Problem 25 - Discrete Mathematics with Applications

Question

There are 15 rabbits in a cage. Five of them are injected with a certain drug. Three of the 15 rabbits are selected successively at random for an experiment. Find the probability that: Only the second rabbit is injected with the drug.

Step-by-Step Solution

Answer

The probability that only the second rabbit is injected with the drug is \(\frac{30}{91}\).

Step 1: Compute the total outcomes

All rabbits can be selected in any order for the experiment. Therefore, the total number of outcomes will be the total ways to select 3 rabbits out of 15, which is calculated as combinations: \[C(15, 3) = \frac{15!}{(15-3)!3!}.\]

Step 2: Compute the favorable outcomes

We want to find the probability that only the second rabbit is injected with the drug. This means that the first and third rabbits selected should not be injected with the drug. So, the number of ways to select the first rabbit (not injected) is 10 (as we have 10 non-injected rabbits). The number of ways to select the second rabbit (injected) is 5 (as we have 5 injected rabbits). Finally, the number of ways to select the third rabbit (not injected) is 9 (as we have only 9 non-injected rabbits left after the first rabbit is selected). To calculate the favorable outcomes, we'll multiply these three numbers: \(10 \times 5 \times 9\).

Step 3: Calculate the probability

Now, we will divide the number of favorable outcomes by the total outcomes to find the probability. Remember that the probability is calculated as: \[P \ = \ \frac{\text{favorable outcomes}}{\text{total outcomes}}.\] Substitute the values from Steps 1 and 2: \(P\ =\ \frac{10 \times 5 \times 9}{C(15, 3)}\).

Step 4: Simplify the expression

Simplify the expression as follows: \(P\ =\ \frac{10 \times 5 \times 9}{\frac{15!}{(15-3)!3!}}\). Next, compute the factorials, and perform the calculations: \(P \ = \ \frac{10 \times 5 \times 9 \times 3! \times (15-3)!}{15!} \ =\ \frac{450 \times 6 \times 12!}{15!}\). Now, cancel the common factorials: \[P\ =\ \frac{450\times 6}{(15\times 14\times 13)}.\] Further simplifying this expression, we get: \[P\ =\ \frac{30}{91}.\] So, the probability that only the second rabbit is injected with the drug is \(\frac{30}{91}\).

Key Concepts

Combinatorics

Combinatorics is a branch of mathematics that deals with counting and arranging. It's crucial when determining the number of possible outcomes in a scenario. In this exercise, combinatorics is used to calculate how many different groups of three rabbits can be selected from a total of fifteen rabbits. This is accomplished using combinations, which is appropriate because the order of selection does not matter. The formula for combinations is given by: \[C(n, k) = \frac{n!}{(n-k)!k!}\], where \(n\) is the total number of items, and \(k\) is the number of items to choose.

• In our scenario, \(n = 15\) and \(k = 3\).
• So, the calculation becomes \(C(15, 3) = \frac{15!}{(15-3)!3!}\).

The factorial function \(!\) indicates a product of all positive integers up to that number, which simplifies this calculation mathematically.

Conditional Probability

Conditional probability refers to the likelihood of an event occurring, given that another event has already occurred. In our rabbit problem, we are interested in finding the chance that only the second rabbit is injected with a drug. This involves the calculation of probability focused on the events surrounding this one condition.

• First, determine the possible selections given this condition: choosing an un-injected rabbit, then an injected one, and another un-injected one.
• Each selection outcome was factored to fit this condition: \(10\) ways for the first, \(5\) for the second, and \(9\) for the third.

By following the conditional steps, we then multiply these particular possibilities: \(10 \times 5 \times 9\). This reflects the number of favorable outcomes given the condition that exactly the second rabbit is injected. This approach is pivotal in narrowing down outcomes to fit specific conditions prescribed by a problem.

Factorials

Factorials are a key component in combinatorics and probability, providing a mathematical way to represent multiplication of a series of descending natural numbers. The notation for a factorial is "\(!\)", and it is essential for calculations involving permutations and combinations. Factorials are used in this problem to discern how many different sets of rabbits we can form when order isn't a concern—hence using the combination formula, which inherently uses factorials.

• The factorial \(15!\) represents \(15 \times 14 \times 13 \times \ldots \times 1\).
• This tool simplifies the process by which we calculate large number combinations.
• Understanding how factorials break down can help simplify complex probability questions.

Recognizing the relationships among the components—such as the total number of ways to select rabbits (\(15!\)) and ways to select particular group formations \((3!)\), empowers students to tackle step-wise probability queries efficiently.