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 January 12th, 2016, 06:26 AM #1 Senior Member     Joined: Oct 2013 From: Far far away Posts: 429 Thanks: 18 Probability question A box contains 4 white and 6 black balls. A second box contains 7 white and 3 black balls. A ball is picked at random from the first box and placed in the second box. A ball is then picked from the second box. What is the probability that it is white? My attempt: There are two possibilities with the first box: picking a black ball or picking a white ball a) P(picking white ball from 1st box) = 4/10 = 2/5 b) P(picking a black ball from 1st box) = 6/10 = 3/5 Each of the two paths a) and b) will be followed by picking a ball from the second box. Path a) leads to P(picking a white ball from 2nd box) = 8/11 Path b) leads to P(picking a white ball from 2nd box) = 7/11 Therefore probability of picking a white ball from 2nd box = (2/5)(8/11)+(3/5)(7/11) = (16/55) + (21/55) = 37/55 Am I correct? I used a probability tree diagram to ''solve''? This problem. Is there a better way to understand this problem? Thanks.
January 12th, 2016, 11:10 AM   #2
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You could start off by figuring out which scenarios could emerge:

"A box contains 4 white and 6 black balls. A second box contains 7 white and 3 black balls. A ball is picked at random from the first box and placed in the second box. A ball is then picked from the second box. What is the probability that it is white?"

First=F

Second=S

SCENARIO 1:

You pick a white ball from the first box and place it in the second box. You then pick a white ball from the second box. [FW, SW]

SCENARIO 2:

You pick a white ball from the first box and place it in the second box. You then pick a black ball from the second box. [FW, SB]

SCENARIO 3:

You pick a black ball from the first box and place it in the second box. You then pick a white ball from the second box. [FB, SW]

SCENARIO 4:

You pick a black ball from the first box and place it in the second box. You then pick a black ball from the second box. [FB, SB]

So... We know that the only scenarios which can emerge are:

[FW, SW], [FW, SB], [FB, SW] and [FB, SB]

Now that we've discovered this logic, we can then go on to think about probabilities...

I'm going to provide two demonstrations about how this problem can be solved. The first demonstration will include squares, whilst the second demonstration will include a tree diagram....

Demonstration 1 (File 1):

You create a 10 x 12 rectangle to figure out the answer. This is because if your first selection is a white ball, you'll get 8 white balls in the second box. If your first selection is a black ball, you'll get 4 black balls in the second box. You black out the areas where a certain scenario cannot possibly emerge.

Demonstration 2 (File 2):

Just an ordinary tree diagram that strengthens the case that you are correct.
Attached Images
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