At a certain gas station, 40% of the customers use regular unleaded gas, 35% use extra unleaded gas and 25% use premium unleaded gas. Of those customers using regular gas, only 30% fill their tanks. Of those customers using extra gas, 60% fill their tanks, whereas of those using premium, 50% fill their tanks.

a. What is the probability that the next customer will request extra unleaded gas and fill the tank? State the rule used.
b. What is the probability that the next customer fills the tank?
c. If the next customer fills the tank, what is the probability that regular gas is not requested?

I calculated parts a and b, but I just want to ensure part c is correct. Would it be the P(A' l F) where A is the regular unleaded, and F the event of a customer filling the tank? If you could do parts a and b, that would be great to compare with what I've got, but not necessary.

You have the right idea. Let us define below events for ease of reference
R : use regular
E : use extra
P : use premium
F : full tank

Given info
P(R) = 0.4
P(E) = 0.35
P(P) = 0.25
P(F|R) = 0.3
P(F|E) = 0.6
P(F|P) = 0.5

a) This can be worked using conditional probability :
[tex]P(E{\cap}F)=P(E)*P(F|E)=0.35*0.6=\boxed{0.21}[/tex]

b) Since [tex]R,E,P[/tex] are pairwise disjoint events, we may use law of total probability to realize [tex]P(F)[/tex] :
[tex]P(F)=P(R{\cap}F)+P(E{\cap}F)+P(P{\cap}F)\\=P(R)*P(F|R)+P(E)*P(F|E)+P(P)*P(F|P)\\=0.4*0.3+0.35*0.6+0.25*0.5\\=\boxed{0.455}[/tex]

c) This can be worked using conditional proability too :
[tex]P(R'|F)=1-P(R|F)\\=1-\dfrac{P(R{\cap}F)}{P(F)}\\=1-\dfrac{P(R)*P(F|R)}{P(F)}\\=1-\dfrac{0.4*0.3}{0.455}\\=1-0.264\\=\boxed{0.736}[/tex]

kvnmurty kvnmurty · Answer 2 · 2015-04-27T01:19:00+05:30

We use the conditional probability rule: Kolomogorov definition.
P (A and B ) = P (A) * P (B | A)

a. 0.35 * 0.60 = 0.21

b. 0.40 * 0.30 + 0.35 * 0.60 + 0.25 * 50 = 0.455

c.
Using the conditional probability theorem :
P ( A | B) = prob that A occurs given that B has occurred.
= P (A Π B) / P (B), A Π B is the intersection of A & B events.

Here B = event that the next customer fills the tank
A = event that regular gas is not filled.

P(B) = prob that a customer fills the tank = 0.455, calculated above.
P(A) = probability that regular gas is not requested = 1 - 0.40 = 0.60

probability that regular gas is not requested and the customer fills the tank
P(A Π B) = 0.35 * 0.60 + 0.25 * 0.50 = 0.335

Required prob = 0.335 / 0.455 = 0.736