Question

A poker hand consisting of 9 cards is dealt from a standard deck of 52 cardsFind the probability that the hand contains exactly 7 face cards. Leave your answer as a reducedfractionThe probability is

Answer 1

A face card is a card that is a King, a Queen or a Jack. In a standard deck, there is a total of 12 face cards.

Suppose that the first 7 cards we take are faces and the remaining two are not.

The probability of taking a face on the first withdrawal, is 12/52, since there are 12 faces and there is a total of 52 cards.

On the second withdrawal, there are 11 faces remaining, and a total of 51 cards remain. Therefore, the probability of the second card also being a face is 11/51.

On the third withdrawal, the probability of taking a face will be 10/50, and so on:

On the fourth withdrawal, the probability will be 9/49.

On the fifth, 8/48.

On the sixth, 7/47.

On the seventh, 6/46.

The next two withdrawals must not be faces. On the eight withdrawal, a total of 45 cards remain, 5 of them are faces. So, the probability of not taking a face, is 40/45.

On the nineth withdrawal, the probability of not taking a face is 39/44.

The combined probability of all this events happening one after another is the product of the individual probabilities. Therefore, the probability of the hand containing exactly 7 face cards, is:

`(12)/(52)\cdot(11)/(51)\cdot(10)/(50)\cdot(9)/(49)\cdot(8)/(48)\cdot(7)/(47)\cdot(6)/(46)\cdot(40)/(45)\cdot(39)/(44)`

Write each factor as a product of powers of prime numbers:

`=(2^2\cdot3)/(2^2\cdot13)\cdot(11)/(3\cdot17)\cdot(2\cdot5)/(2\cdot5^2)\cdot(3^2)/(7^2)\cdot(2^3)/(2^4\cdot3)\cdot(7)/(47)\cdot(2\cdot3)/(2\cdot23)\cdot(2^3\cdot5)/(3^2\cdot5)\cdot(3\cdot13)/(2^2\cdot11)`

Use the laws of exponents to reduce this expression:

`\begin{gathered} =(2^(10)\cdot3^5\cdot5^2\cdot7\cdot11\cdot13)/(2^(10)\cdot3^4\cdot5^3\cdot7^2\cdot11\cdot13\cdot17\cdot23\cdot47) \\ =(3)/(5\cdot7\cdot17\cdot23\cdot47) \\ =(3)/(643195) \end{gathered}`

Therefore, the probability is:

`(3)/(643195)`