Question

Suppose the number of TV's in a household has a binomial distribution with parameters n=8n=8 and p=10p=10%.

Find the probability of a household having:

(a) 4 or 7 TV's

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(b) 5 or fewer TV's

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(c) 4 or more TV's

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(d) fewer than 7 TV's

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(e) more than 5 TV's

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Be sure to give your answers to at least 4 decimal places.

Answer 1

(a)

n =8

p = 0.1

q = 1 - p = 0.9

P(X = 4 or X = 7) = P(X = 4) + P(X = 7)

So,

P(X = 4 or X = 7) = 0.0046

(b)

P(X5) = P(X=0) + P(X = 1) + P( X= 2) + P( X= 3) + P(X =4) + P(X =5)

So,

P(X5) = 0.99998

(c)

P(X4) = P(X=4) + P(X=5) + P(X=6) + P(X=7) + P(X=8)

So,

P(X4) = 0.0050

(d)

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

        = 1.0000

(e)

P(X>5) = P(X=6) + P(X=7) + P(X=8) = 0.00002