Question

A hand of 13 cards is to be dealt at random and without replacement from an ordinary deck of playing cards. Find the conditional probability that there are at least 3 kings in the hand given that the hand contains at least 2 kings.

Answer 1

There are (52/13)ways to select 13 cards from 52 cards.
Consider how you would create a hand of 13 cards with the required number of kings.
There are (4/n) ways of choosing the required number of kings.
There are then (48/13−n)(48/13−n) ways of choosing the remaining (13 - n) cards.

Let A be the event of getting at least 2 kings.
This can be thought of as 3 separate events corresponding to 2, 3 or 4 kings. By the addition principle:

Number of ways of getting A = (4/2)* (48/13−2)+ (4/3) * (48/13−3) + (4/4)* (48/13−4)

So,

P(A)=(4/2)∗(48/13−2)+(4/3)∗(48/13−3)+(4/4)∗(48/13−4)
/(52/13)

Let B be the event of getting at least 3 kings.
This can be thought of as 2 separate events corresponding to 3 or 4 kings. By the addition principle:

P(B)=(4/3)∗(48/13−3)+(4/4)∗(48/13−4)
/(52/13)

Finally, by the conditional probability formula:

P(B|A)=P(B∩A)/P(A)=P(B)/P(A)