A coin comes up heads with probability p and tails with probability q = 1 - p. For the moment, assume that p is a constant. Prove the expected number of coin flips needed to produce the pattern HT for the very first 1 +¹. 1 time is Р q Hint: TTTTTH|HHHHHHHT requires 14 flips to generate the pattern HT. The TTTTTH part before the produces the first H, and the HHHHHHHT part after the produces the first T after that first H.
A coin comes up heads with probability p and tails with probability q = 1 - p. For the moment, assume that p is a constant. Prove the expected number of coin flips needed to produce the pattern HT for the very first 1 +¹. 1 time is Р q Hint: TTTTTH|HHHHHHHT requires 14 flips to generate the pattern HT. The TTTTTH part before the produces the first H, and the HHHHHHHT part after the produces the first T after that first H.
Algebra and Trigonometry (MindTap Course List)
4th Edition
ISBN:9781305071742
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
Chapter14: Counting And Probability
Section14.2: Probability
Problem 3E: The conditional probability of E given that F occurs is P(EF)=___________. So in rolling a die the...
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Question
![A coin comes up heads with probability p and tails with probability q = 1 - p. For the moment, assume that
p is a constant.
Prove the expected number of coin flips needed to produce the pattern HT for the very first
time is +1.
Hint: TTTTTH|HHHHHHHT requires 14 flips to generate the pattern HT. The TTTTTH part before
the produces the first H, and the HHHHHHHT part after the | produces the first T after that first H.
Prove that the expected number of coin flips needed to produce the pattern HH for the first
+2.
time is
1
Hint: Define a random variable X to track the number of flips needed to produce the pattern HH
for the very first time, and use the Law of Total Expectation:
E[X] = E[X|last flip produced H]p + E[X|last flip produced Tjq
Now assume p is no longer a constant but rather a random variable that behaves as
p~ Beta(4, 7)? What's the expected number of flips for part b now?
Recall that B(a, b) is the normalization constant in place so that Beta(a, b) is a valid random variable
with a valid pdf.
B(a, b) = √¹ pª-¹(1 − p)b-¹ dp
This expression for B(a, b) will be useful as you compute values for E[] and E[2], which you'll
need to compute your final answer here. In fact, you should express your final answer as a sum of
fractions involving several B(a, b) constants with different values of a and b.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F95d4e5c4-ff08-4785-b351-19cb8456ee98%2F07c208a0-999c-4ac2-809c-9882c25e68b3%2F05utmsj_processed.jpeg&w=3840&q=75)
Transcribed Image Text:A coin comes up heads with probability p and tails with probability q = 1 - p. For the moment, assume that
p is a constant.
Prove the expected number of coin flips needed to produce the pattern HT for the very first
time is +1.
Hint: TTTTTH|HHHHHHHT requires 14 flips to generate the pattern HT. The TTTTTH part before
the produces the first H, and the HHHHHHHT part after the | produces the first T after that first H.
Prove that the expected number of coin flips needed to produce the pattern HH for the first
+2.
time is
1
Hint: Define a random variable X to track the number of flips needed to produce the pattern HH
for the very first time, and use the Law of Total Expectation:
E[X] = E[X|last flip produced H]p + E[X|last flip produced Tjq
Now assume p is no longer a constant but rather a random variable that behaves as
p~ Beta(4, 7)? What's the expected number of flips for part b now?
Recall that B(a, b) is the normalization constant in place so that Beta(a, b) is a valid random variable
with a valid pdf.
B(a, b) = √¹ pª-¹(1 − p)b-¹ dp
This expression for B(a, b) will be useful as you compute values for E[] and E[2], which you'll
need to compute your final answer here. In fact, you should express your final answer as a sum of
fractions involving several B(a, b) constants with different values of a and b.
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