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Chapter 7: Problem 15

Assume that the probability of a boy being born is the same as the probability of a girl being born. Find the probability that a family with three children will have the given composition. No girls

Short Answer

Expert verified
The probability of a family with 3 children having no girls (i.e., 3 boys) is 0.125 or 12.5%.

Step by step solution

01

Identify the relevant information

- Probability of having a boy: \(P(B) = 0.5\) - Probability of having a girl: \(P(G) = 0.5\) - Number of children: \(n = 3\)
02

Apply the binomial probability formula

The binomial probability formula is given by: \(P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}\) where - \(P(X=k)\) is the probability of having exactly \(k\) boys - \(\binom{n}{k}\) is the number of combinations of selecting \(k\) boys from \(n\) children, which can be calculated as \(\frac{n!}{k!(n-k)!}\) - \(p\) is the probability of having a boy - \(n\) is the number of children - \(k\) is the number of boys In this problem, we need to find the probability of having no girls, which means having all 3 boys: - \(n = 3\) - \(k = 3\) - \(p = 0.5\)
03

Calculate the probability using the formula

Substitute the values into the binomial probability formula: \(P(X=3) = \binom{3}{3} (0.5)^3 (1-0.5)^{3-3}\) First, calculate the number of combinations, \(\binom{3}{3}\): \(\binom{3}{3} = \frac{3!}{3!(3-3)!} = \frac{3!}{3!0!} = 1\) Next, calculate the probability of having 3 boys: \(P(X=3) = 1 \cdot (0.5)^3 \cdot (0.5)^0 = 1 \cdot 0.125 \cdot 1 = 0.125\)
04

Interpret the result

The probability of a family with 3 children having no girls (i.e. 3 boys) is 0.125 or 12.5%.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Theory
Probability theory is a fundamental part of mathematics that deals with the likelihood of events occurring. In the context of the problem we are considering, it helps us determine the likelihood of having boy or girl children. In probability theory, we assign a number between 0 and 1 to represent how likely an event is. A probability of 0 means an event will not happen, while a probability of 1 means it definitely will happen.

In our exercise, the probability of having a boy ( P(B) ) is 0.5, just like the probability of having a girl ( P(G) ). This means each outcome is equally likely. We use these probabilities to solve problems involving possible events, like predicting the gender composition of a family of children. Understanding these basic principles of probability is crucial in many areas of statistics and daily decision-making.
Combinatorics
Combinatorics is the field of mathematics concerned with counting, arranging, and combing items. The binomial probability we used in this exercise involves combinatorics to find the number of possible combinations of certain outcomes. In simpler terms, it's about figuring out how many different ways we can achieve the outcome we're interested in.

In the solution, the term \(\binom{n}{k}\) represents the number of combinations. It tells us the ways to choose k successes (like boys in this case) from n trials (the number of children). For our problem, it was calculated as:\[\binom{3}{3} = \frac{3!}{3!(3-3)!} = 1\]This means there is exactly one way to have all 3 children as boys. Combinatorics is a powerful tool, allowing us to handle complex arrangements and probabilities efficiently.
Statistical Interpretation
Statistical interpretation involves making sense of numerical data from the perspective of probability. In this exercise, once we calculated the probability of a family having three boys as 0.125 or 12.5%, we then interpret what that means in a real-world context.

A 12.5% probability means that, theoretically, if we were to consider many families with three children, only 12.5% of them would likely have all boys and no girls.
  • 12.5% is roughly 1 out of 8, suggesting that a family composed entirely of boys is relatively uncommon given equal chance for boys and girls.
  • This interpretation helps set expectations about the distribution of outcomes we might observe in larger populations.
Understanding statistical interpretation helps in decision-making, predictions, and drawing more profound insights from data.

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Most popular questions from this chapter

The estimated probability that a brand-A, a brand-B, and a brand-C plasma TV will last at least \(30,000 \mathrm{hr}\) is \(.90, .85\), and \(.80\), respectively. Of the 4500 plasma TVs that Ace TV sold in a certain year, 1000 were brand A, 1500 were brand \(\mathrm{B}\), and 2000 were brand \(\mathrm{C}\). If a plasma TV set sold by Ace TV that year is selected at random and is still working after \(30,000 \mathrm{hr}\) of use a. What is the probability that it was a brand-A TV? b. What is the probability that it was not a brand-A TV?

In a survey conducted to see how long Americans keep their cars, 2000 automobile owners were asked how long they plan to keep their present cars. The results of the survey follow: $$ \begin{array}{cc} \hline \text { Years Car Is Kept, } \boldsymbol{x} & \text { Respondents } \\ \hline 0 \leq x<1 & 60 \\ \hline 1 \leq x<3 & 440 \\ \hline 3 \leq x<5 & 360 \\ \hline 5 \leq x<7 & 340 \\ \hline 7 \leq x<10 & 240 \\ \hline 10 \leq x & 560 \\ \hline \end{array} $$ Find the probability distribution associated with these data. What is the probability that an automobile owner selected at random from those surveyed plans to keep his or her present car a. Less than \(5 \mathrm{yr}\) ? b. 3 yr or more?

A study conducted by the Metro Housing Agency in a midwestern city revealed the following information concerning the age distribution of renters within the city. $$\begin{array}{lcc} \hline & & \text { Group } \\ \text { Age } & \text { Adult Population, \% } & \text { Who Are Renters, \% } \\\ \hline 21-44 & 51 & 58 \\ \hline 45-64 & 31 & 45 \\ \hline 65 \text { and over } & 18 & 60 \\ \hline \end{array}$$ a. What is the probability that an adult selected at random from this population is a renter? b. If a renter is selected at random, what is the probability that he or she is in the \(21-44\) age bracket? c. If a renter is selected at random, what is the probability that he or she is 45 yr of age or older?

The personnel department of Franklin National Life Insurance Company compiled the accompanying data regarding the income and education of its employees: \begin{tabular}{lcc} \hline & Income & Income \\ & \(\$ 50,000\) or Relow & Above \$50,000 \\ \hline Noncollege Graduate & 2040 & 840 \\ \hline College Graduate & 400 & 720 \\ \hline \end{tabular} Let \(A\) be the cvent that a randomly chosen cmployee has a college degree and \(B\) the cvent that the chosen cmployec's income is more than \(\$ 50.000\). a. Find cach of the following probabilities: \(P(A), P(B)\), \(P(A \cap B), P(B \mid A)\), and \(P\left(B \mid A^{c}\right)\) b. Are the cvents \(A\) and \(B\) independent events?

Refer to the following experiment: Urn A contains four white and six black balls. Urn B contains three white and five black balls. A ball is drawn from urn A and then transferred to urn B. A ball is then drawn from urn B. What is the probability that the transferred ball was black given that the second ball drawn was black?
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