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प्रश्न
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.
उत्तर
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)
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