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There are two bags. One bag contains 4 white and 2 black balls. Second bad contains 5 white and 4 black balls. 2 balls are transferred from first bag to second bag. Then one ball is taken from the second bag. The probability that the ball is white is ?

- 42/33
- 5/16
- 48/13
- 19/33

+3 votes

Possible cases of balls can be choosen from bag 1 =

P(WW) =(4/6)*(3/5)

P(WB)=(4/6)*(2/5)

P(BW)=(2/6)*(4/5)

P(BB)=(2/6)*(1/5)

So in case WW in bag 2 = 7W+4B

In case WB in bag 2= 6W+5B

In case BW in bag 2= 6W+5B

in case BB in bag 2 = 5W+6B

probability of choosing while ball from bag 2=

(4/3)*(3/5)*(7/11)+(4/6)*(2/5)*(6/11)+(2/6)*(1/5)*(5/11)+(2/6)*(4/5)*(6/11)= **19/33**

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