Thread: statistics problem

Would someone please help me in the solving and checking my following question and an another solution--
Q1. Let X equal the number of people selected at random that you must ask in order to find someone with the same birthday as yours. Assuming each day of the year is equally likely (and ignoring February 29)
(a)what is pmf of X?
(b)give the values of the mean, variance, and standard deviation of X?
(c)find P(X > 400) and P(X < 300).

Q2. An excellent free throw shooter attempts several free throws until she misses.
(a) if p=0.9 is her probability of making a free throw, what is the probability of having the first miss after 12 attempts.
Solution: (0.9)(power 12)*(0.1)
(b) if she continues shooting until she misses three, what is the probability that the third miss occurs on the 30th attempt.
Solution: (0.9)(power 12)*(0.1)(16!/1!15!)(0.9)(power 15)*(0.1)(0.1)
Thanks

ok here are the solutions. Hopefully it would be correct: Q1 (a) (1/365)(364/365)^x-1 where x=1,2,3.......infinity
(b) E(X) = 1/(1/365) = 365
E(X^2) = (364/365)/(1/365)^2 = 132860 (br) SD = sqrt(132860) = 364.5 Q2 (a) 0.9^12 * 0.1 = 0.2824
(b) (29 choose 2) * (0.9^27) * (0.1^2) = 0.2361

Originally Posted by jojo_jojo

Would someone please help me in the solving and checking my following question and an another solution--
Q1. Let X equal the number of people selected at random that you must ask in order to find someone with the same birthday as yours. Assuming each day of the year is equally likely (and ignoring February 29)
(a)what is pmf of X?
(b)give the values of the mean, variance, and standard deviation of X?
(c)find P(X > 400) and P(X < 300).

Q2. An excellent free throw shooter attempts several free throws until she misses.
(a) if p=0.9 is her probability of making a free throw, what is the probability of having the first miss after 12 attempts.
Solution: (0.9)(power 12)*(0.1)
(b) if she continues shooting until she misses three, what is the probability that the third miss occurs on the 30th attempt.
Solution: (0.9)(power 12)*(0.1)(16!/1!15!)(0.9)(power 15)*(0.1)(0.1)
Thanks

I have no time for checking but you might find the following useful:

Q1 Read Birthday problem - Wikipedia, the free encyclopedia

Q2 You can use the negative binomial distribution: Negative binomial distribution - Wikipedia, the free encyclopedia


Edit: Also, see this thread http://www.mathhelpforum.com/math-he...obability.html