- #1
leakin99
- 32
- 0
"A fair coin is flipped repeatedly. What is the probability that the 5th tail occurs before the tenth head?"
What I have so far:
So the 5th tail has to come before 10th head. So if we take getting a tails as success and after the 9th head, we MUST have 5 tails --> we can only have 14 flips or less(but more than 4, since we're looking for 5 Tails).
let X be the # of flips required to get a 5th tail and since X is a negative binomial rv(conditions above, hopefully)
The lower bound on the sum is 5, and the upperbound on the sum is 14. (n-1)C4 is n-1 "CHOOSE" 4
[tex]\sum[/tex]P(X=n) = [tex]\sum[/tex][(n-1)C(4)](0.5)[tex]^{5}[/tex](0.5)[tex]^{n-5}[/tex]
I don't know if I am doing this right because I am not 100% sure if X is a negative binomial RV. Can anyone maybe explain the setting up process or something.
Thanks
What I have so far:
So the 5th tail has to come before 10th head. So if we take getting a tails as success and after the 9th head, we MUST have 5 tails --> we can only have 14 flips or less(but more than 4, since we're looking for 5 Tails).
let X be the # of flips required to get a 5th tail and since X is a negative binomial rv(conditions above, hopefully)
The lower bound on the sum is 5, and the upperbound on the sum is 14. (n-1)C4 is n-1 "CHOOSE" 4
[tex]\sum[/tex]P(X=n) = [tex]\sum[/tex][(n-1)C(4)](0.5)[tex]^{5}[/tex](0.5)[tex]^{n-5}[/tex]
I don't know if I am doing this right because I am not 100% sure if X is a negative binomial RV. Can anyone maybe explain the setting up process or something.
Thanks