Five cards are drawn successively with replacement from a well-shuffled deck of 52 cards, find the probability that only 3 cards are spades
Solution 1
Let X = number of spade cards.
p = probability of drawing a spade card from pack of 52 cards.
Since, there are 13 spade cards in the pack of 52 cards,
∴ p = `13/52 = 1/4 and "q" = 1 - "p" = 1 - 1/4 = 3/4`
Given: n = 5
∴ X ~ B`(5, 1/4)`
The p.m.f. of X is given by
P(X = x) = `"^nC_x p^x q^(n - x)`
i.e. p(x) = `"^5C_x (1/4)^x (3/5)^(5 - x)`, x = 0, 1, 2,...,5
P(only 3 cards are spades) = P(X = 3)
= p(3) = `"^5C_3(1/4)^3(3/4)^(5 - 3)`
`= (5!)/(3! * 2!) (1/4)^3 (3/4)^2`
`= (5* 4* 3!)/(3!* 2* 1) xx 1/64 xx 9/16 = 45/512`
Hence, the probability of only 3 cards are spades = `45/512`
Solution 2
Let X denote the number of spades.
P(getting spade) = p = `(13)/(52) = (1)/(4)`
∴ q= 1 – p = `1 - (1)/(4) = (3)/(4)`
Given, n = 5
∴ X ~ B`(5, 1/4)`
The p.m.f. of X is given by
P(X = x) = `""^5"C"_x(1/4)^x (3/4)^(5 - x), x` = 0, 1, ...,5
P(only 3 cards are spades)
= P(X = 3)
= `""^5"C"_3(1/4)^3(3/4)^2`
= `(5!)/(3! xx 2!) xx (3^2)/(4^3 xx 4^2)`
= `(5 xx 4 xx 3!)/(3! xx 2 xx 1) xx (9)/(4^5)`
= `(90)/(4^5)`.
