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A box contains 7 red and 13 blue balls. Two balls are randomly selected (without replacement) and are discarded without their colors being seen. A third ball is drawn randomly.

(a) Find the probability that the third ball is red.

(b) Given that the third ball is red, find the probability that both discarded balls were blue.

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I've seen that you have placed two questions right after another that look like homework. Generally people on this site don't like to answer homework questions but like to help you with the homework once they know where you are stuck. We will help if you show that you have tried the question first. – kaine Feb 4 '13 at 21:29


(a) After the first draw the box contains one of the following combinations: $5$ red and $13$ blue balls; $6$ red and $12$ blue balls; or $7$ red and $11$ blue balls. Find the probabilities of these outcomes; call them $p_{RR}$, $p_{RB}$, and $p_{BB}$, respectively. For instance, $p_{RR}$ is $$\frac{\binom72}{\binom{20}2}\;,$$ since there are $\binom72$ ways to choose $2$ red balls and $\binom{20}2$ ways to choose $2$ balls altogether.

In each of the cases calculate the probability of getting a red ball once you’re in that case; call these $q_{RR},q_{RB}$, and $q_{BB}$, respectively. Then the overall probability of getting a red ball on the second draw is


Each term is the probability of being in a particular case and then drawing a red ball. For instance, $p_{RR}q_{RR}$ is the probability of drawing and discarding two red balls and then drawing a red ball. The total is the total probability of all ways of getting a red ball on the second draw.

(b) If you know that the third ball was red, then you know that you were in one of the three cases above. The expression $(1)$ is the total probability of that having happened. What fraction of that probability comes having drawn two blue balls in the first draw? That case is covered by the third term of $(1)$, so it accounts for


of the total probability that the third ball is red. That fraction is the probability that you were in case $BB$ given that the third ball is red.

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The answer to (a) is clearly $\dfrac{7}{20}$. This is because every one of the $20$ balls is equally likely to be the third ball drawn.

For (b), let $R$ be the event that the third ball drawn is red. Let $B$ be the event the first two balls drawn are blue. We want the conditional probability $\Pr(B|R)$. By the usual expression for conditional probability, we have $$\Pr(B|R)=\frac{\Pr(B\cap R)}{\Pr(R)}.$$

We already know $\Pr(R)$, so all we need is $\Pr(B\cap R)$. This is not hard to calculate: $$\Pr(B\cap R)=\frac{13}{20}\cdot\frac{12}{19}\cdot \frac{7}{18}.$$

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a. Think about the case of a single discard first. The probability of drawing a red is equal to $P(\mathrm{red\ drawn}\,\cap\,\mathrm{red\ discarded}) + P(\mathrm{red\ drawn}\,\cap \,\mathrm{blue\ discarded})$. What can you say about the sum of these probabilities? How does this generalize?

b. Can you compute $P(\mathrm{red\ drawn}\,|\,\mathrm{two\ blues\ discarded})$? How might you then use part a, and Bayes's theorem, to find $P(\mathrm{two\ blues\ discarded}\,|\,\mathrm{red\ drawn})$?

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