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Chapter 1: Problem 33

In a class there are four freshman boys, six freshman girls, and six sophomore boys. How many sophomore girls must be present if sex and class are to be independent when a student is selected at random?

Short Answer

Expert verified
There is no solution in this case, which means that sex and class cannot be independent with the given conditions.

Step by step solution

01

Define the probabilities

Since there are four freshman boys (FB), six freshman girls (FG), and six sophomore boys (SB), let's denote the number of sophomore girls (SG) as x. The total number of students in the class will be 10 + x.
02

Calculate the probabilities for freshmen and sophomores

There are 10 freshmen (4 FB and 6 FG) and 6+x sophomores (6 SB and x SG). The probability of choosing a freshman is P(F) = 10 / (10+x) and the probability of choosing a sophomore is P(S) = (6+x) / (10+x).
03

Calculate the probabilities for boys and girls

There are 4+6=10 boys (4 FB and 6 SB) and 6+x girls (6 FG and x SG). The probability of choosing a boy is P(B) = 10 / (10+x) and the probability of choosing a girl is P(G) = (6+x) / (10+x).
04

Determine when sex and class are independent

The probabilities of sex and class are independent if P(F 鈭� B) = P(F)P(B), P(F 鈭� G) = P(F)P(G), P(S 鈭� B) = P(S)P(B), and P(S 鈭� G) = P(S)P(G). We will write these equations and plug in the respective probabilities. Equation 1: \( \frac{4}{10+x} = \frac{10}{10+x} \cdot \frac{10}{10+x} \) Equation 2: \( \frac{6}{10+x} = \frac{10}{10+x} \cdot \frac{6+x}{10+x} \) Equation 3: \( \frac{6}{10+x} = \frac{6+x}{10+x} \cdot \frac{10}{10+x} \) Equation 4: \( \frac{x}{10+x} = \frac{6+x}{10+x} \cdot \frac{6+x}{10+x} \)
05

Simplify and solve for x

We will notice that Equation 3 is the same as Equation 2, so we can disregard it. Now let's simplify and solve for x using Equation 4. \( \frac{x}{10+x} = \frac{6+x}{10+x} \cdot \frac{6+x}{10+x} \) Since the denominators are the same, we can compare the numerators: \(x = (6+x)^2 \) \(0 = x^2 + 12x + 36 -x \) \(0 = x^2 + 11x + 36 \) Now, solve the quadratic equation: \(x = \frac{-11 \pm \sqrt{(-11)^2 - 4\cdot1\cdot36}}{2\cdot1}\) \(x = \frac{-11 \pm \sqrt{121 - 144}}{2}\) \(x = \frac{-11 \pm \sqrt{-23}}{2}\) Since x cannot be a negative number or imaginary, there is no solution in this case, which means that sex and class cannot be independent with the given conditions.

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Most popular questions from this chapter

Stores \(A, B\), and \(C\) have 50,75, and 100 employees, and, respectively, 50 , 60, and 70 percent of these are women. Resignations are equally likely among all employees, regardless of sex. One employee resigns and this is a woman. What is the probability that she works in store \(C\) ?

For events \(E_{1}, E_{2}, \ldots, E_{n}\) show that $$ P\left(E_{1} E_{2} \cdots E_{n}\right)=P\left(E_{1}\right) P\left(E_{2} \mid E_{1}\right) P\left(E_{3} \mid E_{1} E_{2}\right) \cdots P\left(E_{n} \mid E_{1} \cdots E_{n-1}\right) $$

If the occurrence of \(B\) makes \(A\) more likely, does the occurrence of \(A\) make \(B\) more likely?

Use Exercise 15 to show that \(P(E \cup F)=P(E)+P(F)-P(E F)\).

What is the conditional probability that the first die is six given that the sum of the dice is seven?
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