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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

A box contains three marbles: one red, one green, and one blue. Consider an experiment that consists of taking one marbie from the box then replacing it in the box and drawing a second marble from the box. What is the sample space? If, at all times, each marble in the box is equally likely to be selected, what is the probability of each point in the sample space?

Let \(E, F, G\) be three events. Find expressions for the events that of \(E, F, G\) (a) only \(F\) occurs, (b) both \(E\) and \(F\) but not \(G\) occurs, (c) at least one event occurs, (d) at least two events occur, (c) all three events occur, (f) none occurs, (g) at most one occurs, (h) at most two occur.

A fair coin is continually flipped. What is the probability that the first four flips are (a) \(H, H, H, H ?\) (b) \(T, H, H, H ?\) (c) What is the probability that the pattern \(T, H, H, H\) occurs before the pattern \(\boldsymbol{H}, \boldsymbol{H}, \boldsymbol{H}, \boldsymbol{H} ?\)

Let \(E\) and \(F\) be mutually exclusive events in the sample space of an experiment. Suppose that the experiment is repeated until either event \(E\) or event \(F\) occurs. What does the sample space of this new super experiment look like? Show that the probability that event \(E\) occurs before event \(F\) is \(P(E) /[P(E)+P(F)]\) Hint: Argue that the probability that the original experiment is performed \(n\) times and \(E\) appears on the \(n\) th time is \(P(E) \times(1-p)^{n-1}\) \(n=1,2, \ldots\), where \(p=P(E)+P(F) .\) Add these probabilities to get the desired answer.

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