Problem 24 - 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 first rabbit is injected with the drug.

Step-by-Step Solution

Answer

The probability that only the first rabbit is injected with the drug is \(\dfrac{18}{546}\).

Step 1: Determine the total number of rabbits with and without the drug injection.

We are told that there are 15 rabbits in total, out of which 5 are injected with the drug. This means that there are 10 rabbits without the drug injection.

Step 2: Find the successful outcomes.

A successful outcome is defined as a situation where only the first rabbit selected has the drug injection, while the second and third rabbits selected do not have the drug injection. To find the number of successful outcomes, consider the following: - The probability of choosing a drugged rabbit first: There are 5 drugged rabbits and 15 rabbits in total, so the probability is \(\dfrac{5}{15}\). - The probability of choosing a non-drugged rabbit second: There are 10 non-drugged rabbits remaining, and 14 rabbits in total, so the probability is \(\dfrac{10}{14}\). - The probability of choosing a non-drugged rabbit third: There are 9 non-drugged rabbits remaining, and 13 rabbits in total, so the probability is \(\dfrac{9}{13}\).

Step 3: Calculate the probability of successful outcomes.

To find the probability of successful outcomes, multiply the probabilities found in Step 2: \(\dfrac{5}{15} \times \dfrac{10}{14} \times \dfrac{9}{13} = \dfrac{5 \times 10 \times 9}{15 \times 14 \times 13}\)

Step 4: Simplify the expression from Step 3.

To simplify the expression, cancel out common factors, such as 5, in the numerator and the denominator, resulting in: \(\dfrac{5 \times 10 \times 9}{15 \times 14 \times 13} = \dfrac{1 \times 2 \times 9}{3 \times 14 \times 13} = \dfrac{18}{546}\)

Step 5: Compute the required probability.

After simplification, the probability of having only the first rabbit injected with the drug is: \(P(\text{only first rabbit injected}) = \dfrac{18}{546}\) The probability that only the first rabbit is injected with the drug is \(\dfrac{18}{546}\).

Key Concepts

Combinatorics

The concept of combinatorics is all about counting and arranging. It helps us figure out how many different ways we can select objects from a larger group. This is essential in probability because it allows us to determine the total possible outcomes. In our exercise, we had 15 rabbits, and we needed to calculate the probability of choosing the first rabbit with a specific trait.
We used a simple method for this, which involves the following steps:

• Counting the total rabbits: 15 in this case.
• Classifying them into categories, such as those with or without injections.

This scenario didn't require complicated permutations or combinations, but understanding how to categorize and count possibilities is key. Combinatorics enables us to simplify complicated problems by breaking them down into manageable parts.

Conditional Probability

Conditional probability measures how likely an event is to happen, given that another event has already occurred. In our rabbit example, the probability was dependent on the selection order.

We started by choosing a drugged rabbit first. But the second selection depended on the first, specifically that the first rabbit was drugged. So our probability changed each time.

• Probability for the first rabbit: 5 drugged out of 15 rabbits.
• Conditional probability for the second rabbit being non-drugged: 10 out of 14 remaining rabbits.
• Conditional probability for the third rabbit still being non-drugged: 9 out of 13 remaining rabbits.

Each selection affected the probabilities of the next chooses. We multiplied these conditional probabilities to get the result. This process illustrates how conditional amounts factor into overall predictions, creating a stepwise calculation adjusting as previous conditions modify the scenario.

Sampling without Replacement

Sampling without replacement means once an item is selected, it can't be chosen again for the same round of selections. It impacts the odds and probabilities in very meaningful ways, as we saw in the rabbit problem.

Once we pick a rabbit, both the total number of available rabbits and specific categories change:

• Start with 15 rabbits, pick one, now we have 14 left.
• If the first was drugged, only 9 non-drugged remain from the original 10.

Sampling without replacement recalculates the pool for each step. This ensures that each choice accurately reflects the altered circumstances, preventing repeated choices and keeping the results fair and accurate.
This method reflects real-world difficulties where items or subjects can't be reused, ensuring each selection is unique and adheres to natural diminishing options.