Respuesta :
Answer:
The probability that a driver needs to be concerned if the warning light goes on is 0.8462.
Step-by-step explanation:
The conditional probability of an X given that another event Y has already occurred is:
[tex]P(X|Y)=\frac{P(Y|X)P(X)}{P(Y)}[/tex]
The law of total probability states that:
[tex]P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})[/tex]
Denote the events as follows:
A = the light flashes
B = the oil pressure is low.
Given:
[tex]P(A|B) =0.99\\P(A|B^{c})=0.02\\P(B) =0.10[/tex]
Compute the probability that the light flashes using the law of total probability as follows:
[tex]P(A)=P(A|B)P(B)+P(A|B^{c})P(B^{c})\\=(0.99\times0.10)+(0.02\times(1-0.10))\\=0.099+0.018\\=0.117[/tex]
Compute the probability that the oil pressure is low given that the light flashes as follows:
[tex]P(B|A)=\frac{P(A|B)P(B)}{P(A)}=\frac{0.99\times0.10}{0.117}=0.8462[/tex]
Thus, the probability that a driver needs to be concerned if the warning light goes on is 0.8462.
Answer:
Probability that a driver needs to be concerned if the warning light goes on is 0.8462 .
Step-by-step explanation:
We are given that a dashboard warning light is supposed to flash red if a car’s oil pressure is too low.
Let the Probability that the oil pressure really is low = P(L) = 0.10
Probability that the oil pressure really is not low = P(L') = 1 - P(L) = 0.90
Also, let F = event that light flashing goes on
So, Probability that the light flashing is on when it should be = P(F/L) = 0.99
Probability that the light flashing is on when it shouldn't be = P(F/L') = 0.02
So, probability that a driver needs to be concerned if the warning light goes on = P(L/F)
Using Bayes' theorem for above probability we get;
P(L/F) = [tex]\frac{P(L)*P(F/L)}{P(L)*P(F/L) + P(L')*P(F/L')}[/tex] = [tex]\frac{0.10*0.99}{0.10*0.99 + 0.90*0.02}[/tex] = [tex]\frac{0.099}{0.117}[/tex] = 0.8462
Therefore, Required probability = 0.8462 .
