Question

Suppose the number of children in a household has a binomial distribution
with parameters n = 16, and p = 80 %.
Find the probability of a household having:
(a) 6 or 15 children
(b) 13 or fewer children
(c) 10 or more children
(d) fewer than 15 children
(e) more than 13 children

Answer 1

Here

n = 16

p = 0.8

q = 1 - p = 0.2

Formula for Binomial distribution:

a.)

In this case,

P[x = 6 or x = 15] = P[x = 6] + P[x = 15]

Putting in above formula we get,

P[x = 6] = 0.000215

P[x = 15] = 0.1126

P[x = 6 or x = 15] = 0.000215 + 0.1126 = 0.112815

b.)

Here,

P[X 13] = 1 - P[X > 13]

Putting in formula for x = 14,15,16 one by one we get and adding all of them we get

P[X > 13] = 0.3518

P[X 13] = 1 - 0.3518

P[X 13] = 0.6482

c.)

10 or more

We need to put x = 10,11,12,13,14,15,16 one by one in above formula and all of them

So, P[X 10] = 0.9733

d.)

Here we have to find less than 15 children which means

P[X < 15] = 1 - P[ X 15]

Now put x = 15 and 16 in above formula and adding the probabilities of both we get

P[ X 15] = 0.1407

P[X < 15] = 1 - 0.1407 = 0.8593

e.)

Here we need to find more than 13 children which means

P[X >13] = P[X = 14] + P[X = 15] + P[X = 16]

P[X >13] = 0.3518

Note: I am using binomial calculator available in below link. You can use tht as well

https://stattrek.com/online-calculator/binomial.aspx