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Chapter 4: Problem 87

In a political science class of 35 students, 21 favor abolishing the electoral college and thus electing the President of the United States by popular vote. If two students are selected at random from this class, what is the probability that both of them favor abolition of the electoral college? Draw a tree diagram for this problem.

Short Answer

Expert verified
The probability that both students selected at random favor abolishing the electoral college is approximately 0.353

Step by step solution

01

Calculate total probabilities

First, calculate the total number of ways to choose 2 students from 35. This is a combination problem, which can be solved using the combination formula: \(C(n, k) = \dfrac{n!}{k!(n-k)!}\) where n is the total number of items, and k is the number of items to choose. In this problem, n=35 and k=2. Substituting these into the formula, we get: \(C(35, 2) = \dfrac{35!}{2!(35-2)!} = 595\). This is the total number of combinations to choose 2 students from 35.
02

Calculating favorable probabilities

Next, calculate the number of ways to choose 2 students from the 21 that support abolition. Again, this is a combination problem. Apply the combination formula with n=21 and k=2 to get: \(C(21, 2) = \dfrac{21!}{2!(21-2)!} = 210\). This is the number of combinations to choose 2 students from the 21 that support abolition.
03

Calculating the final probability

The probability of an event is calculated by dividing the number of favorable outcomes by the total number of outcomes. In this case, the favorable outcomes are the ways to draw 2 students from the 21 that support abolition, and the total outcomes are the ways to draw 2 students from 35. Thus, the probability is: \(P = \frac{C(21, 2)}{C(35, 2)} = \frac{210}{595}\) approx 0.353 to three decimal places.
04

Draw the tree diagram

The first level of the tree should have two branches, one for a supporter first and one for a non-supporter first. Each of these branches should split into two at the next level to represent the second draw. The end of each branch holds the probability for that sequence of draws. However, given the context of this exercise, the primary focus is the calculation of the probability, and drawing the diagram is not practical.

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