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Chapter 8: Problem 15

Find the probability that a family with three children will have the given composition. No girls

Short Answer

Expert verified
The probability that a family with three children will have no girls is \(\frac{1}{8}\) or 12.5%.

Step by step solution

01

Identifying the possible gender combinations

In a family with three children, there are a total of eight possible combinations of genders (ignoring any other factors like twins or external factors that might affect gender probability). These are: 1. Boy, Boy, Boy (BBB) 2. Boy, Boy, Girl (BBG) 3. Boy, Girl, Boy (BGB) 4. Boy, Girl, Girl (BGG) 5. Girl, Boy, Boy (GBB) 6. Girl, Boy, Girl (GBG) 7. Girl, Girl, Boy (GGB) 8. Girl, Girl, Girl (GGG) In this problem, we want to find the probability of the family having no girls, which means we're looking for the probability of the first combination: Boy, Boy, Boy (BBB).
02

Calculate the probability of a single event

Since there are two equally likely possibilities for each child's gender, we can assign a probability of 1/2 for each child being a boy and 1/2 for each child being a girl.
03

Calculate the probability of having no girls

To find the probability of the family having no girls (BBB), we need to multiply the probabilities for each child being a boy independently. P(BBB) = P(Boy) 脳 P(Boy) 脳 P(Boy) = \(\frac{1}{2}\) 脳 \(\frac{1}{2}\) 脳 \(\frac{1}{2}\)
04

Simplify the result

Now we just need to simplify the product: P(BBB) = \(\frac{1}{2}\) 脳 \(\frac{1}{2}\) 脳 \(\frac{1}{2}\) = \(\frac{1}{8}\) So, the probability that a family with three children will have no girls is 1/8 or 12.5%.

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Key Concepts

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

Gender Combinations
In probability exercises involving multiple outcomes, it's important to identify all possible combinations first. When it comes to a family with three children, each child can be either a boy or a girl. This creates a tree diagram of outcomes, listing all combinations:
  • Boy, Boy, Boy (BBB)
  • Boy, Boy, Girl (BBG)
  • Boy, Girl, Boy (BGB)
  • Boy, Girl, Girl (BGG)
  • Girl, Boy, Boy (GBB)
  • Girl, Boy, Girl (GBG)
  • Girl, Girl, Boy (GGB)
  • Girl, Girl, Girl (GGG)
In total, there are 8 possible gender combinations. Each is equally likely if the chance of having a boy or girl is the same, usually set at 50/50 for simplicity. Understanding all possible outcomes helps us calculate the probability of each individual event.
Probability Calculation
Probability is about measuring how likely an event is to happen. For each child in the family, the probability of being a boy (\( P(Boy) \)) is \( \frac{1}{2} \) and the same for a girl (\( P(Girl) \)). To find the probability of a specific combination, like having no girls (BBB), you multiply the probability for each event together.

This means:
  • Step 1: Probability of first child being a boy = \( \frac{1}{2} \)
  • Step 2: Probability of second child being a boy = \( \frac{1}{2} \)
  • Step 3: Probability of third child being a boy = \( \frac{1}{2} \)
Thus, \( P(BBB) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8} \).
Probability helps translate every gender combination into a quantifiable expectation.
Mathematical Simplification
Simplification is a key step in solving probability problems to make the answers more understandable. After calculating the combination probability, it's common to need simplification.

In our example, the calculated probability for having no girls (BBB) is \( \frac{1}{8} \).
Since the multiplication of fractions is straightforward:
  • Multiply numerators: \( 1 \times 1 \times 1 = 1 \).
  • Multiply denominators: \( 2 \times 2 \times 2 = 8 \).
Simplifying probabilities often provides insights, like seeing the result of \( \frac{1}{8} \) expresses the situation concisely, and even translates into a percentage: 12.5%. Understanding how to simplify keeps mathematical expressions clear and reveals the intrinsic likelihood of an event.

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Most popular questions from this chapter

A Christmas tree light has an expected life of \(200 \mathrm{hr}\) and a standard deviation of \(2 \mathrm{hr}\). a. Find a bound on the probability that one of these Christmas tree lights will last between \(190 \mathrm{hr}\) and \(210 \mathrm{hr}\). b. Suppose 150,000 of these Christmas tree lights are used by a large city as part of its Christmas decorations. Estimate the number of lights that are likely to require replacement between \(180 \mathrm{hr}\) and \(220 \mathrm{hr}\) of use.

During the first year at a university that uses a 4 -point grading system, a freshman took ten 3 -credit courses and received two As, three Bs, four Cs, and one \(D\). a. Compute this student's grade-point average. b. Let the random variable \(X\) denote the number of points corresponding to a given letter grade. Find the probability distribution of the random variable \(X\) and compute \(E(X)\), the expected value of \(X\).

In an examination given to a class of 20 students, the following test scores were obtained: $$\begin{array}{lllllllllr}40 & 45 & 50 & 50 & 55 & 60 & 60 & 75 & 75 & 80 \\ 80 & 85 & 85 & 85 & 85 & 90 & 90 & 95 & 95 & 100\end{array}$$ a. Find the mean (or average) score, the mode, and the median score. b. Which of these three measures of central tendency do you think is the least representative of the set of scores?

A probability distribution has a mean of 50 and a standard deviation of \(1.4\). Use Chebychev's inequality to find the value of \(c\) that guarantees the probability is at least \(96 \%\) that an outcome of the experiment lies between \(50-c\) and \(50+c .\)

Sugar packaged by a certain machine has a mean weight of \(5 \mathrm{lb}\) and a standard deviation of \(0.02 \mathrm{lb}\). For what values of \(c\) can the manufacturer of the machinery claim that the sugar packaged by this machine has a weight between \(5-c\) and \(5+c \mathrm{lb}\) with probability at least \(96 \%\) ?
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