Sum

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.

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#### 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`.

Concept: Conditional Probability

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