P(a family has a car) = p = 65% = 0.65
P(a family has no car) = q = 35% = 0.35
For seven families and using the binomial probability distribution, we have
$(p+q{)}^7=p^7+7p^{6}q+21p^5{q}^2+35p^4{q}^3+35p^3{q}^4+21p^2{q}^5+7p{q}^6+q^7$
(a) P(exactly 5 have cars) = $21p^5{q}^2=21(0.65{)}^5(0.35{)}^2$
= $21\times 0.116029\times 0.1225=0.29848$
$\approxeq 0.298$ (to 3 d.p)
(b) P(3 or 4 of them have cars) = $35p^4{q}^3+35p^3{q}^4$
= $35(0.65{)}^4(0.35{)}^3+35(0.65{)}^3(0.35{)}^4$
= $35(0.27463)(0.015006)+35(0.178506)(0.042875)$
= $0.144238+0.267871$
$\approxeq 0.412$ (to 3 d.p)
(c) P(at most 2 of them have cars) = P(0 cars) + P(1 car) + P(2 cars)
= $q^7+7p{q}^6+21p^2{q}^5$
= $(0.35{)}^7+7(0.65)(0.35{)}^6+21(0.65{)}^2(0.35{)}^5$
= $0.000643+0.00836+0.04658$
= $0.5558\approxeq 0.556$ (3 d.p)