Hi can anybody help with this probability question?

i have 8 purple socks 4 green socks and 4 red socks in a drawer. two socks are chosen without replacement. What is the probability that a pair the same colour is chosen?

thanks

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- Oct 28th 2008, 10:28 AMmath_leteprobability
Hi can anybody help with this probability question?

i have 8 purple socks 4 green socks and 4 red socks in a drawer. two socks are chosen without replacement. What is the probability that a pair the same colour is chosen?

thanks - Oct 28th 2008, 11:21 AMSoroban
Hello, math_lete!

There are at least two ways to approach this problem . . .

Quote:

There are 8 purple socks 4 green socks and 4 red socks in a drawer.

Two socks are chosen without replacement.

What is the probability that a pair the same colour is chosen?

$\displaystyle \begin{array}{ccccc}P(\text{2 purple}) &=& \frac{8}{16}\cdot\frac{7}{15} &=& \frac{14}{60} \\ \\[-4mm]

P(\text{2 green}) &=& \frac{4}{16}\cdot\frac{3}{15} &=& \frac{3}{60} \\ \\[-4mm]

P(\text{2 red}) &=&\frac{4}{16}\cdot\frac{3}{15} &=& \frac{3}{60} \end{array}$

Therefore: .$\displaystyle P(\text{2 purple or 2 green or 2 red}) \;=\;\frac{14}{60} + \frac{3}{60} + \frac{3}{60} \;=\;\frac{20}{30} \;=\;\frac{1}{3} $

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There are: $\displaystyle {16\choose2} = 120$ possible pairs.

. . There are: $\displaystyle {8\choose2} = 28$ ways to get two purple socks.

. . There are: $\displaystyle {4\choose2} = 6$ ways to get two green socks.

. . There are: $\displaystyle {4\choose2} = 6$ ways to get two red socks.

Hence, there are: .$\displaystyle 28 + 6 + 6 \:=\:40$ ways to get a matching pair.

Therefore: .$\displaystyle P(\text{matching color}) \;=\;\frac{40}{120} \;=\;\frac{1}{3}$