Question

Three cards are drawn from a deck of 52 cards without replacement.

(a) What is the probability that the third card is a spade (♠) given that the first card is a spade?

(b) What is the probability that all cards are spades given that at least one of them is a spade?

(c) Let Y be the number of black cards drawn. What is the probability that all 3 cards are black given that the first card is a spade? An ace?

(d) Let X be the number of aces drawn. Find P[Y = 2|X = 2]

Answer 1

There are 52 cards in a deck.

●26 of them are black and 26 of them are red.

●13 cards are kept in clubs spades hearts and diamonds in which clubs and spades are black and hearts and diamonds are red.

●each of ace king queen and Jack are present one in each group.

●9 cards numbered 2 to 10 are also present in each group.

a)probability of first card being a spade =13/52.

Probability of 2nd card not being a spade=39/51 (1st card is spade--->52-1=51)

Probability of 3rd card being a spade=12/50 (13-1=12)

Probability that 3rd card is a spade given that 1st card is a spade=(13/52)×(39/51)×(12/50) =39/850     =0.045882

b)using formula for conditional probability

P(A/B)= P(AB)/P(B)

Here P(B)=probability of atleast one of them being a spade. =((13C1×39C2)+(13C2×39C1)+ (13C3))/52C3

=0.586

P(A/B)=(13/52 × 12/51 × 11/50)/ 0.586

=0.22

C) probability that 1st card being a spade =(13C1×39C2)/52C3

=0.435~0.44

Let the above be the P(B)

P(A/B)=P(AB)/P(B)

=0.1176/0.44(probability of all 3 being black=0.1176)

=0.267