If getting 5 or 6 in a throw of an unbiased die is a success and the random variable `X` denotes the number of successes in six throws of the die, find `P(X >=4)` .

Explanation

Let `X ` denote the number of successes, i.e. of getting `5 or 6` in a throw of die in `6` throws. Then,
` X` follows a binomial distribution with `n=6`;

`p=` of getting` 5 or 6=frac{1}{6}+frac{1}{6}=frac{1}{3} ; q=1-p=frac{2}{3};`


`P(X=r)=^{6} C_{r}(frac{1}{3})^{r}(frac{2}{3})^{6-r}`


`P(X geq 4)=P(X=4)+P(X=5)+P(X=6)`

`=^{6} C_{4}(frac{1}{3})^{4}(frac{2}{3})^{6-4}+^{6} C_{5}(frac{1}{3})^{5}(frac{2}{3})^{6-5}+^{6} C_{6}(frac{1}{3})^{6}(frac{2}{3})^{6-6} `

`=frac{1}{3^{6}}(60+12+1) `

...