A survey indicated that 65% of the families in an area have cars. Find, correct to three decimal places, the probability that among 7 families selected at random in the area (a) exactly 5 ; (b) 3 or 4 ; (c) at most 2 of them have cars.
Explanation
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)
