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Question
A die is tossed twice. A ‘success’ is getting an even number on a toss. Find the variance of the number of successes.
Solution
Let E be the event of getting even number on tossing a die.
∴ P(E) = `3/6 = 1/2` and `"P"(bar"E") = 1 - 1/2 = 1/2`
Here X = 0, 1, 2
P(X = 0) = `"P"(bar"E")*"P"(bar"E")`
= `1/2*1/2`
= `1/4`
P(X = 1) = `"P"("E")*"P"(bar"E") + "P"(bar"E")*"P"("E")`
= `1/2*1/2 + 1/2*1/2`
= `1/4 + 1/4`
= `2/4`
P(X = 2) = P(E).P(E)
= `1/2*1/2`
= `1/4`
∴ Probability distribution table is
| X | 0 | 1 | 2 |
| P(X) | `1/4` | `2/4` | `1/4` |
E(X) = `0 xx 1/4 + 1 xx 2/4 + 2 xx 1/4`
= `2/4 + 2/4`
= 1
E(X2) = `0 xx 1/4 + 1 xx 2/4 + 4 xx 1/4`
= 3/2`
∴ Variance (X) = E(X2) – [E(X)]2
= `3/2 - 1`
= `1/2`
= 0.5
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