91Ó°ÊÓ

Six people hope to be selected as a contestant on a TV game show. Two of these people are younger than 25 years old. Two of these six will be chosen at random to be on the show. a. What is the sample space for the chance experiment of selecting two of these people at random? (Hint: You can think of the people as being labeled \(\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}, \mathrm{E},\) and \(\mathrm{F}\). One possible selection of two people is \(\mathrm{A}\) and \(\mathrm{B}\). There are 14 other possible selections to consider.) b. Are the outcomes in the sample space equally likely? c. What is the probability that both the chosen contestants are younger than \(25 ?\) d. What is the probability that both the chosen contestants are not younger than \(25 ?\) e. What is the probability that one is younger than 25 and the other is not?

Step by step solution

01

Identify the possible combinations

First, we will find the sample space, i.e., the number of possible ways to select two contestants from the six people. Let's label the people as A, B, C, D, E, and F, where A, B are the ones younger than 25 years old. Use combinations to find the possible ways: \[\binom{6}{2} = \frac{6!}{(2!(6-2)!)} = \frac{6 \times 5}{2} = 15\] hence, there are 15 possible ways to select two contestants.

02

Enumerate the Sample Space

Now let's find the actual selections (pairs) possible: {AB, AC, AD, AE, AF, BC, BD, BE, BF, CD, CE, CF, DE, DF, EF}

03

Assess the Equally Likely Outcomes

As every person has the same chance of being selected, the sample space outcomes are equally likely. Each pair has a probability of \(\frac{1}{15}\) of being chosen.

04

Calculate the Probability of Both Contestants Being Younger Than 25

Since A and B are the contestants who are younger than 25, there is only 1 out of the 15 possible pairs that have both contestants younger than 25. So, the probability will be: \[\frac{1}{15}\]

05

Calculate the Probability of Both Contestants NOT Being Younger Than 25

There are four people older than 25, and we need to find the pairs with those people only. Use combinations here as well. \[\binom{4}{2} = \frac{4!}{(2!(4-2)!)} = \frac{4 \times 3}{2} = 6\] So, there are 6 possible pairs of contestants older than 25, and hence, the probability of both contestants being not younger than 25 is: \[\frac{6}{15}\]

06

Calculate the Probability of One Contestant Being Younger Than 25 and the Other Being Not Younger Than 25

This case can be seen as complementary to the other two cases, so the probability can be calculated using the complementary rule: \[P(\text{one younger, one not}) = 1 - P(\text{both younger}) - P(\text{both older})\] \[P(\text{one younger, one not}) = 1 - \frac{1}{15} - \frac{6}{15}\] \[P(\text{one younger, one not}) = \frac{8}{15} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Combinatorics

Combinatorics is a branch of mathematics that deals with counting, arranging, and ordering objects. In our exercise, we use combinatorics to determine how many ways we can select two contestants from a group of six people. The formula we use is the combination formula, which is given by \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \). Here, \( n \) signifies the total number of items (or people, in this case), and \( r \) is the number of items to choose.
We applied this formula to find \( \binom{6}{2} = 15 \), indicating that there are 15 possible combinations of choosing two people from six. This foundational step is crucial as it lays down the sample space for further probability calculations.

Sample Space

In probability theory, the sample space represents the set of all possible outcomes of a probability experiment. In the exercise at hand, the sample space involves combinations of selecting two people from a group of six.
By labeling the people as A, B, C, D, E, and F, and employing the concept of combinations, we derive the complete sample space as: {AB, AC, AD, AE, AF, BC, BD, BE, BF, CD, CE, CF, DE, DF, EF}.

• *Sample Space*: Collection of all possible pairs {AB, AC, AD, AE, AF, BC, BD, BE, BF, CD, CE, CF, DE, DF, EF}
• Size of Sample Space = 15 pairs, reflecting all possible contestant selections.

With the sample space defined, we can move forward to determining probabilities of finding specific arrangements within these 15 combinations.

Equally Likely Outcomes

When discussing probability, an equally likely outcome means each event in the sample space has the same chance of occurring. For example, in the TV game show contestant selection, each pair of two people has the same chance of being picked.
Since we're randomly selecting the contestants without bias, each outcome in our 15-member sample space has a probability of \( \frac{1}{15} \).

• This reflects the unbiased nature of random selection.
• Ensures fairness in the selection process, as every possible pair stands the same chance.

Understanding this concept helps in accurately calculating the probabilities of specific events occurring in our problem.

Complementary Rule

The complementary rule in probability is a powerful tool that simplifies calculating the probability of specific events. It involves understanding that the probability of an event not happening is 1 minus the probability of it happening.
In our exercise, we calculate the probability that one contestant is younger than 25 and the other is not using this rule. Since we know the probabilities for both being younger and both being not younger, the complementary probability is:

• \[P(\text{one younger, one not}) = 1 - P(\text{both younger}) - P(\text{both older})\]
• Calculated as \( \frac{8}{15} \), ensuring we cover all other possible combinations.

The complementary rule is valuable as it helps derive probabilities efficiently and accurately, especially when multiple situations blend into one exhaustive probability division.