Question #6665

Two bags A and B contain 4 white 3 black balls and 2 white and 2 black balls respectively. From bag A two balls
are transferred to bag B. Find the probability of drawing
(a) 2 white balls from bag B ?
(b) 2 black balls from bag B ?
(c) 1 white & 1 black ball from bag B ?

Expert's answer

Initial conditions are:

A: 4w 3b

B: 2w 2b

There are 3 possible situations: we choose 2 white balls from bag A, 1 white and 1 black or 2 black balls.

1. 2 white balls from A, probability of that is p1 = C(4,2)/C(7,2) = 6/21

& Now we have in B: 4w 2b

& Let's calculate probabilities of choosing 2 white, 2 black or 1 white and 1 black balls from B:

& p1(2w) = C(4,2)/C(6,2) = 6/15 = 2/5

& p1(2b) = C(2,2)/C(6,2) = 1/15

& p1(1b1w) = (C(4,1)*C(2,1))/C(6,2) = 4*2/15 = 8/15

2. 1w 1b, p2 = (C(4,1)*C(3,1))/C(7,2) = (4*3)/21 = 12/21.

& Now we have in B: 3w 3b

& p2(2w) = C(3,2)/C(6,2) = 3/15 = 1/5

& p2(2b) = C(3,2)/C(6,2) = 3/15 = 1/5

& p2(1b1w) = (C(3,1)*C(3,1))/C(6,2) = 9/15

3. 2b, p3 = C(3,2)/C(7,2) = 3/21

& Now we have in& B: 2w 4b

& p3(2w) = C(2,2)/C(6,2) = 1/15

& p3(2b) = C(4,2)/C(6,2) = 6/15 = 2/5

& p3(1b1w) = (C(2,1)*C(4,1))/C(6,2) = 8/15

The total probabilities are:

P(2w) = p1*p1(2w) + p2*p2(2w) + p3*p3(2w) = 6/21*2/5 + 12/21*1/5 + 3/21*1/15 = 5/21,

P(2b) = p1*p1(2b) + p2*p2(2b) + p3*p3(2b) = 6/21*1/15 + 12/21*1/5 + 3/21*2/5 = 4/21,

P(1w1b) = p1*p1(1w1b) + p2*p2(1w1b) + p3*p3(1w1b) = 6/21*8/15 + 12/21*9/15 + 3/21*8/15 = 4/7.

Really, P(2w) + P(2b) + P(1w1b) = 5/21+4/21+4/7 = 1.

A: 4w 3b

B: 2w 2b

There are 3 possible situations: we choose 2 white balls from bag A, 1 white and 1 black or 2 black balls.

1. 2 white balls from A, probability of that is p1 = C(4,2)/C(7,2) = 6/21

& Now we have in B: 4w 2b

& Let's calculate probabilities of choosing 2 white, 2 black or 1 white and 1 black balls from B:

& p1(2w) = C(4,2)/C(6,2) = 6/15 = 2/5

& p1(2b) = C(2,2)/C(6,2) = 1/15

& p1(1b1w) = (C(4,1)*C(2,1))/C(6,2) = 4*2/15 = 8/15

2. 1w 1b, p2 = (C(4,1)*C(3,1))/C(7,2) = (4*3)/21 = 12/21.

& Now we have in B: 3w 3b

& p2(2w) = C(3,2)/C(6,2) = 3/15 = 1/5

& p2(2b) = C(3,2)/C(6,2) = 3/15 = 1/5

& p2(1b1w) = (C(3,1)*C(3,1))/C(6,2) = 9/15

3. 2b, p3 = C(3,2)/C(7,2) = 3/21

& Now we have in& B: 2w 4b

& p3(2w) = C(2,2)/C(6,2) = 1/15

& p3(2b) = C(4,2)/C(6,2) = 6/15 = 2/5

& p3(1b1w) = (C(2,1)*C(4,1))/C(6,2) = 8/15

The total probabilities are:

P(2w) = p1*p1(2w) + p2*p2(2w) + p3*p3(2w) = 6/21*2/5 + 12/21*1/5 + 3/21*1/15 = 5/21,

P(2b) = p1*p1(2b) + p2*p2(2b) + p3*p3(2b) = 6/21*1/15 + 12/21*1/5 + 3/21*2/5 = 4/21,

P(1w1b) = p1*p1(1w1b) + p2*p2(1w1b) + p3*p3(1w1b) = 6/21*8/15 + 12/21*9/15 + 3/21*8/15 = 4/7.

Really, P(2w) + P(2b) + P(1w1b) = 5/21+4/21+4/7 = 1.

## Comments

## Leave a comment