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A die is thrown 100 times. Getting an even number is considered a success. What is the variance of the number of successes?

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  • $\begingroup$ Calculate the probabilities of no of even numbers outcomes $N$ from 0 to 100. Then use the formula $$\sigma ^2 = \text{E}(N^2) - (\text{E}(N))^2$$ $\endgroup$ – samjoe Apr 6 '17 at 6:03
  • $\begingroup$ By a die, do you mean a six sided die? $\endgroup$ – Benji Altman Apr 6 '17 at 6:20
  • $\begingroup$ Yes, Benji Altman $\endgroup$ – DireVish Apr 6 '17 at 7:31
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p(getting even number) $= \frac 36 = \frac 12$

q(not even number $= \frac 36 = \frac 12$

Variance $= npq$

$= 100 \times \frac 12 \times \frac 12$

$= 25$

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  • $\begingroup$ Sorry sir I am unaware of this result. If we use the standard formula will we get the same result? $\endgroup$ – samjoe Apr 6 '17 at 6:44
  • $\begingroup$ Yes if you use standard formula you get same result. $\endgroup$ – Kanwaljit Singh Apr 6 '17 at 9:42
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What you are considering is actually a binomial distribution, like exactly in the case of a fair coin. You have a $p=50\% $ probability of getting an even number at each throw. The probability distribution of having $k$ successes out of $n$ tries is: $$f(k;n,p)=\binom{n}{k}p^k(1-p)^{n-k}$$

The expected number of successes over $n=100$ tries is: $$E[X]=\sum_{k=0}^n k\binom{n}{k}p^k(1-p)^{n-k} =np=100\cdot0.5=50$$ The proof is available on every probability textbook.

The variance is: $$\sigma^2=E[X^2]-E[X]^2=np(1-p)=25$$

The article on wikipedia is well done. https://en.wikipedia.org/wiki/Binomial_distribution

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