Question

A fast-food restaurant has determined that the chance a customer will order a soft drink is 0.90. The proba- bility that a customer will order a hamburger is 0.60. The probability that a customer will order french fries is 0.50. a) If a customer places an order, what is the probability that the order will include a soft drink and no fries if those two events are independent? b) The restaurant has also determined that if a customer orders a hamburger, the probability the customer will order fries is 0.80. Determine the probability that the order will include a hamburger and fries.

Answer 1

Answer:

(a) The probability that the order will include a soft drink and no fries is 0.45.

(b) The probability that the order will include a hamburger and fries is 0.48.

Explanation:

Let the events be denoted as follows:

S = an order of soft drink

H = an order of hamburger

F = an order of french fries.

Given:

P (S) = 0.90

P (H) = 0.60

P (F) = 0.50

(a)

It is provided that the event of ordering a soft drink and fries are independent.

If events A and B are independent then the probability of event (A ∩ B) is:

$P(A\cap B)=P(A)\times P(B)$

Compute the probability that the order will include a soft drink and no fries as follows:

$P(S\cap \bar F)=P(S)\times P(\bar F)\\=P(S)\times[1-P(F)]\\=0.90\times (1-0.50)\\=0.45$

Thus, the probability that the order will include a soft drink and no fries is 0.45.

(b)

It is provided that the conditional probability that a customer will order fries given that he/she has already ordered a hamburger as, P (F|H) = 0.80.

The conditional probability of an event B given another event A has already occurred is:

$P(B|A)=\frac{P(A\cap B}{P(A)}$

Compute the probability that the order will include a hamburger and fries as follows:

$P(F|H)=\frac{P(H\cap F)}{P(H)}\\P(H\cap F)=P(F|H)\times P(H)\\=0.80\times 0.60\\=0.48$

Thus, the probability that the order will include a hamburger and fries is 0.48.