   

samoon  (8 years ago)

choosing one box=1/2
prob of black=32c1
tot prob=63c1
req prob=1/2*(32c1)/(63c1)
=.25

Is this solution Helpfull? Yes (3) | No (1)  

MOON SINHA  (8 years ago)

0.50 ans

Is this solution Helpfull? Yes (2) | No (5)  

praveen  (8 years ago)

you have choose 1 box from 2 boxes so prob is 1/2
among 63 ball u r going to choose 1 ball hence prob is 63c1
for getting black ball prob is 32c1
so 32c1/(63c1*0.5)=0.254

Is this solution Helpfull? Yes (2) | No (3)  

kolimi neelufar  (8 years ago)

Number of Black balls=32
Number of Red balls=31
Total balls=66
Probalility to pick one black ball from total balls=32/66=>0.50

Is this solution Helpfull? Yes (2) | No (8)  

Bijit Baral  (8 years ago)

max prob= prob of choosing 1 box (1/2) + prob of choosing black ball ( 32c1/63C1)=0.75 ans

Is this solution Helpfull? Yes (2) | No   

kratika gupta  (8 years ago)

some one can tell which ans is correct? please

Is this solution Helpfull? Yes (1) | No (3)  

Khilesh  (7 years ago)

they asked to move the boxes and not the balls..so I guess we are not supposed to interchange the balls..keeping this in mind max probability of putting the hand in box containing black balls is 0.50

Is this solution Helpfull? Yes (1) | No   

ABHISHEK  (7 years ago)

Probability of chosen the box is =1/2
and now we have to calculate the selecting red ball from the box
i,e. (1/2)*31/62=1/4
now the probability of selecting black ball is 1-1/4
i.e. 3/4=.75

Is this solution Helpfull? Yes (1) | No   

sonal agrawal  (7 years ago)

Move 31 balls from black ball box to other box having 31 red balls.
now one box has one black ball and other box is having 31 black balls and 31 red balls.

then prob of getting black ball = 1/2 (1+31/62) = 1/2 *3/2 =3/4 =.75

Is this solution Helpfull? Yes (1) | No   

Ritesh kumar  (8 years ago)

Ans- 1/2 * 1 + 1/2 * 31/ 63 = 0.75

Is this solution Helpfull? Yes | No (2)  

sarathkumar reddy  (8 years ago)

Move 31 balls from red ball box to other box having 31 green balls.

now one box has one red ball
and
other box is having 31 red balls and 31 green balls.

then prob of getting red ball = 1/2 (1+31/62) = 1/2 *3/2 =3/4

Is this solution Helpfull? Yes | No   

malliakrjuna  (8 years ago)

so ,total outcomes =63,favourable outcomes of black=32,probability=0.50 the answer is d

Is this solution Helpfull? Yes | No   

Subhradip  (7 years ago)

A : B : C = 3 : 2 : 5
(33) Means 3, Therefore 33/3 = 11 .
3*11=33
2*11=22
5*11=55
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+ 110 (Ans)

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