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 be the random variable such that X = 0, 1, 2

And E = The event of drawing an ace

And F = The event of drawing non-ace.

∴ P(E) = `4/52` and `"P"(bar"E") = 48/52`

Now P(X = 0) = `"P"(bar"E")*"P"(bar"E")`

= `48/52*47/51`

= `188/221`

P(X = 1) = `"P"("E")*"P"(bar"E") + "P"(bar"E")*"P"("E")`

= `4/52 xx 48/51 xx 48/52 xx 4/51`

= `32/221`

P(X = 2) = P(E).P(E)

= `4/52*3/51`

= `1/221`

We have Distribution Table:

X	0	1	2
P(X)	`188/221`	`32/221`	`1/221`

Now, Mean E(X) = `0 xx 188/221 + 1 xx 32/221 + 2 xx 1/221`

= `32/221 + 2/221`

= `34/221`

= `2/13`

E(X²) = `0 xx 188/221 + 1 xx 32/221 + 4 xx 1/221`

= `32/221 + 4/221`

= `36/221`

∴  Variance = E(X²) – [E(X)]²

= `36/221 - (2/13)^2`

= `36/221 - 4/169`

= `(468 - 68)/(13 xx 221)`

= `400/2873`

Standard deviation = `sqrt(400/2873)`

= 0.377  .....(Approx)