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  • The probability of a man hitting a target is 0.25. He shoots 7 times. What is the probability of his hitting at least twice?

The probability of a man hitting a target is 0.25. He shoots 7 times. What is the probability of his hitting at least twice?

Answers (1)

Given-

The man shoots 7 times, therefore, n=7

And probability of hitting target is

\\ \mathrm{P}=0.25 \mathrm{or } \frac{1}{4} \\ \mathrm{q}=1-\frac{1}{4}=\frac{3}{4}, \mathrm{r} \geq 2 \\ \text { Where, } \mathrm{P}(\mathrm{X})=^\mathrm{n}{\mathrm{c_r}}(\mathrm{p})^{\mathrm{r}} \mathrm{q}^{\mathrm{n}-\mathrm{r}} \\ \therefore \mathrm{P}(\mathrm{X}=\mathrm{r} \geq 2)=1-[\mathrm{p}(\mathrm{r}=0)+\mathrm{P}(\mathrm{r}=1)]

\\ \Rightarrow 1-\left[7 c_{0}\left(\frac{1}{4}\right)^{0}\left(\frac{3}{4}\right)^{7-0}+7 c_{1}\left(\frac{1}{4}\right)^{1}\left(\frac{3}{4}\right)^{7-1}\right] \\ =1-\left[\left(\frac{3}{4}\right)^{7}+{7}\left(\frac{1}{4}\right)\left(\frac{3}{4}\right)^{6}\right] \\ =1-\left[\left(\frac{3}{4}\right)^{6}\left(\frac{3}{4}+\frac{7}{4}\right)\right] \\ =1-\left[\frac{3^{6}}{4^{6}} \times \frac{10}{4}\right] \\ =\quad \\ =1-\left[\frac{7290}{16384}\right] \\ =\frac{4547}{8192}

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infoexpert22

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