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?

Explanation

Let hitting the target be a sucess in a shoot.

We have

`p=`Probability of hitting the target `=0.25=frac{1}{4}`

also, `q=1-p=1-frac{1}{4}=frac{3}{4}`

Let `X` denotes the number of succes in a sample of `7` trials. Then,

`X` follows binomial distribution with parameters `n=7\and\p=frac{1}{7}`

`therefore P(X=r)=^{7} C_{r} p^{r} q^{(7-r)}={ }^{7} C_{r}(frac{1}{4})^{r}(frac{3}{4})^{(7-r)}=frac{{ }^{7} C_{r} 3^{(7-r)}}{4^{7}}, where r=0,1,2,3,4`

Now,

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