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  • A die is tossed twice. A ‘success’ is getting an even number on a toss. Find the variance of the number of successes.

A die is tossed twice. A ‘success’ is getting an even number on a toss. Find the variance of the number of successes.

Answers (1)

Let X be the random variable for a ‘success’ for getting an even number on a toss.

∴ X = 0, 1, 2

n = 2
Even number on dice = 2, 4, 6
∴ Total possibility of getting an even number = 3
Total number on dice = 6
p = probability of getting an even number on a toss

\begin{aligned} &=\frac{3}{6}\\ &=\frac{1}{2}\\ &q=1-p\\ &=1-\frac{1}{2}\\ &=\frac{1}{2}\\ &\text { The probability of x successes in n-Bernoulli trials is }{ }^{n} \mathrm{C}_{r} \mathrm{p}^{r} \mathrm{q}^{\text {n-r }}\\ &P(x=0)=^2C_{0}\left(\frac{1}{2}\right)^{0}\left(\frac{1}{2}\right)^{2-0}=1 \times 1 \times \frac{1}{4}=\frac{1}{4}\\ &P(x=1)=^2C_{1}\left(\frac{1}{2}\right)^{1}\left(\frac{1}{2}\right)^{2-1}=2 \times \frac{1}{2} \times \frac{1}{2}=\frac{1}{2}\\ &P(x=2)=^2C_{2}\left(\frac{1}{2}\right)^{2}\left(\frac{1}{2}\right)^{2-2}=1 \times \frac{1}{4} \times 1=\frac{1}{4} \end{aligned}

\begin{aligned} &\begin{array}{|c|c|c|c|} \hline \mathrm{X} & 0 & 1 & 2 \\ \hline \mathrm{P}(\mathrm{X}) & 1 / 4 & 1 / 2 & 1 / 4 \\ \hline \end{array}\\ &E(X)=\Sigma X P(X)=0 \times \frac{1}{4}+1 \times \frac{1}{2}+2 \times \frac{1}{4}\\ &=\frac{1}{2}+\frac{1}{2}\\ &=1\\ &\text { And, } \operatorname{Var}(\mathrm{X})=\mathrm{E}\left(\mathrm{X}^{2}\right)-[\mathrm{E}(\mathrm{X})]^{2}\\ &=\Sigma X^{2} P(X)-[E(X)]^{2} \end{aligned}

\\ =\left[0 \times \frac{1}{4}+1^{2} \times \frac{1}{2}+2^{2} \times \frac{1}{4}\right]-(1)^{2} \\ =\left[\frac{1}{2}+1\right]-1 \\ =\frac{3}{2}-1 \\ =1.5-1 \\ =0.5

 

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