Supose that 80 % of all families own a elevision set. If 10 families are interviewed at random, find the probability that seven famileis own a television set .

Explanation

Let x be the no. of families who own a television set.
P (families who own a television set) `=P=80%=(80)/(100)=0.8`.
`q=1-p=1-0.8=0.2`
Given : n=10
`X:~B(n,p):.X~B(10,0.8)`
The p.m.f of X is given by
`P(X=x)=""^(n)C_(x).P^(x).q^(n-x),x=1,2,.........n,`
`:.P(X=x)=""^(10)C_(x).(0.8)^(x).(0.2)^(10-x),x=1,2,.........10,`
P (at most 3 families own a television set)
`impliesP(xle3)=P(x=0)+P(x=1)+P(x=2)+P(x=3)`
`=""^(10)C_(0).(0.8)^(0)(0.2)^(10)+""^(10)C_(1)(0.8)^(1)(0.2)^(9)+""^(10)C_(2)(0.8)^(0)(0.2)^(10)+""^(10)C_(2)(0.8)^(2)(0.2)^(8)+""^(10)C_(3)(0.8)^(3)(0.2)^(7)`
`=1xx1xx(1024)/(10^(10))+10xx0.8xx(512)/(10^(9))+45xx0.64xx(256)/(10^(8))+120xx(0.512)xx(128)/(10^(7))`
`=(1024+40960+737280+7864320)/(10^(10))=(8643584)/(10^(10))=0.00086`