Answer:
Step-by-step explanation:
Total number of people = 130
Number of people who use the gym = 73
Number of people who use the pool = 62
Number of people who use the track = 58
Number of people who use the gym and the pool = 22
Number of people who use the pool and the track = 29
Number of people who use the gym and the track = 25
Number of people who use all three facilities = 11
Total number of people who use at least two facilities = 22 + 29 + 25 + 11 = 87
The probability that the randomly selected person uses all three facilities = number of those who use all three facilities ÷ total number of people who use at least two facilities.
==> 11 ÷ 87
==>
Answer:
Step-by-step explanation:
24 hope it helps you!!
1/3=2/6
-1/6= -1/6
then I guessing you add them?
2/6+-1/6=1/6
I believe the answer is B. 3
A fleet of nine taxis is to be dispatched to three airports in such a way that three go to airport A, five go to airport B, and one goes to airport C. In how many distinct ways can this be accomplished?
2.44) Refer to Exercise 2.43. Assume that taxis are allocated to airports at random.
a) If exactly one of the taxis is in need of repair, what is the probability that it is dispatched to airport C?
b) If exactly three of the taxis are in need of repair, what is the probability that every airport receives one of the taxis requiring repairs?
So, my answer to 2.44a is 1/9. Hopefully this is correct at least :)
For 2.44b, my guess was
(3C1)(1/3)(2/3)2 * (5C1)(1/3)(2/3)4 * 1/3
The solutions manual on chegg (which seems to be riddled with errors) says something completely different. Is my calculation correct?
