Question

At a particular vehicle service station, tire pressure is routinely checked. Based on past experience, the following distribution has been determined regard the number of underinflated tires a 4-tire vehicle has.

# under inflated tires(X) 0 ___1__ _2 ___3__ _4

P(X) _______________0.2 _0.4 _0.25 _0.1_ 0.05

a. If 4 random cars enter the service station (random implies the # underinflated tires is independent car-to-car) what is the probability that at least one of the cars has at least one underinflated tire?

b. If 4 random cars enter the service station, what is the probability that 2 of the 4 cars has at least one underinflated tire?

c. Cars enter the service station throughout the day. What is the probability that the 10th car to enter is the first car that doesn’t have any underinflated tires?

Answer #1

(a)

Binomial Distribution:

P( Al at least one of the cars has at least one underinflated
tire) =1 -P(All the 4 cars have no underinflated tire) = 1 -
0.2^{4} = 1 - 0.0016 = 0.0.9984

So,

Answer is:

**0.9984**

(b)

Binomial Distribution

n = 4

p = 1 - 0.2 =0.8

q = 1 - 0.8 = 0.2

So,

Answer is:

**0.1536**

(c)

Geometric Distribution

P( 10th car to enter is the first car that doesn’t have any
underinflated tires ) = q^{9} X p = 0.8^{9} X 0.2 =
0.1342 X 0.2 = 0.0268

So

Answer is:

**0.0268**

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