A First Course in Probability (10th Edition)
10th Edition
ISBN: 9780134753119
Author: Sheldon Ross
Publisher: PEARSON
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Answer the following question about PMFs. Be sure to explicitly state which distribution you are using in each of the following cases.
1. In a fair coin toss, what is the probability of getting exactly 3 heads in 5 tosses?
Work
![### Step 1: Write the given information
Let \( X \) be the number of heads and \( n \) be the number of tosses.
Given, the coin is fair. The probability of getting a head is \( \frac{1}{2} = 0.5 \) and \( n = 5 \).
### Step 2: Find the probability of getting exactly 3 heads in 5 tosses
The probability of getting a head is constant for all tosses. The success is a head, and failure is a tail. Each toss is independent. Then, \( X \) follows a binomial distribution with \( n = 5 \) and \( p = 0.5 \).
The Probability Mass Function (PMF) of \( X \) is:
\[
P(x) = \binom{n}{x} p^x (1-p)^{n-x}, \quad x = 0, 1, 2, \ldots, n
\]
Then, the probability of getting exactly 3 heads in 5 tosses is:
\[
P(X = 3) = \binom{5}{3} (0.5)^3 (1-0.5)^{5-3}
\]
Calculating this, we have:
\[
= 10 \times 0.5^3 \times 0.5^2
\]
\[
= 0.3125
\]
### Solution
The probability of getting exactly 3 heads in 5 tosses is 0.3125.](https://content.bartleby.com/qna-images/question/2b62c17d-5c05-48d6-ac2b-a3a7c31907f7/440e7536-de3e-434c-852c-192ff43f42b6/nfcs5oj_processed.jpeg)
Transcribed Image Text:### Step 1: Write the given information
Let \( X \) be the number of heads and \( n \) be the number of tosses.
Given, the coin is fair. The probability of getting a head is \( \frac{1}{2} = 0.5 \) and \( n = 5 \).
### Step 2: Find the probability of getting exactly 3 heads in 5 tosses
The probability of getting a head is constant for all tosses. The success is a head, and failure is a tail. Each toss is independent. Then, \( X \) follows a binomial distribution with \( n = 5 \) and \( p = 0.5 \).
The Probability Mass Function (PMF) of \( X \) is:
\[
P(x) = \binom{n}{x} p^x (1-p)^{n-x}, \quad x = 0, 1, 2, \ldots, n
\]
Then, the probability of getting exactly 3 heads in 5 tosses is:
\[
P(X = 3) = \binom{5}{3} (0.5)^3 (1-0.5)^{5-3}
\]
Calculating this, we have:
\[
= 10 \times 0.5^3 \times 0.5^2
\]
\[
= 0.3125
\]
### Solution
The probability of getting exactly 3 heads in 5 tosses is 0.3125.
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