If there are 7 cards left in a deck of cards, and 5 cards in the other players hand and you've seen all the cards except for the following:
Spades: ♠ 10 K
Hearts:♥ 10 Q K A
Clubs: ♣ 10 J Q K A
Diamonds: ♦ A
- What is the probability that the Ace of Diamonds is in the deck?
- What is the probability that the other player's hand contains at most 1 spade?
- What is the probability that the other player's hand contains at most 1 spade, given that the ace of diamonds is in the deck?
My first thought for number 1 is simply P(A) = 7/12, but I have a feeling that is wrong.
For number 2, wouldn't the answer be 762/792?
Since $\frac{\binom{2}{2}\binom{10}{3}}{\binom{12}{5}}$ represents the number of hands where both spades are drawn,
P(B) = 1- $\frac{\binom{2}{2}\binom{10}{3}}{\binom{12}{5}}$ = $\frac{762}{792}$ where A is the event that at most one spade is drawn.
For number 3, I know in order to compute P(B|A), you need to know the intersection of A $\cap$ B, which I am confused about how to compute because they are aren't disjoint.
How do I go about solving these?