Question

A typical deck of cards has 4 suits ( ♡,♢, ♡,♢, ♣,♠ ♣,♠ : and the following 13 denominations: Ace, 2 thru 10, Jack, Queen, and king). A full deck contains 52 cards. A single card has three characteristics: suit, denomination and colour. examples: [King of ♡ ♡ , is red] or [4 of ♣ ♣ , is black] etc.

(a) Say 4 cards are selected from the deck of 52, without replacement. Let X X be the number of ♢ ♢ 's drawn. What is the standard deviation of X,SD[X]=? (use at least four digits after the decimal point if rounding)

(b) If 30 cards are selected from the deck of 52, with replacement and you are also told the number of ♣ ♣ 's drawn is at most 10 but at least 3, what is the probability there will be exactly 6 ♣ ♣ 's drawn. Probability = equation editor (use at least four digits after the decimal point if rounding

Answer 1

(a) Here 4 cards are selected from the deck of 52.

Here X is the number of diamonds out of 4.

so here the distribution of number of diamonds are hypergeometric

f(x) = ¹³C_X³⁹C_(4-X)/ ⁵²C₄

f(x = 0) = ¹³C₀³⁹C₄/ ⁵²C₄ = 0.3038

f(x = 1) = ¹³C₁³⁹C₃/ ⁵²C₄ = 0.4388

f(x = 0) = ¹³C₂³⁹C₂/ ⁵²C₄ = 0.2135

f(x = 0) = ¹³C₃³⁹C₃/ ⁵²C₄ = 0.0412

f(x = 0) = ¹³C₄³⁹C₀/ ⁵²C₄ = 0.3038

Mean = 4 * 13/52 = 1

Variance = 0.3038 * (0 - 1)² + 0.4388 * (1 -1)² + 0.2135 * (2 - 1)² + 0.0412 * (3 - 1)² + 0.0026 * (4 -1)²  = 0.705882

SD[X] = sqrt(0.705882) = 0.8402

(b) Here 30 cards are drawn from standard deck of 52 so n = 30

Number of Clubs = 13

Pr(CLubs) = 1/4 = 0.25

so here it is given that number of clubs drawn are in between 3 to 10.

so we have to find that

Pr(X = 6 l 3 ≤ X ≤ 10) = BIN( x = 6 ; 30 ; 0.25) / BIN (3 ≤ X ≤ 10 ; 30 ; 0.25)

Pr(3 ≤ X ≤ 10) = BIN (3 ≤ X ≤ 10 ; 30 ; 0.25) = BIN (x ≤ 10 ; 30 ; 0.25) - BIN (x < 3 ; 10 ; 0.25)

= 0.8943 - 0.0106 = 0.8837

Pr(X = 6) = 0.1455

Pr(X = 6 l 3 ≤ X ≤ 10) = BIN( x = 6 ; 30 ; 0.25) / BIN (3 ≤ X ≤ 10 ; 30 ; 0.25) = 0.1646