Question

Can Any One Help!

A deck of cards consists of 8 blue cards and 5 white cards. A simple random sample (random draws without replacement) of 6 cards is selected. What is the chance that one of the colors appears twice as many times as the other?

P(one card appearing Blue card)=8/13 P(second card appearing blue card)=7/12 P(One of the Color appear twice)=8/13*7/12?

Answer 1

Your total sample space would be the number of ways you can select 6 cards out of the 13 total, right, so... let's just make note of that...
\[\huge \frac{}{\left(\begin{matrix}13\\6\end{matrix}\right)}\]

Answer 2

By the way
\[\huge \left(\begin{matrix}n\\r\end{matrix}\right)=_nC_r\]

Answer 3

Catch me so far?

Answer 4

@Grad2013 These always sucked for me too I would take @terenzreignz advice on this and just follow through on what he's doing. He's obviously a G for Genius.

Answer 5

There are two ways of having one of the colors appearing twice as many times as the other:
2 blue and 4 white
4 blue and 2 white
\[P(2\ blue)=\frac{8C2\times 5C4}{13C6}\]
\[P(4\ blue)=\frac{8C4\times 5C2}{13C6}\]
The probability that one of the colors appears twice as many times as the other is given by:
\[P(2\ blue)+P(4\ blue)\]