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Svetach [21]
3 years ago
12

At a certain gas station 40% of the customers request regular gas, 35% request unleaded gas, and 25% request premium gas. Of tho

se customers requesting regular gas, only 30% fill their tanks all the way up, while the remaining 70% only fill up part of their tank. Of those customers requesting unleaded gas, 60% fill their tanks all the way up, while of those requesting premium, 50% fill their tanks all the way up. If the next customer does not fill the tank all the way up (only fills it up part of the way), what is the probability that they requested regular gas?
Mathematics
1 answer:
Temka [501]3 years ago
5 0

Answer:

51.38% probability that they requested regular gas

Step-by-step explanation:

Bayes Theorem:

Two events, A and B.

P(B|A) = \frac{P(B)*P(A|B)}{P(A)}

In which P(B|A) is the probability of B happening when A has happened and P(A|B) is the probability of A happening when B has happened.

In this question, we have that:

Event A: Not filling the tank

Event B: Regular gas

40% of the customers request regular gas

This means that P(B) = 0.4

Of those customers requesting regular gas, 70% only fill up part of their tank.

This means that P(A|B) = 0.7

Probability of not filling the tank:

70% of 40%(regular gas)

100 - 60 = 40% of 35%(unleaded gas).

100 - 50 = 50% of 25%(premium gas).

So

P(A) = 0.7*0.4 + 0.4*0.35 + 0.5*0.25 = 0.545

What is the probability that they requested regular gas?

P(B|A) = \frac{0.4*0.7}{0.545} = 0.5138

51.38% probability that they requested regular gas

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