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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

Three prisoners are informed by their jailer that one of them has been chosen at random to be executed, and the other two are to be freed. Prisoner \(A\) asks the jailer to tell him privately which of his fellow prisoners will be set free, claiming that there would be no harm in divulging this information, since he already knows that at least one will go free. The jailer refuses to answer this question, pointing out that if \(A\) knew which of his fellows were to be set free, then his own probability of being executed would rise from \(\frac{1}{3}\) to \(\frac{1}{2}\), since he would then be one of two prisoners. What do you think of the jailer's reasoning?

Argue that \(E=E F \cup E F^{c}, E \cup F=E \cup F E^{c}\).

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

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

Suppose we have ten coins which are such that if the \(i\) th one is flipped then heads will appear with probability \(i / 10, i=1,2, \ldots, 10\). When one of the coins is randomly selected and flipped, it shows heads. What is the conditional probability that it was the fifth coin?
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