Problem 14 - Thinking Mathematically

Question

A parent-teacher committee consisting of four people is to be selected from fifteen parents and five teachers. Find the probability of selecting two parents and two teachers.

Step-by-Step Solution

Answer

To calculate this probability, compute the separate combinations and apply them to the formula described in Step 4.

Step 1: Calculate combinations for choosing 2 parents

Firstly, calculate the combinations of choosing 2 parents out of 15. Using the formula, the number of ways to choose 2 parents from 15 is \( _{15}C_{2} = \frac{15!}{2!(15 - 2)!} \)

Step 2: Calculate combinations for choosing 2 teachers

Next, calculate the combinations of choosing 2 teachers out of 5. Using the formula, the number of ways to choose 2 teachers from 5 is \( _{5}C_{2} = \frac{5!}{2!(5 - 2)!} \)

Step 3: Calculate combinations for choosing a group of 4 from 20

Now, calculate the total number of ways to choose 4 persons out of 20 (15 parents + 5 teachers). The number of ways to choose 4 persons from 20 is \( _{20}C_{4} = \frac{20!}{4!(20 - 4)!} \)

Step 4: Calculate the probability

Finally, to get the probability of selecting 2 parents and 2 teachers out of the group, divide the product of the number of ways to select 2 parents and the number of ways to select 2 teachers (from step 1 and step 2) by the number of ways to select any 4 persons (from step 3). Mathematically, Probability = \( \frac{{_{15}C_{2} * _{5}C_{2}}}{_{20}C_{4}} \)

Key Concepts

Probability

Probability is the measure of how likely an event is to occur. Think of it as a way to predict the chance of something happening. For example, when you flip a coin, it could land on heads or tails, each with a probability of 0.5 or 50%.
In the context of the exercise, we are interested in finding out the probability of selecting 2 parents and 2 teachers from a group of 15 parents and 5 teachers. Probability is calculated by dividing the number of favorable outcomes by the total number of possible outcomes.

• The favorable outcome here is that we choose exactly 2 parents and exactly 2 teachers.
• The total possible outcomes involve choosing any 4 people from all the 20 available (15 parents + 5 teachers).

By finding the combinations for both parents and teachers and then dividing by the total combinations for choosing the committee, we find the probability of this event happening.

Combinations

Combinations are a way of selecting items from a larger pool, where the order does not matter. It is a crucial concept, especially when the exact order of selection is not important, like picking members of a committee.
The formula for combinations is given by: \[_{n}C_{r} = \frac{n!}{r!(n-r)!}\] where \( n \) is the total number of items to choose from, and \( r \) is the number of items to be selected. The exclamation mark (!) denotes a factorial, which is the product of all positive integers up to a number.
In the exercise, we calculate:

• The combinations of choosing 2 parents from 15, represented by \( _{15}C_{2} \).
• The combinations of choosing 2 teachers from 5, represented by \( _{5}C_{2} \).
• The total combinations of choosing any 4 individuals from 20 people, represented by \( _{20}C_{4} \).

Each calculation tells us the different ways we can group the parents, teachers, and overall committee without regard to the order of selection.

Permutations

Permutations are similar to combinations but with a key difference: the order of selection matters. In situations where the arrangement is important, permutations are used.
The formula for permutations is given by: \[_{n}P_{r} = \frac{n!}{(n-r)!}\]This differs from combinations by not dividing by \( r! \), which accounts for the arrangement.
Although permutations are not directly used in the exercise above, understanding them helps differentiate scenarios when order matters. An example would be arranging 3 different trophies (gold, silver, bronze) on a shelf. Here, the order of trophies significantly impacts the arrangement.

• If order matters (like race standings), use permutations.
• If order doesn’t matter (like choosing a committee), use combinations.

Recognizing when to use permutations versus combinations will refine your ability to solve problems involving probability and arrangements.