problem in computing probability

Question

Find the probability that a poker hand of 5 cards will contain no card smaller than 7, given that it contains at least 1 card over 10, where aces are treated as high cards.

Answer 1

Let $ A $ be the event "poker hand of 5 cards contains no card $ < $ than 7," and let $ B $ be the event "poker hand of 5 card contains at least 1 card $ > $ 10."

You need to use the conditional probability rule, i.e. $ P(A|B) = \frac{P(A \cap B)}{P(B)} $.

Note: there are 52 cards in total, including: 1) 16 cards $ \geq 10 $, 2) 32 cards $ \geq 7 $, and 3) 16 cards $ \geq 7 $ and $ \leq 10 $

Part 1: $ P(B) $

Let $ X $ be the number of cards over 10 in a poker hand of 5 cards. Then $ X \sim Hypergeometric(N=52,n=5,k=16). $

$ P(B) = P(X \geq 1 ) = 1 - P(X=0) = 1 - \frac{{{16}\choose{0}} \times {{52-16}\choose{5-0}} } {{52}\choose{5}} = 1 - \frac{{{36}\choose{5}}}{{{52}\choose{5}}} = \frac{{{{52}\choose{5}}} - {{36}\choose{5}}}{{{52}\choose{5}}} $

Part 2: $ P(A \cap B) $

For this part, it is easiest to use combinatorics.

You can start by looking at the different possible 5-card hands that contain no card $ < $ 7. Since there are 32 cards $ \geq $ 7, then there are $ {32}\choose{5} $ such hands.

To account for the second condition (i.e. that you must also have at least one card $ > $ 10), you must subtract all of the hands that do NOT have at least one card $ > $ 10. How many hands within the above subset don't have at least one card $ > $ 10? Since there are 16 cards $ \geq 7 $ and $ \leq 10 $ , we find that there are $ {16}\choose{5} $ such hands.

Putting this all together, we get $ P(A \cap B) = \frac{ {{32}\choose{5}} - {{16}\choose{5}} }{{52}\choose{5}} $.

Final answer: $ P(A|B) $

$$ P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{{{{52}\choose{5}}} - {{36}\choose{5}}}{{{52}\choose{5}}}}{\frac{ {{32}\choose{5}} - {{16}\choose{5}} }{{52}\choose{5}}} = \frac{{{{52}\choose{5}}} - {{36}\choose{5}}} {{{32}\choose{5}} - {{16}\choose{5}}} $$