One of the bartleby experts helped me with this Intro to Probability and Statistics homework problem. Below is the work that was provided by the expert. What I need to know is which distribution is being used? 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
One of the bartleby experts helped me with this Intro to Probability and Statistics homework problem. Below is the work that was provided by the expert. What I need to know is which distribution is being used? 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
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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One of the bartleby experts helped me with this Intro to
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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2b62c17d-5c05-48d6-ac2b-a3a7c31907f7%2F440e7536-de3e-434c-852c-192ff43f42b6%2Fnfcs5oj_processed.jpeg&w=3840&q=75)
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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