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.

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

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

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

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

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

Well, the number of ways to choose exactly 2 wild cards out of 8 is given by the combination formula: C(8, 2) = 28.

The number of remaining cards to choose from is 108 - 8 = 100.

So, the number of ways to choose the remaining 5 cards out of 100 is given by the combination formula: C(100, 5) = 75,287,520.

The total number of possible 7-card hands is given by the combination formula: C(108, 7) = 153,858,295.

Therefore, the probability P(a hand containing exactly two wild cards) = (28 * 75,287,520) / 153,858,295.

Calculating this, we get P(a hand containing exactly two wild cards) ≈ 0.137857.

So the answer, rounded to six decimal places, is approximately 0.137857.

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

The number of ways to choose exactly 2 wild cards out of 8 is C(8, 2) = 28.

The number of ways to choose exactly 2 red cards out of 25 is C(25, 2) = 300.

The number of ways to choose exactly 3 blue cards out of 25 is C(25, 3) = 2,300.

The number of remaining cards to choose from is 108 - 8 - 25 - 25 = 50.

So, the number of ways to choose the remaining 0 cards out of 50 is C(50, 0) = 1.

The total number of possible 7-card hands is given by the combination formula: C(108, 7) = 153,858,295.

Therefore, the probability P(a hand containing two wild cards, two red cards, and three blue cards) = (28 * 300 * 2,300 * 1) / 153,858,295.

Calculating this, we get P(a hand containing two wild cards, two red cards, and three blue cards) ≈ 0.000440.

So the answer, rounded to six decimal places, is approximately 0.000440.

To solve these probability problems, we need to use some basic concepts of probability and combinations.

Definition: The probability of an event occurring is the number of favorable outcomes divided by the number of possible outcomes.

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

To calculate this probability, we need to determine the total number of possible hands and then find the number of hands that contain exactly two wild cards.

Total number of possible hands: There are 108 cards in the deck, and each player is dealt a hand of 7 cards. So, the total number of possible hands is given by the combination formula:

C(108, 7) = 108! / (7! * (108 - 7)!)

Number of hands containing exactly two wild cards: There are 8 wild cards in the deck. We need to choose 2 of them for our hand. The remaining 5 cards can be any non-wild cards, which are the remaining 100 cards. So, the number of hands containing exactly two wild cards is given by:

C(8, 2) * C(100, 5) = (8! / (2! * (8 - 2)!) ) * (100! / (5! * (100 - 5)!) )

Now, we can calculate the probability:

P(a hand containing exactly two wild cards) = (Number of hands containing exactly two wild cards) / (Total number of possible hands)

P(a hand containing exactly two wild cards) = [ C(8, 2) * C(100, 5) ] / [ C(108, 7) ]

Calculate the values using a calculator or software, and round the final answer to six decimal places.

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

To calculate this probability, we need to determine the total number of possible hands that satisfy the given condition and then find the number of those hands.

Total number of possible hands: Same as in the previous question, the total number of possible hands is given by:

C(108, 7) = 108! / (7! * (108 - 7)!)

Number of hands containing two wild cards, two red cards, and three blue cards:

We need to choose 2 wild cards, 2 red cards, and 3 blue cards. The remaining 0 cards can be any color other than wild, red, or blue, which means they can be yellow or green. So, we need to choose 0 cards from the remaining 25 yellow and green cards.

The number of hands containing two wild cards, two red cards, and three blue cards is given by:

C(8, 2) * C(25, 2) * C(25, 3) * C(25, 0) = (8! / (2! * (8 - 2)!) ) * (25! / (2! * (25 - 2)!)) * (25! / (3! * (25 - 3)!)) * (25! / (0! * (25 - 0)!))

Now, we can calculate the probability:

P(a hand containing two wild cards, two red cards, and three blue cards) = (Number of hands containing two wild cards, two red cards, and three blue cards) / (Total number of possible hands)

P(a hand containing two wild cards, two red cards, and three blue cards) = [ C(8, 2) * C(25, 2) * C(25, 3) * C(25, 0) ] / [ C(108, 7) ]

Calculate the values using a calculator or software, and round the final answer to six decimal places.

1.) 0.004846?

2.) 3.462816?

Are those the right answers?

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