Thread: Probability-taking from one box to another then drawing

Hello, good day,
Here is a thread to go about.
The problem I faced is that no box has been specified please help.

Question.
The contents of two boxes are as follows; box A contains 3 red and 2 white balls, box B contains 2 red and 5 white balls. a box is selected at random, a ball is drawn and put into second box, and then a ball is drawn from the second box. Find the probability that both balls drawn are of the same colour.

Originally Posted by Brain Exploder

Hello, good day,
Here is a thread to go about.
The problem I faced is that no box has been specified please help.

Question.
The contents of two boxes are as follows; box A contains 3 red and 2 white balls, box B contains 2 red and 5 white balls. a box is selected at random, a ball is drawn and put into second box, and then a ball is drawn from the second box. Find the probability that both balls drawn are of the same colour.

Suppose the first ball is chosen from A now you can calculate the probability that they are both the same colour.

Now instead; suppose the first ball is chosen from B, now you can calculate the probability that they are both the same colour.

These two cases occur with equal probability so you can combine the results in the appropriate manner to get the final answer.

CB

I'm not 100% certain, but l'll take a crack at it.

r_A = red chosen from bucket A
w_A = white chosen from bucket A
r_B = red chosen from bucket B
w_B = white chosen from bucket B

P(same color)
= (1/2) * P(a ball is chosen from A first and a ball from B is same color) +
(1/2) * P(a ball is chosen from B first and a ball from A is same color)

P(a ball is chosen from A first and a ball from B is same color)
= P(r_A and r_B) + P(w_A and w_B)
= P(r_A) * P(r_B) + P(w_A) * P(w_B)
= (3/5)(3/8) + (2/5)(6/8)

P(a ball is chosen from B first and a ball from A is same color)
= P(r_B and r_A) + P(w_B and w_A)
= P(r_B) * P(r_A) + P(w_B) * P(w_A)
= (2/7)(4/6) + (5/7)(3/6)

=> P(same color)
= (1/2)((3/5)(3/8) + (2/5)(6/8)) + (1/2)((2/7)(4/6) + (5/7)(3/6))
= .53631

Originally Posted by doolindalton

I'm not 100% certain, but l'll take a crack at it.

r_A = red chosen from bucket A
w_A = white chosen from bucket A
r_B = red chosen from bucket B
w_B = white chosen from bucket B

P(same color)
= (1/2) * P(a ball is chosen from A first and a ball from B is same color) +
(1/2) * P(a ball is chosen from B first and a ball from A is same color)

P(a ball is chosen from A first and a ball from B is same color)
= P(r_A and r_B) + P(w_A and w_B)
= P(r_A) * P(r_B) + P(w_A) * P(w_B)
= (3/5)(3/8) + (2/5)(6/8)

P(a ball is chosen from B first and a ball from A is same color)
= P(r_B and r_A) + P(w_B and w_A)
= P(r_B) * P(r_A) + P(w_B) * P(w_A)
= (2/7)(4/6) + (5/7)(3/6)

=> P(same color)
= (1/2)((3/5)(3/8) + (2/5)(6/8)) + (1/2)((2/7)(4/6) + (5/7)(3/6))= .53631

Although I did not do the calculation, the part in blue is correct.