Respuesta :
Answer:
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)
= [tex]1-(0.6)^{7} = 0.972[/tex]
[tex]\therefore[/tex] The probability that at least one of them will require repairs in the first six months is 0.972
Answer:
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;
[tex]P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....[/tex]
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 ~ [tex]Binom(n=7, p=0.40)[/tex]
Now, Probability that at least one of them will require repairs in the first six months is given by = P(X [tex]\geq[/tex] 1)
P(X [tex]\geq[/tex] 1) = 1 - P(X = 0)
= [tex]1- \binom{7}{0}\times 0.40^{0} \times (1-0.40)^{7-0}[/tex]
= [tex]1-( 1 \times 1 \times 0.60^{7})[/tex]
= [tex]1-0.60^{7}[/tex] = 0.972
Therefore, Probability that at least one of them will require repairs in the first six months is 0.972.
