Answer:
At a certain pizza parlor,36 % of the customers order a pizza containing onions,35 % of the customers order a pizza containing sausage, and 66% order a pizza containing onions or sausage (or both). Find the probability that a customer chosen at random will order a pizza containing both onions and sausage.
Step-by-step explanation:
Hello!
You have the following possible pizza orders:
Onion ⇒ P(on)= 0.36
Sausage ⇒ P(sa)= 0.35
Onions and Sausages ⇒ P(on∪sa)= 0.66
The events "onion" and "sausage" are not mutually exclusive, since you can order a pizza with both toppings.
If two events are not mutually exclusive, you know that:
P(A∪B)= P(A)+P(B)-P(A∩B)
Using the given information you can use that property to calculate the probability of a customer ordering a pizza with onions and sausage:
P(on∪sa)= P(on)+P(sa)-P(on∩sa)
P(on∪sa)+P(on∩sa)= P(on)+P(sa)
P(on∩sa)= P(on)+P(sa)-P(on∪sa)
P(on∩sa)= 0.36+0.35-0.66= 0.05
I hope it helps!
Answer:
Pretty sure its 40
Step-by-step explanation:
There is only 3 odd numbers on a 6 sided dice (1, 3, 5) and you would just divide 120 and 3
Answer:
0.554
Step-by-step explanation:
In the figure attached, the table is shown
Total people: 1156
Nonsmoker men + occasional smoker men + regular smoker men + heavy smoker men + heavy smoker women = men + heavy smoker women = 601 + 39 = 640
Probability that the person is a man or a heavy smoker = 640/1156 = 0.554
(Notice that if you directly add heavy smoker and total men you will count twice heavy smoker men)
Answer:
20
Step-by-step explanation:
20
