# For binomial distribution X, with n = 7 and p = 0.6, what is P(X>3)?

##### 1 Answer
Oct 19, 2015

Pr{X>3} = f(4) + f(5) + f(6) + f(7) = ${\left(0.6\right)}^{4}$ [35$\times$ ${\left(0.4\right)}^{3}$
+ 21$\times$ 0.6$\times$${\left(0.4\right)}^{2}$ + 7$\times$${\left(0.6\right)}^{2}$ $\times$ 0.4 + ${\left(0.6\right)}^{3}$ ]
=0.710208

#### Explanation:

f (x) = $7 {C}_{x}$ )${\left(0.6\right)}^{x}$ ${\left(0.4\right)}^{7 - x}$ , x = 0, 1, ... 7 $\implies$
X follows Binomial Distribution with parameters p = 0.6 and n=7.
$7 {C}_{4}$ =$7 {C}_{3}$ = 7$\times$ 6$\times 5$/ 1$\times 2$ $\times 3$ = 7$\times 5$ = 35
$7 {C}_{5}$= $7 {C}_{2}$ = 7$\times 6$ /1$\times 2$ = 21, $7 {C}_{6}$ = $7 {C}_{1}$ = 7 and
$7 {C}_{7}$ = 1.