Answer:
(a)
The probability that you stop at the fifth flip would be
[tex]p^4 (1-p) + (1-p)^4 p[/tex]
(b)
The expected numbers of flips needed would be
[tex]\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1} = 1/p[/tex]
Therefore, suppose that [tex]p = 0.5[/tex], then the expected number of flips needed would be 1/0.5 = 2.
Step-by-step explanation:
(a)
Case 1
Imagine that you throw your coin and you get only heads, then you would stop when you get the first tail. So the probability that you stop at the fifth flip would be
[tex]p^4 (1-p)[/tex]
Case 2
Imagine that you throw your coin and you get only tails, then you would stop when you get the first head. So the probability that you stop at the fifth flip would be
[tex](1-p)^4p[/tex]
Therefore the probability that you stop at the fifth flip would be
[tex]p^4 (1-p) + (1-p)^4 p[/tex]
(b)
The expected numbers of flips needed would be
[tex]\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1} = 1/p[/tex]
Therefore, suppose that [tex]p = 0.5[/tex], then the expected number of flips needed would be 1/0.5 = 2.
