this helps... but the question is asking what is the expected times we would flip a head. My thinking is that we use the same approach as above but replace the [1..9] with 0 for (1/2)^10 and 1 for (1/2) .. (1/2)^9. I'm also not sure how to handle if the 10th flip is a head? Thanks!

this helps... but the question is asking what is the expected times we would flip a head. My thinking is that we use the same approach as above but replace the [1..9] with 0 for (1/2)^10 and 1 for (1/2) .. (1/2)^9. I'm also not sure how to handle if the 10th flip is a head? Thanks!

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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Question

I'm also not sure how to handle if the 10th flip is a head?

Thanks!

Expert Solution

Step 1: Explaining in steps:

To solve this problem, we can consider the cases where we flip a head in the first, second, third, and so on, up to the tenth attempt, and calculate the probabilities for each case. Then we can find the expected value, which is the sum of the products of the number of attempts and the probability of flipping a head in that number of attempts.

Let's calculate the probabilities for each case:

Flipping a head on the first attempt: Probability = $1 half$
Flipping a head on the second attempt: Probability = $1 half cross times 1 half equals 1 fourth$
Flipping a head on the third attempt: Probability = $1 half cross times 1 half cross times 1 half equals 1 over 8$

And so on, up to the tenth attempt.

Step by step

Solved in 3 steps with 10 images

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