Each child born to a particular set of parents has probability 0.35 of having a blood type B. Suppose these parents have 5 children. Let X= number of children who have type B blood. a) Make a probability distribution table for X b) Find E(X) ( Hint: Note that it is a binomial distribution)
Each child born to a particular set of parents has probability 0.35 of having a blood type B. Suppose these parents have 5 children. Let X= number of children who have type B blood. a) Make a probability distribution table for X b) Find E(X) ( Hint: Note that it is a binomial distribution)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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![### Problem Statement
Each child born to a particular set of parents has a probability of 0.35 of having blood type B. Suppose these parents have 5 children. Let \( X \) be the number of children who have type B blood.
#### a) Create a Probability Distribution Table for \( X \)
The probability distribution table represents the likelihood of each possible outcome for the number of children (from 0 to 5) having blood type B.
#### b) Find \( E(X) \)
*(Hint: Note that it is a binomial distribution)*
To find \( E(X) \), which is the expected value or mean of \( X \) for a binomial distribution, use the formula:
\[ E(X) = n \times p \]
Where:
- \( n \) = number of trials (5 children)
- \( p \) = probability of success (0.35)
### Explanation
Since the task involves a binomial distribution, each child is an independent trial with two possible outcomes: having blood type B or not. The binomial distribution models the number of successes in a fixed number of independent Bernoulli trials.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4e6094c0-ff73-4780-aa2d-7b0dc41eb8c7%2Fc61255fc-d993-4b56-8a61-22c3d1ea9237%2Fa78bfc_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement
Each child born to a particular set of parents has a probability of 0.35 of having blood type B. Suppose these parents have 5 children. Let \( X \) be the number of children who have type B blood.
#### a) Create a Probability Distribution Table for \( X \)
The probability distribution table represents the likelihood of each possible outcome for the number of children (from 0 to 5) having blood type B.
#### b) Find \( E(X) \)
*(Hint: Note that it is a binomial distribution)*
To find \( E(X) \), which is the expected value or mean of \( X \) for a binomial distribution, use the formula:
\[ E(X) = n \times p \]
Where:
- \( n \) = number of trials (5 children)
- \( p \) = probability of success (0.35)
### Explanation
Since the task involves a binomial distribution, each child is an independent trial with two possible outcomes: having blood type B or not. The binomial distribution models the number of successes in a fixed number of independent Bernoulli trials.
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