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?
Answer:
the answer will be D.
Step-by-step explanation:
Answer:
Step-by-step explanation:
<u>Total number of outcomes:</u>
<u>Number of combinations of getting 6 heads:</u>
- 15C6 = 15!/6!(15-6)! = 5005
<u>Required probability is:</u>
- P(6 heads out of 15 flips) = 5005/32768 = 0.1527... ≈ 15.3%
Correct choice is D
Answer:
XZ=12 in
Step-by-step explanation:
area of a rhombus is A=pq/2
where p=WY=15.5 and A=93 and q=XZ
93=15.5(XZ)/2
93(2)=15.5(XZ)
186/15.5=XZ
XZ=12 in
Need more info in order to answer the question. Ill edit my answer once you've added more info.
