Problem 43
Question
Genetics A family has five children. Assume that the probability of having a boy is \(0.5 .\) Write the term in the expansion of \((b+g)^{5}\) for each outcome described. Then evaluate each probability. \(\begin{array}{llll}{\text { a. exactly } 3 \text { boys }} & {\text { b. exactly } 4 \text { boys }} & {\text { c. exactly } 4 \text { girls }}\end{array}\)
Step-by-Step Solution
Verified Answer
The probability of having exactly 3 boys is 0.3125; for having exactly 4 boys it is 0.15625; and for having exactly 4 girls it is also 0.15625.
1Step 1: Apply the binomial theorem for exactly 3 boys
We need to find the coefficient of \(b^3g^2\) in the expansion of \((b+g)^5\) for exactly 3 boys & 2 girls. According to the binomial theorem, this is given by \({5 \choose 3} b^{3} g^{2} = 10 * b^{3} * g^{2}\). The probability of this outcome is \(10 * (0.5)^{3} * (0.5)^{2}=0.3125\).
2Step 2: Apply the binomial theorem for exactly 4 boys
We need to find the coefficient of \(b^4g^1\) in the expansion of \((b+g)^5\) for exactly 4 boys & 1 girl. According to the binomial theorem, this is given by \({5 \choose 4} b^{4} g^{1} = 5 * b^{4} * g^{1}\). The probability of this outcome is \(5 * (0.5)^{4} * (0.5)^{1}=0.15625\).
3Step 3: Apply the binomial theorem for exactly 4 girls
We need to find the coefficient of \(b^1g^4\) in the expansion of \((b+g)^5\) for exactly 1 boy & 4 girls. According to the binomial theorem, this is given by \({5 \choose 1} b^{1} g^{4} = 5 * b^{1} * g^{4}\). The probability of this outcome is \(5 * (0.5)^{1} * (0.5)^{4}=0.15625\).
Key Concepts
ProbabilityProbability DistributionBinomial Expansion
Probability
Probability is the measure of how likely an event is to occur. It is a fundamental concept in statistics and mathematics.
For this exercise, we are looking at the probability of having a certain number of boys among five children. Probability values range between 0 and 1, where 0 means the event will not happen and 1 means it will certainly happen.
The probability for a single event, like having a boy, can be understood simply as:
For this exercise, we are looking at the probability of having a certain number of boys among five children. Probability values range between 0 and 1, where 0 means the event will not happen and 1 means it will certainly happen.
The probability for a single event, like having a boy, can be understood simply as:
- If the chance of having a boy child is 0.5, the probability of having a girl is also 0.5. This is because these two outcomes (boy or girl) are equally likely.
- To find the probability of multiple independent events happening, like having 3 boys and 2 girls in five births (independent events), we multiply the probabilities for each event.
Probability Distribution
Probability distribution shows all the possible outcomes of a random experiment and assigns a probability to each outcome. It provides a complete description of the randomness of an event.
For example, when dealing with genetic outcomes like boys and girls, probability distributions tell us all the different ways these children could be boys or girls, and how likely each scenario is.
Consider this:
For example, when dealing with genetic outcomes like boys and girls, probability distributions tell us all the different ways these children could be boys or girls, and how likely each scenario is.
Consider this:
- When we say there is a distribution for having boys in a family of five, each possible number of boys (from 0 to 5) has a specific probability attached to it.
- The most common statistical distributions include normal distribution, binomial distribution, and uniform distribution. Here, we use a binomial distribution.
Binomial Expansion
Binomial expansion is a way to expand expressions that are raised to a power, such as \( (b + g)^5 \). It is instrumental in solving probability problems involving combinations of two possibilities, like flipping a coin or using scenarios in genetics.
The binomial expansion formula is given by: \[ (a + b)^n = \sum_{k=0}^{n} {n \choose k} a^{(n-k)} b^{k} \] Here, \( {n \choose k} \) is a binomial coefficient that tells you how many ways you can choose \( k \) successes from \( n \) trials, and \( a^{(n-k)} b^{k} \) are the terms of the expansion.
In this context:
The binomial expansion formula is given by: \[ (a + b)^n = \sum_{k=0}^{n} {n \choose k} a^{(n-k)} b^{k} \] Here, \( {n \choose k} \) is a binomial coefficient that tells you how many ways you can choose \( k \) successes from \( n \) trials, and \( a^{(n-k)} b^{k} \) are the terms of the expansion.
In this context:
- For 3 boys and 2 girls, we look at the term \( b^3g^2 \), and the binomial coefficient \( {5 \choose 3} \) yields 10 corresponding scenarios.
- Each term's coefficient multiplies with the respective probabilities \( (0.5)^3 \, (0.5)^2 \).
Other exercises in this chapter
Problem 42
Determine whether each binomial is a factor of \(x^{3}+x^{2}-16 x-16\). $$ x+2 $$
View solution Problem 42
Simplify. Classify each result by number of terms. $$ \left(-12 x^{3}+5 x-23\right)-\left(4 x^{4}+31-9 x^{3}\right) $$
View solution Problem 43
Solve each equation using the Quadratic Formula. $$ 2\left(x^{2}+2\right)=3 x $$
View solution Problem 43
Write each function in factored form. Check by multiplication. $$ y=3 x^{3}-27 x^{2}+24 x $$
View solution