Answer:
3/13
Step-by-step explanation:
Without going into probability theory and formulae, it is easier to solve this problem logically but I have given the theory explanation
P(A|B) is the probability of drawing a face card given you have drawn a diamond card
Once you have drawn a diamond card, you are left with 13 cards to deal with and only 3 of them are face cards
*** Hence P(A|B) = 3/13 ***
The long and theoretical explanation... :)
In this situation
We are asked to compute the conditional probability of one event happening given the other event has already happened
P(A|B) is the conditional probability of A happening given B has already occurred
The formula for conditional probability is (Wikipedia and other sources)
[tex]P(A|B) = \dfrac{P(A \cap B)}{P(A)}[/tex]
where P(A ∩ B) represents the probability of events A and B happening together
With these we can tackle the problem as follows:
Given
A = draw face card : {J, Q, K}
B = draw a diamond suit
Let's compute some of the probabilities
P(A) = probability of drawing a face card
=> since there are 3 face cards and 4 suits, total number of face cards = 3 * 4 = 12 and therefore
P(A) = total number of face cards/total number of cards
= 12/52
P(B) = probability of drawing a diamond suit
= number of diamond suit cards/total number of cards
= 13/52
P(A ∩ B) = probability of drawing a diamond which is also a face card
There are only 3 face cards of each suit so only 3 diamond face cards
Hence P(A∩B) = 3/52
We have already computed P(B) = 13/52
Using the formula
[tex]P(A|B) = \dfrac{P(A \cap B)}{P(B)}[/tex]
[tex]P(A|B) = \dfrac{3/52}{13/52} = \dfrac{3}{13}[/tex]