# Five cards are drawn successively with replacement from a well-shuffled deck of 52 cards, find the probability that only 3 cards are spades - Mathematics and Statistics

Sum

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(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).

Concept: Binomial Distribution
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#### APPEARS IN

Balbharati Mathematics and Statistics 2 (Commerce) 12th Standard HSC Maharashtra State Board
Chapter 8 Probability Distributions
Exercise 8.3 | Q 1.04 | Page 150