A consumer organization estimates that over a 1-year period 17% of cars will need to be repaired...

Question:

A consumer organization estimates that over a 1-year period 17% of cars will need to be repaired once, 7% will need repairs twice, and 1% will require three or more repairs. What is the probability that a car chosen at random will need

a) no repairs?

b) no more than one repair?

c) some repairs?

a) The probability that a car will require no repairs is _____. (Do not round.)

b) The probability that a car will require no more than one repair is _____. (Do not round.)

c) The probability that a car will require some repairs is _____. (Do not round.)

Probability:

The probability is the ratio of the number of favored outcomes by the total number of outcomes of an experiment. The number of favored outcomes cannot be more than the total number of outcomes in that experiment. So, the total probability in an experiment is always equal to one.

Given:

If X denotes the number of repairs a car needs in a 1-year period, then

X P(X)
0 0.75
1 0.17
2 0.07
>= 3 0.01
Total 1

a) The probability that a car chosen at random will need no repair:

{eq}\begin{align*} P(X = 0) & = 1 - P(X \geq 1)\\[1ex] & = 1 - P(X = 1) - P(X = 2) - P(X \geq 3)\\[1ex] & = 1 - 0.17 - 0.07 - 0.01\\[1ex] & = 0.75 \end{align*} {/eq}

b) The probability that a car chosen at random will need no more than one repair:

{eq}\begin{align*} P(X \leq 1) & = P(X = 0) + P(X = 1)\\[1ex] & = 0.75 + 0.17\\[1ex] & = 0.92 \end{align*} {/eq}

c) The probability that a car chosen at random will need some repairs:

{eq}\begin{align*} P(X > 0) &= 1 - P(X = 0)\\[1ex] & = 1 - 0.75\\[1ex] & = 0.25 \end{align*} {/eq}