A restaurant in a fast food franchise has determined that the chance a customer

Question

A restaurant in a fast food franchise has determined that the chance a customer will order a soft drink is 0.87. The probability that a customer will order a hamburger is

0.52. The probability that a customer will order french fries is 0.41.

Complete parts a and b below.

a. If a customer places an order, what is the probability that the order will include a soft drink and no fries if these two events are independent?

The probability is ? (Round to four decimal places as needed.)

b. The restaurant has also determined that if a customer orders a hamburger, the probability the customer will order fries is 0.82 ?

Determine the probability that the order will include a hamburger and fries.The probability is ?

(Round to four decimal places as needed.)

Explanation / Answer

From now on let`s represent soft drinks, hamburger and frenchfries with S, H, F respectievely.

P(S) = 0.87

P(H) = 0.52

P(F) = 0.41

a)

what is the probability that the order will include a soft drink and no fries if these two events are independent =

P(S n F')

As S and F are independent S and F' are also independent.

Thus, P(S n F') = P(S) * P(F') = P(S) * (1 - P(F)) = 0.87 * (1 - 0.41)

= 0.87 * 0.59

= 0.5133

b)

P(F | H) = 0.82

This gives,

P(F n H) / P(H) = 0.82

P(F n H) = 0.82 * P(H)

P(F n H) = 0.82 * 0.87

P(F n H) = 0.7134.

Hence, probability that the order will include a hamburger and fries = 0.7134.

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