Probability questons!!?

Question

How would you find the probability of
1) an ace given that it is a diamond (standard deck 52 cards)
2) a diamond given it is an ace (standard deck of cards)

the answer my professor gave us was
1) 4/52
2) 13/52

i dont know how he got this so any help would be appreciated!

Answer 1

The definition for conditional probability is:

For any two events A and B where P(B) ≠ 0, the probability of A given B is:

P( A | B) = P(A ∩ B) / P(B)

P(ace and diamond) = 1/52
P(diamond) = 13/52
P(ace) = 4/52 = 1/13

P(Ace | diamond)
= P(Ace ∩ diamond) / P(diamond)
= (1/52) / (13 / 52) = 1 / 13 = 4 / 13

P(diamond | Ace)
= P( ace ∩ diamond) / P(ace)
= (1/52) / (1/13) = 13 / 52 = 1/4

Answer 2

There are four aces in a pack of cards and there are 13 diamonds. These overlap, so that one ace is a diamond.

Given you have a diamond, there is a one in 13 chance that it is an ace 1/13 = 4/52 (standard notation for cards is to put odds in terms of 52)

Given you have an ace, there is a one in 4 chance that it is a diamond 1/4 = 13/52 (standard notation)

Answer 3

notice that 4/52 reduces to 1/13
1) there are 13 diamond cards, out of which 1 is an ace that is a diamond.
so P = 1/13


2) 13/52 = 1/4

there are 4 aces. 1 is diamond
so P = 1/4

Answer 4

1) There are 4 aces in a deck and there is one ace per suit. So then you have a 4 (how many aces in the deck) in 52 (the total number of cards) chance that an ace is a diamond.
2) The same principle is used here. There are 13 cards in each suit and only one ace per suit. So then you have a 13 (how many cards in a suit) in 52 (total number of cards in a deck) chance that a diamond is an ace.

Answer 5

If one parent is NN, which is homozygous dominant, then the probability is zero. If both parents are Nn, then there is a 25 percent chance.