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)](https://tex.z-dn.net/?f=P%28A%5Ccap%20B%29%3DP%28A%29%5Ctimes%20P%28B%29)
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](https://tex.z-dn.net/?f=P%28S%5Ccap%20%5Cbar%20F%29%3DP%28S%29%5Ctimes%20P%28%5Cbar%20F%29%5C%5C%3DP%28S%29%5Ctimes%5B1-P%28F%29%5D%5C%5C%3D0.90%5Ctimes%20%281-0.50%29%5C%5C%3D0.45)
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)}](https://tex.z-dn.net/?f=P%28B%7CA%29%3D%5Cfrac%7BP%28A%5Ccap%20B%7D%7BP%28A%29%7D)
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](https://tex.z-dn.net/?f=P%28F%7CH%29%3D%5Cfrac%7BP%28H%5Ccap%20F%29%7D%7BP%28H%29%7D%5C%5CP%28H%5Ccap%20F%29%3DP%28F%7CH%29%5Ctimes%20P%28H%29%5C%5C%3D0.80%5Ctimes%200.60%5C%5C%3D0.48)
Thus, the probability that the order will include a hamburger and fries is 0.48.