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A bag has seven black balls and three white balls. Two balls are drawn in sequence without replacement. Find the probability that:

a) the second ball is white. b) the second ball is white given that the first one is black. c) the second ball is white given that the first one is white.

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    $\begingroup$ Found, now what? $\endgroup$ – barak manos Jan 6 '16 at 13:41
  • $\begingroup$ oh i'm sorry that should be black and white also not green and red $\endgroup$ – user303108 Jan 6 '16 at 13:50
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Hint on a)

There are $7$ black balls and $3$ white balls that can be chosen to be the second ball. This with equal probability.

Hint on b)

If the first ball drawn is black then the second is drawn from a bag that contains $6$ black balls and $3$ white balls.

Hint on c)

If the first ball drawn is white then the second is drawn from a bag that contains $7$ black balls and $2$ white balls.

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  • $\begingroup$ thank you i'm sorry about the green and red it's also black and white though $\endgroup$ – user303108 Jan 6 '16 at 13:50
  • $\begingroup$ I only looked at a) so wasn't bothered by other colors. $\endgroup$ – drhab Jan 6 '16 at 13:51
  • $\begingroup$ in letter b my answer is 3/9 or 1/3. am i correct? $\endgroup$ – user303108 Jan 6 '16 at 14:43
  • $\begingroup$ Yes, you are correct in that. $\endgroup$ – drhab Jan 6 '16 at 15:38
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It is highly desirable to produce your own working when asking questions.

In this case, since you sample without replacement, you should use the following approach: $$ P(S) = \frac{ \text{total number of ways to get the result}}{\text{total number of ways to get the outcome}} $$

For example, in the first question, there are 2 ways to get the result (WW, BW): $\frac{\binom{3}{2} + \frac{1}{2} \cdot \binom{3}{1} \cdot \binom{7}{1}}{\binom{10}{2}}$

The $\frac{1}{2}$ in the second term is because we are specifically interested in BW, BW and WB

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A simple way to handle cases (b) and (c) is to say that if you know the first ball taken was white, there remain 2 white and 7 black ones, and you are back to asking what the probability is of taking a white out of that.

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For a) another way is to use the law of total probability: $$ P(2W) = P(2W|1W)P(1W) + P(2W|1B)P(1B) = \frac{2}{9} \cdot \frac{3}{10} + \frac{3}{9} \cdot \frac{7}{10} $$

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