Question

Two cards are drawn successively without replacement from a well shuffled deck of cards . Find the mean and standard variation of the random variable X , where X is the number of aces .

Answer 1

Let X denotes a random variable of number of aces .
`therefore " " X = 0,1,2`
Now , `" " P(X = 0) = (48)/(52) * (47)/(51) = (2256)/(2652)`
`P(X = 1) = (48)/(52) * (4)/(51) + (4)/(52) * (48)/(51) = (384)/(2652)`
`P (X = 2) = (4)/(52) * (3)/(51) = (12)/(2652)`

We know that , Mean `(mu) = E(X) = sumXP(X)`
= `0 + (384)/(2652) + (24)/(2652)`
= `(408)/(2652) = (2)/(13)`
Also , Var(X) = `E(X^(2)) - [E(X)]^(2) = sumX^(2) P(X) - [E(X)]^(2)`
= `[0 + (384)/(2652) + (48)/(2652)] - ((2)/(13))^(2) " " [because E(X) = (2)/(13)]`
= `(432)/(2652) - (4)/(169) = 0.1628 - 0.0236 = 0.1391`
Standard deviation = `sqrt("Var"(X)) = sqrt(0.139) = 0.373` (approx)