A committee of three has to be chosen form a group of 4 men and 5 women. If the selection is made at random, what is the probability that exactly two members are men?
5/14
Understanding the Probability Problem
We need to find the probability of selecting a committee of exactly three members from a group consisting of 4 men and 5 women such that the committee has exactly two men.
To solve this, we'll use the concepts of combinations. The total number of ways to choose a committee will be our denominator, and the number of ways to choose a committee with exactly two men will be our numerator.
Calculating Total Possible Outcomes
The total number of people in the group is the sum of men and women:
\( \text{Total people} = 4 \text{ men} + 5 \text{ women} = 9 \)
We need to choose a committee of 3 members from these 9 people. The number of ways to do this is given by the combination formula \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \), where \( n \) is the total number of items, and \( k \) is the number of items to choose.
Total number of ways to choose a committee of 3 from 9 people:
\( \binom{9}{3} = \frac{9!}{3!(9-3)!} = \frac{9!}{3!6!} \)
\( \binom{9}{3} = \frac{9 \times 8 \times 7 \times 6!}{ (3 \times 2 \times 1) \times 6! } \)
\( \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} \)
\( \binom{9}{3} = \frac{504}{6} = 84 \)
So, there are 84 total possible ways to form the committee of three.
Calculating Favorable Outcomes (Exactly Two Men)
We want the committee to have exactly two men and, since the committee size is 3, it must also have exactly one woman.
- Number of ways to choose exactly 2 men from 4 men: \( \binom{4}{2} \)
- Number of ways to choose exactly 1 woman from 5 women: \( \binom{5}{1} \)
Using the combination formula:
\( \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{4 \times 3 \times 2!}{ (2 \times 1) \times 2! } = \frac{4 \times 3}{2} = \frac{12}{2} = 6 \)
\( \binom{5}{1} = \frac{5!}{1!(5-1)!} = \frac{5!}{1!4!} = \frac{5 \times 4!}{1 \times 4!} = 5 \)
To get exactly two men and one woman, we multiply the number of ways to choose men by the number of ways to choose women:
\( \text{Favorable outcomes} = \binom{4}{2} \times \binom{5}{1} = 6 \times 5 = 30 \)
There are 30 ways to form a committee with exactly two men and one woman.
Calculating the Probability
The probability of an event is defined as the ratio of the number of favorable outcomes to the total number of possible outcomes.
\( \text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \)
\( \text{Probability (Exactly 2 men)} = \frac{30}{84} \)
Now, we simplify the fraction:
\( \frac{30}{84} = \frac{6 \times 5}{6 \times 14} = \frac{5}{14} \)
The probability that exactly two members of the committee are men is 5/14.
Summary of Calculations
| Description | Calculation | Result |
|---|---|---|
| Total number of people | 4 men + 5 women | 9 |
| Committee size | - | 3 |
| Total possible committees (\( \binom{9}{3} \)) | \(\frac{9 \times 8 \times 7}{3 \times 2 \times 1}\) | 84 |
| Ways to choose 2 men from 4 (\( \binom{4}{2} \)) | \(\frac{4 \times 3}{2 \times 1}\) | 6 |
| Ways to choose 1 woman from 5 (\( \binom{5}{1} \)) | \(\frac{5}{1}\) | 5 |
| Favorable committees (2 men, 1 woman) | \(6 \times 5\) | 30 |
| Probability (Favorable / Total) | \(\frac{30}{84}\) | \(\frac{5}{14}\) |
Revision Table: Probability and Combinations
| Concept | Definition | Formula/Use Case |
|---|---|---|
| Probability | Measure of the likelihood of an event occurring. | \(\text{P(Event)} = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}}\) |
| Combination | A way of selecting items from a collection, where the order of selection does not matter. | \(\binom{n}{k} = C(n, k) = \frac{n!}{k!(n-k)!}\) for choosing \(k\) items from \(n\) distinct items. |
| Favorable Outcome | An outcome that satisfies the specific condition of the event we are interested in. | Calculated based on the criteria given in the problem (e.g., exactly 2 men). |
| Total Outcome | The total number of possible outcomes for an experiment or selection process. | Calculated based on the total pool of items and the number being selected. |
Additional Information: Committee Selection Probability
Committee selection problems are common applications of combinations in probability. The key is to identify:
- The total group size.
- The size of the committee or subgroup being selected.
- The specific composition required for the 'favorable' outcome (e.g., number of men and women).
Once these are identified, use combinations to calculate:
- The total number of ways to select the committee from the entire group.
- The number of ways to select the committee with the desired composition.
The probability is then the ratio of the second number to the first. Remember that 'exactly' a certain number of individuals of one type often implies a specific number of individuals of the other type(s) to meet the total size requirement.
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