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  • Ten coins are tossed. What is the probability of getting at least 8 heads?

Ten coins are tossed. What is the probability of getting at least 8 heads?

Answers (1)

Let X = the random variable for getting a head.

Here, n=10, r≥8

 r=8,9,10

\begin{aligned} &\mathrm{p}=\frac{1}{2}, \mathrm{q}=\frac{1}{2}\\ &\text { It is known to us that, }\\ &\mathrm{P}(\mathrm{X}=\mathrm{r})=\mathrm{n}_{\mathrm{c}_{\mathrm{r}}}(\mathrm{p})^{\mathrm{r}} \mathrm{q}^{\mathrm{r}} \mathrm{q}^{\mathrm{n}-\mathrm{r}}\\ &\therefore P(X=r)=P(r=8)+P(r=9) + P(r=10)\\ &=^{10}C_{8}\left(\frac{1}{2}\right)^{10} (\frac{1}{2})^{10-8}+^{10}{c_{9}}\left(\frac{1}{2}\right)^{9}\left(\frac{1}{2}\right)^{10-9}+^{10}c_{10}\left(\frac{1}{2}\right)^{10} \frac{1}{2}^{10-10}\\ &=\frac{10 !}{8 ! 2 !}\left(\frac{1}{2}\right)^{10}+\frac{10 !}{9 ! 1 !}\left(\frac{1}{2}\right)^{10}+\frac{10 !}{0 ! 10 !}\left(\frac{1}{2}\right)^{10}\\ &=\left(\frac{1}{2}\right)^{10}\left[\frac{10 \times 9}{2}+10+1\right]\\ &=\left(\frac{1}{2}\right)^{10} \times 56=\frac{1}{2^{7} \times 2^{3}} \times 56\\ &=\frac{7}{128} \end{aligned}

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