# Thread: [SOLVED] Two Conditional Probability Problems

1. ## [SOLVED] Two Conditional Probability Problems

Hello, I just don't get how to solve the two following problems.

1. A drawer has 8 different pairs of socks. Six socks are taken randomly and without replacement, compute the probability that there is at least one matching pair among these socks.

Hint: Compute the probability that there is not a matching pair.
Answer: $\displaystyle \frac {111} {143}$

2.
Bowl I contains 6 red chips and 4 blue chips. Five of these 10 chips are selected at random and without replacement and put in bowl II, which was originally empty. One chip is then drawn at random from bowl II. Given that this chip is blue, find the conditional probability that 2 red chips and 3 blue chips are transferred from bowl I to bowl II.

Answer:$\displaystyle \frac {5}{14}$

2. Originally Posted by akolman
Hello, I just don't get how to solve the two following problems.

1. A drawer has 8 different pairs of socks. Six socks are taken randomly and without replacement, compute the probability that there is at least one matching pair among these socks.

Hint: Compute the probability that there is not a matching pair.
Answer: $\displaystyle \frac {111} {143}$

[snip]
It's a good hint. But also a crummy one since you should already know this technique.

Since the color of the first sock doesn't matter (why?), the probability of no matching pair is $\displaystyle \left( \frac{14}{15}\right) \, \left( \frac{12}{14}\right) \, \left( \frac{10}{13}\right) \, \left( \frac{8}{12}\right) \, \left( \frac{6}{11}\right)$.

Do you see why?

3. Originally Posted by akolman
[snip]
2. Bowl I contains 6 red chips and 4 blue chips. Five of these 10 chips are selected at random and without replacement and put in bowl II, which was originally empty. One chip is then drawn at random from bowl II. Given that this chip is blue, find the conditional probability that 2 red chips and 3 blue chips are transferred from bowl I to bowl II.

Answer:$\displaystyle \frac {5}{14}$
$\displaystyle \Pr(\text{3 B taken from bowl I} \, | \, \text{B chip drawn from bowl II})$ $\displaystyle = \frac{ \Pr(\text{3 B taken from bowl I and B chip drawn from bowl II})}{\Pr(\text{B chip drawn from bowl II})}$.

$\displaystyle \Pr(\text{3 B taken from bowl I and B chip drawn from bowl II}) = \frac{^6C_2 \, ^4C_3}{^{10}C_5} \times \frac{3}{5}$.

$\displaystyle \Pr(\text{B chip drawn from bowl II}) = \frac{^6C_4 \, ^4C_1}{^{10}C_5} \times \frac{1}{5} ~ + ~ \frac{^6C_3 \, ^4C_2}{^{10}C_5} \times \frac{2}{5} ~ + ~ \frac{^6C_2 \, ^4C_3}{^{10}C_5}$ $\displaystyle \times \frac{3}{5} ~ + ~ \frac{^6C_1 \, ^4C_4}{^{10}C_5} \times \frac{4}{5}$.

Do you see why?

4. Thank you very much!!

5. Thanks.

I ran into these problems reading Hogg & Tanis' Probability and Statistical Inference, 1977: 1-6.5 and 1-7.5.
The answer to 1-7.5 given by the book is 5/16 (and is obviously wrong).

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### a drawn contains eight pairs of socks.if six socks are taken at random and without replacement comute the probability that there is atleast one matching pair among these six socks condotional probability

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