Hello, I\'m struggling with these probability problems.I asked this before and r
ID: 3355778 • Letter: H
Question
Hello, I'm struggling with these probability problems.I asked this before and received incorrect answers, I have the answer sheet I need to know how it was done.Please try to give as clear of an explanation as possible since I need to study for a test.Thanks.
The last person claimed the sample space for a) was 1/15! but it is 15! If i am wrong please explain why.
2. Experment is the HAT experiment with 15 people. Thus, the 15 hats are oollected and then iandomly redistributed - one hat to a person each outoome equally likey. (a) (5 pts) Carefiully describe the sample space S, and give its size, IS,explaining your answer. (b) (6 pts) Find the probability of A, the event that person 1 does NOT get his own hat, and of the event B, that per1 and 2 get each others hats. (c (6 points) Are A and B independt? Carefully explain your answer (d) (5 pts) Five of the people are women (the rest are NOT). Find the piobability of C, the event that the women ALL get their own hats. Explain your answer (e) (8pts As above firl the probability of D, the event that the5 wome get their own hats back but NONE of the other ten Ixorkpt their own hats.Explanation / Answer
Solution-
(a) Sample space would have- 15! possibilities.
(b) When peron 1 doesn't get his own hat, his hat can go to 14 people with 14 ways and remaining 14 hats can be arranged in 14! ways. So required number of ways = 14 * 14!
So, P(A) = 14* 14! / 15!
= 14 / 15
When person 1 and person 2 get each other hats, then remaining 13 hats can be arranged in 13! ways, So
P(B) = 13! / 15!
= 1/(14*15)
(c) A and B are not independent as A already include the case when his hat goes to B and B gets his. Thus the occurrence of event B must affect the probability of occurrence of event A.
(d) When women get their own hats with just one possibility then remianing 10 hats can be arranged in 10! ways
Thus the P(C) = 10! / 15!
= 1/( 15*14*13*12*11)
Answers
TY!
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