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.

Answer 1

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