Question #218342

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?


1
Expert's answer
2021-07-19T05:50:15-0400

12+8=2012+8=20

(i)


P(3A & 2F)=(123)(82)(205)=220(28)15504P(3A \ \&\ 2F)=\dfrac{\dbinom{12}{3}\dbinom{8}{2}}{\dbinom{20}{5}}=\dfrac{220(28)}{15504}

=3859690.3973=\dfrac{385}{969}\approx0.3973

(ii)


P(at least 3A)=(123)(82)(205)+(124)(81)(205)P(at\ least\ 3A)=\dfrac{\dbinom{12}{3}\dbinom{8}{2}}{\dbinom{20}{5}}+\dfrac{\dbinom{12}{4}\dbinom{8}{1}}{\dbinom{20}{5}}

+(125)(80)(205)=220(28)+495(8)+792(1)15504+\dfrac{\dbinom{12}{5}\dbinom{8}{0}}{\dbinom{20}{5}}=\dfrac{220(28)+495(8)+792(1)}{15504}=6829690.7038=\dfrac{682}{969}\approx0.7038

(iii)


P(5 same)=(125)(80)(205)+(120)(85)(205)P(5\ same)=\dfrac{\dbinom{12}{5}\dbinom{8}{0}}{\dbinom{20}{5}}+\dfrac{\dbinom{12}{0}\dbinom{8}{5}}{\dbinom{20}{5}}

=792(1)+1(56)15504=539690.0547=\dfrac{792(1)+1(56)}{15504}=\dfrac{53}{969}\approx0.0547



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