Question

two bags A and B contain 4 white and 3 black balls,2 white and 2 black balls .from bag A two balls are drawn at random and then transferred to bag B.a ball is then drawn from B is found to be black.what is the probability that the tranffered balls were 1 white and 1 black.

Answer 1

We choose 2 balls from A,
We can get 2 black, 2 white or 1 white 1 black

P(2 black balls drawn) = 3/7 * 2/6 = 6/42
P(2 white balls drawn)= 4/7 * 3/6 = 12/42
P(1 white and 1 black) = (4/7 * 3/6) + (3/7 * 4/6) = 24/42

Suppose we had drawn 2 black balls from A and transferred to B, then B would have had 2 white and 4 black balls.
P(Black ball drawn) = 4/6

2nd case, when 2 white balls had been transferred to B, then B would have had 4 white and 2 black balls
P(black ball drawn) = 2/6

3rd case, when 1 white and 1 black ball been transferred to B, B would have had 3 white and 3 black balls
P(black ball) = 3/6

Favorable probability = P(1 white and 1 black) * P(black ball) = 24/42 * 3/6

Total probability = P(1 white and 1 black) * P(black ball) + P(2 white) * P(black ball) + P(2 black) * P(black ball)
= (24/42 * 3/6) + (12/42 * 2/6)+ (6/42 * 4/6)

Hence required probability = ( 24/42 * 3/6) / ((24/42 * 3/6) + (12/42 * 2/6)+ (6/42 * 4/6))

Answer 2

Just simplify. Hope I didn't make a mistake ?

Answer 3

have you felt of using nC2=n(n-1)/2

Answer 4

@shubhamsrg

Answer 5

What do you mean felt ?

Answer 6

The same probability can be derived using combinations also. Its no big deal.

Answer 7

then please help me by using that method.

Answer 8

What is the problem in this method ?

Answer 9

i know this one. i wan know that too.

Answer 10

Though I doubt what you say, still I may try to explain it using combinations.

Answer 11

i'd be thankful

Answer 12

From combinations point of view, P(selecting 2 black balls from A) = C(3,2) /C(7,2) = 3/21
P(2 white balls) = C(4,2)/C(7,2) = 6/21
P(1 white 1 Black) = C(3,1)*C(4,1) /C(7,2) = 12/21

Rest all remains same.

AND AM QUITE SURE YOU HAVEN'T GOT THE FIRST METHOD, THEN YOU WOULDN'T HAVE ASKED FOR THE SECOND ONE.

Answer 13

i dont know how to find using combinations thats why i just asked!thx

Answer 14

hmm..