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Question
A card is drawn at random and replaced four times from a well shuftled pack of 52 cards. Find the probability that -
(a) Two diamond cards are drawn.
(b) At least one diamond card is drawn.
Solution
Let a random variable
X = number of diamond cards.
Here X follows Binomial distribution n = 4.
P = P(success) = `13/52 = 1/4`
q = 1 - p
= 1 - `1/4 = 3/4`
X - B `(4, 1/4)`
`therefore` p.m.f is
P(X = x) = `""^n"C"_xp^xq^(n-x)`
P(X = x) = `""^4"C"_x (1/4)^x (3/4)^(4 -x)`
(a) P (Two diamonds cards are drawn)
= P(X = 2)
= `""^4"C"_2 (1/4)^2 (3/4)^(4-2)`
= `(4 xx 3)/(2 xx 1) xx 1/16 xx 9/16`
= `27/128`
= 0.2109
(b) P (at least one diamond card drawn)
= 1 - (X = 0)
= 1 - `""^4"C"_0 (1/4)^0 (3/4)^(4 -0)`
= `0- (3/4)^4 = 1 - 81/256`
= `175/256` = 0.6835
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