Question

Suppose the number of children in a household has a binomial distribution with parameters n=25 and p=15%.

Find the probability of a household having:

(a) 1 or 24 children  
(b) 22 or fewer children  
(c) 12 or more children  
(d) fewer than 24 children  
(e) more than 22 children

Answer 1

Let X be the number of children

X ~ Binomial(25,0.15)

Here are the individual probabilities:

X	P(X=x)
0	0.0171978098522079000000
1	0.0758726905244466000000
2	0.1606715799341220000000
3	0.2173791963814590000000
4	0.2109856906055340000000
5	0.1563776295076310000000
6	0.0919868408868419000000
7	0.0440609237861344000000
8	0.0174947785621416000000
9	0.0058315928540471800000
10	0.0016465673940839100000
11	0.0003962327953677880000
12	0.0000815773402227797000
13	0.0000143960012157846000
14	0.0000021775464023875900
15	0.0000002818001226619240
16	0.0000000310808958818298
17	0.0000000029037515183716
18	0.0000000002277452171272
19	0.0000000000148069645810
20	0.0000000000007838981249
21	0.0000000000000329368960
22	0.0000000000000010567988
23	0.0000000000000000243253
24	0.0000000000000000003577
25	0.0000000000000000000025

To solve all these problems would use the above probability values:-

a)

P(1 or 24 children) = P(1 child) + P(24 child) = 0.0758726905244466 + 0.0000000000000000003577 =

0.0758726905244466000000

b)

P(22 or fewer children)

=

0.9999999999999990000000

c)

P(12 or more children) =

0.0000984669159811122000

d)

P(fewer than 24 children)

= 0.9999999999999990000000

e)

P(more than 22 children)

= 0.0000000000000000246855

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