Question

suppose that 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?

Answer 1

if you do: x+2x=6 you get that x=2 and 2x=4 is the only possible combination of colors that fulfills "one of the colors appears twice as many times as the other",then you need to analyze:
2 BLUE & 4 WHITE (case 1) or 2 WHITE & 4 BLUE (case 2)
We need to apply hypergeometric distribution here:\[p(k)=\frac{ \left(\begin{matrix}K \\ k\end{matrix}\right) \left(\begin{matrix}N-K \\ n-k\end{matrix}\right)}{ \left(\begin{matrix}N \\ n\end{matrix}\right)}\] Where (I will put the numbers for case 1):
K=Maximum number of successes (number of BLUE cards=8)
k=number of expected successes (2 BLUE)
N=Population size (13 cards=8 BLUE+5 WHITE)
n=number of draws(6).
Then you have for the first case:.\[p_1(k=2)=\frac{ \left(\begin{matrix}8\\ 2\end{matrix}\right) \left(\begin{matrix}5 \\4\end{matrix}\right)}{ \left(\begin{matrix}13\\6\end{matrix}\right)}=0.0815\]
Now you do it for second case

Answer 2

I would like to know the final answer?

Answer 3

I understand that you want the final answer. Let us do something, you solve the case "2 WHITE+4 BLUE" with the indications I have given and I check your results.

Answer 4

case "2 WHITE+4 BLUE" P(k2=2) = 0.4079. And now...what to do?...

Answer 5

It is case 1 or case 2, then the total probability is 0.0815+0.4079=0.4895, that is a 48.95%

Answer 6

Ok...thank you.