Answer to Question #218342 in Statistics and Probability for rene

Question

A taxi cab company has 12 Ambassadors and 8 Fiats. If 5 of these taxi cabs are in the shop for repairs and Ambassador is as likely to be in for repairs as a fiat. what is the probability that (i) 3 of them are Ambassadors and 2 are fiats? (ii) at least 3 of them are Ambassadors? And (iii) all 5 of them are of the same make?

Answer 1

$$12+8=20$$

(i)

$$P(3A \ \&\ 2F)=\dfrac{\dbinom{12}{3}\dbinom{8}{2}}{\dbinom{20}{5}}=\dfrac{220(28)}{15504}$$

$$=\dfrac{385}{969}\approx0.3973$$

(ii)

$$P(at\ least\ 3A)=\dfrac{\dbinom{12}{3}\dbinom{8}{2}}{\dbinom{20}{5}}+\dfrac{\dbinom{12}{4}\dbinom{8}{1}}{\dbinom{20}{5}}$$

$$+\dfrac{\dbinom{12}{5}\dbinom{8}{0}}{\dbinom{20}{5}}=\dfrac{220(28)+495(8)+792(1)}{15504}$$ $$=\dfrac{682}{969}\approx0.7038$$

(iii)

$$P(5\ same)=\dfrac{\dbinom{12}{5}\dbinom{8}{0}}{\dbinom{20}{5}}+\dfrac{\dbinom{12}{0}\dbinom{8}{5}}{\dbinom{20}{5}}$$

$$=\dfrac{792(1)+1(56)}{15504}=\dfrac{53}{969}\approx0.0547$$