Question

In: Statistics and Probability

draw six cards at random from a deck of 52 playing cards 60 times with replacement....

draw six cards at random from a deck of 52 playing cards 60 times with replacement. Let X be the number of queen cards.

Find the probability distribution of X and Var (x)

Expert Solution

Probability of drawing a queen at one random draw = 4/52 = 1/13

Probability of not drawing a queen at one random draw = 1 - 1/13 = 12/13

Mass points of X are :- 0,1,2,3,4,5,6

P(X=0) = Probability of not drawing a queen at 6 random draws = (12/13)⁶ = 2985984/4826809

P(X=1) = Probability of drawing a queen at 1 random draw and not drawing a queen at 5 random draws = (1/13)(12/13)⁵ = 248832/4826809

P(X=2) = Probability of drawing a queen at 2 random draws and not drawing a queen at 4 random draws =(1/13)²(12/13)⁴

= 20736/4826809

P(X=3) = Probability of drawing a queen at 3 random draws and not drawing a queen at 3 random draws = (1/13)³(12/13)³

= 1728/4826809

P(X=4) = Probability of drawing a queen at 4 random draws and not drawing a queen at 2 random draws = (1/13)⁴(12/13)²

= 144/4826809

P(X=5) = Probability of drawing a queen at 5 random draws and not drawing a queen at 1 random draw = (1/13)⁵(12/13)

= 12/4826809

P(X=6) = Probability of drawing a queen at 6 random draws = (1/13)⁶ = 1/4826809

E(X) = 0P(X=0) + 1P(X=1) + 2P(X=2) + 3P(X=3) + 4P(X=4) + 5P(X=5) + 6P(X=6)

= 248832/4826809 + 220736/4826809 + 31728/4826809 + 4144/4826809 + 512/4826809 + 61/4826809

= 296130/4826809

E(X²) = 0²P(X=0) + 1²P(X=1) + 2²P(X=2) + 3²P(X=3) + 4²P(X=4) + 5²P(X=5) + 6²P(X=6)

= 248832/4826809 + 420736/4826809 + 91728/4826809 + 16144/4826809 + 2512/4826809 + 361/4826809

= 349968/4826809

Var(X) = E(X²) - [E(X)]² = 296130/4826809 - (349968/4826809)² = 1306885348146/23298085122481 = 0.0561

orchestra answered 1 year ago

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