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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

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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)`.

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
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