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alance07 · May 9, 2009, 11:27 AM

You are given a standard deck of playing cards (52). Assume five cards are randomly chosen from the deck.

(a) How many hands contain four aces?

(b) How many hands contain two aces and two kings?

(c) How many hands contain exactly two 7's?

Thanks in advance!

N0help4u · May 9, 2009, 11:40 AM

You are suppose to do one post per question topic
These need combined
https://www.askmehelpdesk.com/mathem...ds-351703.html

galactus · May 9, 2009, 01:25 PM

Originally Posted by alance07

You are given a standard deck of playing cards (52). Assume five cards are randomly chosen from the deck.

(a) How many hands contain four aces?

I will write these as a probability. Can you count the number then?

We choose 4 Aces from the 4 in the deck. We need 1 of the remaining 48 cards.

In all, we draw 5 from 52.

$P(\text{2 Aces})=\frac{C(4,4)C(48,1)}{C(52,5)}=\frac{1}{5414 5}$

is the probability of being dealt 4 Aces.

(b) How many hands contain two aces and two kings?

We choose 2 Aces from the 4 and 2 Kings form the 4. 1 card from the remaining 44.

$P(\text{2 Aces and 2 Kings})=\frac{C(4,2)C(4,2)C(44,1)}{C(52,5)}=\frac{ 33}{54145}$

(c) How many hands contain exactly two 7's?

We choose 2 7's from the 4 and 3 from the remaining 48

$P(\text{2 7's})=\frac{C(4,2)C(48,3)}{C(52,5)}=\frac{2162}{54 145}\approx .04$

See how that works now?

I know I used probability instead of counting the number. Can you do that now?