Question

The rental car agency has 30 cars on the lot. 10 are in great shape, 16 are in good shape, and 4 are in poor shape. Four cars are selected at random to be inspected. Do not simplify your answers. Leave in combinatorics form. What is the probability that:

a. Every car selected is in poor shape
b. At least two cars selected are in good shape.
c. Exactly three cars selected are in great shape.
d. Two cars selected are in great shape and two are in good shape.
e. One car selected is in good shape but the other 3 selected are in poor shape.

Answer 1

a.
`\binom{4}{1}(2)/(15) ^1(1-(2)/(15) )^(3)`+
`\binom{4}{2}(2)/(15) ^2(1-(2)/(15) )^(2)`+
`\binom{4}{3}(2)/(15) ^3(1-(2)/(15) )^(1)`+
`\binom{4}{4}(2)/(15) ^4(1-(2)/(15) )^(0)`

b.
`\binom{4}{1}(8)/(15) ^1(1-(8)/(15) )^(3)`+
`\binom{4}{2}(8)/(15) ^2(1-(8)/(15) )^(2)`

c.
`\binom{4}{3}(1)/(3) ^3(1-(1)/(3) )^(1)`

d.
`(\binom{4}{2}(1)/(3) ^2(1-(1)/(3) )^(2))` ×
`(\binom{4}{2}(8)/(15) ^2(1-(8)/(15) )^(2))`

e.
`(1-\binom{4}{0}(1)/(3) ^0(1-(1)/(3) )^(4))`×
`\binom{4}{3}(2)/(15) ^3(1-(2)/(15) )^(1)`

Explanation:

Number of cars in lot = 30

Number of cars in great shape = 10

Number of cars in good shape = 16

Number of cars in poor shape = 4

The probability that a car selected is in great shape = 10/30

The probability that a car selected is in good shape = 16/30

The probability that a car selected is in poor shape = 4/30

Therefore, we have

When 4 cars are selected

P(Car is in poor shape) =
`\binom{n}{r}p^r(1-p)^(n-r)`

Where:

p = Probability of success = 2/15

n = Total number of count = 30

r = Number selected = 4

P =
`\binom{4}{1}(2)/(15) ^1(1-(2)/(15) )^(3)`+
`\binom{4}{2}(2)/(15) ^2(1-(2)/(15) )^(2)`+
`\binom{4}{3}(2)/(15) ^3(1-(2)/(15) )^(1)`+
`\binom{4}{4}(2)/(15) ^4(1-(2)/(15) )^(0)`

b. Here we have

P(At least two cars are in good shape) =
`\binom{4}{1}(8)/(15) ^1(1-(8)/(15) )^(3)`+
`\binom{4}{2}(8)/(15) ^2(1-(8)/(15) )^(2)`

c. P(Exactly 3 cars are in great shape) =
`\binom{4}{3}(1)/(3) ^3(1-(1)/(3) )^(1)`

d. P(2 Great and 2 Good) =
`(\binom{4}{2}(1)/(3) ^2(1-(1)/(3) )^(2))` ×
`(\binom{4}{2}(8)/(15) ^2(1-(8)/(15) )^(2))`

e. P(1 Great and 3 poor) =
`(1-\binom{4}{0}(1)/(3) ^0(1-(1)/(3) )^(4))`×
`\binom{4}{3}(2)/(15) ^3(1-(2)/(15) )^(1)`