At a local restaurant, 40% of the lunch customers order a hamburger, 70% of the lunch customers order French fries, and 30% of lunch customers order both a hamburger and French fries. If a lunch customer is randomly selected from this restaurant, what is the probability that the customer order fries if we know they ordered a hamburger?

0.40
0.40

0.43

0.75

0.30

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chappellp chappellp · Answer 1 · 2020-12-15T20:03:14+01:00

Answer:

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Step-by-step explanation:

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MrRoyal MrRoyal · Answer 2 · 2022-01-20T10:34:08+01:00

The probability that a customer orders fries given that the customer orders hamburger is 0.75

To solve this, we make use of the following representations

So, we have:

[tex]P(A) = 40\%[/tex]

[tex]P(B) = 70\%[/tex]

[tex]P(A\ n\ B) = 30\%[/tex]

The probability that a customer orders fries given that the customer orders hamburger is calculated as:

[tex]P(B | A) =\frac{P(A\ n B)}{P(A)}[/tex]

So, we have:

[tex]P(B | A) =\frac{30\%}{40\%}[/tex]

[tex]P(B | A) =0.75[/tex]

Hence, the probability that a customer orders fries given that the customer orders hamburger is 0.75