Suppose the number of children in a household has a binomial

Suppose the number of children in a household has a binomial distribution
with parameters n = 11, and p = 70 %.
Find the probability of a household having:
(a) 2 or 7 children
(b) 5 or fewer children
(c) 4 or more children
(d) fewer than 7 children
(e) more than 5 children

Solution

given

Bionomial distribution n=11   p=0.7

letX is number of children

A) we have to find P(X=2 or X=7)

P(X=2 or X=7) =P(X=2) +P(X=7)

                       =¹¹C₂ (0.7)² (0.3)⁹ + ¹¹C₇ (0.7)⁷ (0.3)⁴

                       =0.00053 +0.2201 =0.22063   answer

B) 5 or fewer children

P(X<5)=?

P(X<5) =P(X=0) +P(X=1) +p(X=2) +P(X=3) +P(X=4) +P(X=5)

               =¹¹C₀ (0.7)⁰ (0.3)¹¹+.....................+¹¹C₅ (0.7)⁵ (0.3)⁶

                   ⁼0.0000018+0.0000455+0.0005304+0.0037132 +0.0173282 +0.0566056 =0.0782247

C) 4 or more children

P(X>4) =?

since

P(X>4) =1-P(X<4)

also

P(X<4) =P(X<3) =P(X=0)+P(X=1) +P(X=2) +P(X=3)

                          =¹¹C₀ (0.7)⁰ (0.3)¹¹ +....................+¹¹C₃ (0.7)³ (0.3)⁸

                           =0.0000018 +0.0000455+0.0005304+0.0037132=0.0042909

so

required probability

P(X>4) =1 - 0.0042909 =0.9957

D) fewer than 7 children

P(X<7)=?

so P(X<7)=P(X<6) =P(X=0)+P(X=1)+P(X=2) +P(X=3)+P(X=4)+P(X=5)+P(X=6)

                               =¹¹C₀ (0.7)⁰ (0.3)¹¹ +.........................+¹¹C₆ (0.7)⁶ (0.3)⁵

                               =0.0000018 +0.0000455+0.0005304 +0.0037132 +0.0173282+0.0566056+0.1320798

                             =0.2103045

E)

more than 5 children

P(X>5)=1-P(X< 5)

from B

we get

P(X<5) =0.0782247

P(X>5) =1-0.0782247 =0.921775