- #1
Jameson
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I have checked and rechecked my thought process, but can't seem to figure this one out.
There are n socks, 3 of which are red. What is the value of n if 2 of the socks are chosen at random, the probability that both are red is 0.5?
This seems like a fairly straightforward combination problem. Here's my work.
\(\displaystyle \left( \frac{\binom{3}{2}}{\binom{n}{2}}=\frac{1}{2} \right) \rightarrow \left( \frac{3}{\frac{n(n-1)}{2!}}=\frac{1}{2} \right) \rightarrow n(n-1)=3\)
Solving that quadratic leads to non-integer solutions so we have a problem. Where's my error?
EDIT: Even approaching it another way I get the same thing. Let's not use combinations. Instead say the probability of choosing 1 red sock is \(\displaystyle \frac{3}{n}\) and after that the probability of choosing another red sock is \(\displaystyle \frac{2}{n-1}\). We get the same thing \(\displaystyle \frac{3}{n} \frac{2}{n-1} = \frac{1}{2}\)
There are n socks, 3 of which are red. What is the value of n if 2 of the socks are chosen at random, the probability that both are red is 0.5?
This seems like a fairly straightforward combination problem. Here's my work.
\(\displaystyle \left( \frac{\binom{3}{2}}{\binom{n}{2}}=\frac{1}{2} \right) \rightarrow \left( \frac{3}{\frac{n(n-1)}{2!}}=\frac{1}{2} \right) \rightarrow n(n-1)=3\)
Solving that quadratic leads to non-integer solutions so we have a problem. Where's my error?
EDIT: Even approaching it another way I get the same thing. Let's not use combinations. Instead say the probability of choosing 1 red sock is \(\displaystyle \frac{3}{n}\) and after that the probability of choosing another red sock is \(\displaystyle \frac{2}{n-1}\). We get the same thing \(\displaystyle \frac{3}{n} \frac{2}{n-1} = \frac{1}{2}\)