Question

2. A car rental agency has 8 compact cars and S mid-size cars. If four of the cars are selected at random and without replacement, what is the probability of getting a) two cars of each type b) at least three mid-size cars

Answer 1

A car rental agency has 8 compact cars and 5 Mid-size cars.

Total rental cars = 8 + 5 = 13

Four cars are randomly selected without replacement.

Total number of ways to select 4 cars from total cars 13 = ¹³C₄ = 715

{ Using Excel function , =COMBIN(13,4) = 715 }

a) Two cars of each type.

Number of ways to select two compact cars = ⁸C₂ = 28     { Using Excel , =COMBIN(8,2) = 28 }

Number of ways to select two Mid-size cars = ⁵C₂ = 10 { Using Excel , =COMBIN(5,2) = 10 }

Total number of ways to select two cars of each type =⁸C₂ *⁵C₂ = 28*10 = 280

So, P(Two cars of each type) = 280/715 = 0.3916

b ) At least three mid-sized cars.

P( At least three mid-sized cars )

= P( 1 compact , 3 mid-size ) + P( 0 compact , 4 mid-size )

$=\frac{^{8}C_{1}\times^{5}C_{3} }{^{13}C_{4}} + \frac{^{8}C_{0}\times^{5}C_{4} }{^{13}C_{4}}$

$=\frac{8\times10 }{715} +\frac{1\times5 }{715}$

$=\frac{80}{715} +\frac{5 }{715}$

$=0.111888 +0.006993$

$=0.1189$

P( At least three mid-sized cars ) = 0.1189