A deck from a card game is made up of about 108 cards. Twenty-five of them are red, yellow, blue, and green, and eight are wild cards. Each player is randomly dealt a seven-card hand. Write the final answer as a decimal rounded to six decimal places.

Question

1.) P(a hand containing exactly two wild cards)

2.) P(a hand containing two wild cards, two red cards, and three blue cards)

Answer 1

p(exactly two wild)=8/108*7/107

P(two wild,two red, 3bluej)=8/108*7/107*25/105*24/104*25/103*24/102*23/101

check that

Answer 2

1.) 0.004846?
2.) 3.462816?

Are those the right answers?

Answer 3

The second is obviously wrong, a probability cannot be greater than one.

Answer 4

In the first one I think Bob forgot about the other 5 cards that have to be drawn

I am going to do these by using combinations
P(exactly 2 wild, 5 non-wild)
= ( C(8,2) x C(100,5) / C(108,7)
= .0756

or using bob's method

P(exactly 2 wild, 5 non-wild
= 8/108 * 7/107 * 100/106 *99/105 * 98/104 * 97/103 * 96/102
= .003600135
but that is the specific order of WWNNNNN
which can be permutated in 7!/(2!5!) or 21 ways

so .003600135(21) = .0756 , the same as the C(n,r) method

in the 2nd, the answer obtained by bob has to be multiplied by 7!/(2!2!3!) to get .0006929 , for the same reason I stated above

using the C(n,r) way
prob = (C(8,2) x C(25,2) x C(25,3) / C(108,7)
= .006929