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Chapter 11: Problem 13

A city council consists of six Democrats and four Republicans. If a committee of three people is selected, find the probability of selecting one Democrat and two Republicans.

Short Answer

Expert verified
The probability of selecting one Democrat and two Republicans from the city council is \(\frac{C(6, 1) * C(4, 2)}{C(10, 3)}\), where \(C(n, k)\) is the combination formula

Step by step solution

01

Calculate the number of ways to select one Democrat

Use the combination formula to determine how many ways you can choose one Democrat from a group of six. This is denoted as \( C(6, 1) \) where the first value is the total number (6), and the second value is the number you want to choose (1). Calculate this using the formula \[ C(n, k) = \frac{n!}{k!(n - k)!} \] where \(n\) is the total number and \(k\) is the number you want to choose.
02

Calculate the number of ways to select two Republicans

Similarly, use the combination formula to determine the number of ways you can choose two Republicans from a group of four. This is denoted as \( C(4, 2) \). Use the same formula as in step 1.
03

Calculate the total number of ways to select a committee of three people

Now, determine the total number of ways a committee of three people can be selected from the city council, which has ten members. This is denoted as \( C(10, 3) \). Again, use the same combination formula.
04

Multiply the number of ways to select one Democrat and two Republicans

Next, multiply the values obtained from step 1 and 2. This will give the total number of ways to select one Democrat and two Republicans.
05

Calculate the probability

The probability is the ratio of the number of favorable outcomes to the total number of outcomes. The number of favorable outcomes is the value from step 4, and the total number of outcomes is the value from step 3. Use the formula \[ P(A) = \frac{number of favorable outcomes}{total number of outcomes} \]

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Most popular questions from this chapter

Nine cards numbered from 1 through 9 are placed into a box and two cards are selected without replacement. Find the probability that both numbers selected are odd, given that their sum is even.

An ice chest contains six cans of apple juice, eight cans of grape juice, four cans of orange juice, and two cans of mango juice. Suppose that you reach into the container and randomly select three cans in succession. Find the probability of selecting no grape juice.

The probability that a region prone to flooding will flood in any single year is \(\frac{1}{10}\). a. What is the probability of a flood two years in a row? b. What is the probability of flooding in three consecutive years? c. What is the probability of no flooding for ten consecutive years? d. What is the probability of flooding at least once in the next ten years?

We return to our box of chocolates. There are 30 chocolates in the box, all identically shaped. Five are filled with coconut, 10 with caramel, and 15 are solid chocolate. You randomly select one piece, eat it, and then select a second piece. Find the probability of selecting a coconut-filled chocolate followed by a caramel-filled chocolate.

Make Sense? Determine whether each statement makes sense or does not make sense, and explain your reasoning. I must have made an error calculating probabilities because \(P(A \mid B)\) is not the same as \(P(B \mid A)\).
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