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Question
Suppose that 80% of all families own a television set. If 5 families are interviewed at random, find the probability that
a. three families own a television set.
b. at least two families own a television set.
Solution
X = Number of families who own a television set.
P = Probability of families who own a television set.
`P=80%=80/100=4/5`
`q=1-p=1-4/5=1/5`
Given `n=5, X~B(5,4/5)`
The p.m.f. or X is given as
`P(X=x)=""^nC_xp^xq^(n-x)`
`=""^nC_xp^xq^(5-x)`
a. P(families own television set)
`=P(X=3)`
`=""^5C_3(4/5)^3(1/5)^(5-3)`
`=128/625`
`=0.2048`
b. P(At least two families own television set)
`P(X>=2)=1-P(X<2)`
`=1-[P(X=0)+P(X=1)]`
`=1-[""^5C_0(4/5)^0(1/5)^5+""^5C_1(4/5)^1xx(1/5)^4]`
`=1-(1/55+20/55)`
`=1-(21/55)=34/55`.
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