The probability that a certain make of car will need repairs in the first six months is 0.4. A dealer sells seven such cars. What is the probability that at least one of them will require repairs in the first six months?

0.972

Step-by-step explanation:

This problem can be solved binomialy with n = 7

The probability that the car will need repair ; p(need repair) = 0.4

The probability that the car will not need repair ; p(no repair) = 1 - 0.4 = 0.6

Probability at least 1 will need repair = 1 - Probability none will need repair

Probability at least 1 will need repair = 1 - p(no repair)

=

The probability that at least one of them will require repairs in the first six months is 0.972

Probability that at least one of them will require repairs in the first six months is 0.972.

Step-by-step explanation:

We are given that the probability that a certain make of car will need repairs in the first six months is 0.4. A dealer sells seven such cars.

The above situation can be represented through Binomial distribution;

where, n = number of trials (samples) taken = 7 cars

            r = number of success = at least one

           p = probability of success which in our question is probability that a

                 make of car will need repairs in the first six months, i.e; 0.40

LET X = Number of cars that require repairs in the first six months

So, it means X ~

Now, Probability that at least one of them will require repairs in the first six months is given by = P(X 1)

       P(X 1)  = 1 - P(X = 0)

                     =  

                     =

                     = = 0.972

Therefore, Probability that at least one of them will require repairs in the first six months is 0.972.