# Thread: Probability Question causing me a bit of bother

1. ## Probability Question causing me a bit of bother

There's probably some really easy way to do this question but I spent about an hour at study this evening trying to do this question but I couldn't get ny head around it.... and yes it is annoying me at this point lol

there are 11 black socks in a drawer and n white socks... if the probability of picking 2 n socks is 1/12 what is the value for n.

Thanking you

2. The probability of choosing 2 white socks from the drawer is:

$\displaystyle \frac{_nC_2}{_{11 + n}C_2} = \frac{1}{12}$

$\displaystyle \frac{\frac{n(n-1)}{2}}{\frac{(11 + n)(10 + n)}{2}} = \frac{1}{12}$

$\displaystyle \frac{n(n-1)}{(11 + n)(10 + n)} = \frac{1}{12}$

$\displaystyle 12n(n-1) = (11 + n)(10 + n)$

$\displaystyle 12n^2 - 12n = 110 + 21n + n^2$

$\displaystyle 11n^2 - 33n - 110 = 0$

$\displaystyle 11(n - 5)(n + 2) = 0$

$\displaystyle n = 5$

3. thaning you

could you possibibly help explain line 2 to me..

EDIT
i understand it now i got it out now

thanking you

thread can be locked now