Problem 32 - Calculus for Biology and Medicine

Question

Assume that \(20 \%\) of a very common insect species in your study area is parasitized. Assume that insects are parasitized independently of each other. If you collect 10 specimens of this species, what is the probability that no more than 2 specimens in vour sample are parasitized?

Step-by-Step Solution

Answer

The probability that no more than 2 specimens are parasitized is approximately 0.6778.

Step 1: Understanding the Problem

We need to find the probability that no more than 2 out of 10 specimens are parasitized. This can include 0, 1, or 2 parasitized specimens.

Step 2: Define the Random Variable

Let the random variable \( X \) represent the number of parasitized specimens in a sample of 10. \( X \) follows a binomial distribution with parameters \( n = 10 \) (the number of trials, i.e., the number of specimens) and \( p = 0.20 \) (the probability that a single specimen is parasitized).

Step 3: Probability Mass Function

The probability that \( X = k \) for a binomial distribution is given by the formula: \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \). Here, \( \binom{n}{k} \) is a binomial coefficient.

Step 4: Calculate \( P(X = 0) \)

\( P(X = 0) = \binom{10}{0} (0.20)^0 (0.80)^{10} = 1 \times 1 \times 0.1074 = 0.1074 \).

Step 5: Calculate \( P(X = 1) \)

\( P(X = 1) = \binom{10}{1} (0.20)^1 (0.80)^9 = 10 \times 0.20 \times 0.1342 = 0.2684 \).

Step 6: Calculate \( P(X = 2) \)

\( P(X = 2) = \binom{10}{2} (0.20)^2 (0.80)^8 = 45 \times 0.04 \times 0.1678 = 0.3020 \).

Step 7: Sum Probabilities for \( X = 0, 1, 2 \)

Add the probabilities from Steps 4, 5, and 6: \( P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.1074 + 0.2684 + 0.3020 = 0.6778 \).

Key Concepts

Probability

Probability is the measure of how likely an event is to occur. It's a fundamental concept in statistics and is often expressed between 0 and 1, where 0 indicates impossibility and 1 indicates certainty.
When we talk about the probability that no more than 2 specimens in a sample are parasitized, we're interested in a specific event occurring among many possibilities. In this context, the event involves several potential outcomes like 0, 1, or 2 specimens being parasitized out of 10.

• The probability of zero specimens being parasitized is calculated using the formula: \( P(X = 0) \) based on the chances of each failing to be parasitized.
• The probability of exactly one specimen being parasitized is given by \( P(X = 1) \).
• Similarly, \( P(X = 2) \) computes the likelihood of exactly 2 parasites.

Finally, to find out the total probability of having no more than 2 parasitized specimens, you sum the individual probabilities: \( P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.1074 + 0.2684 + 0.3020 = 0.6778 \). This means there's about a 67.78% chance of having 2 or fewer parasitized.

Random Variable

A random variable is a numerical representation of possible outcomes of a random phenomenon. In our problem, the random variable \( X \) refers to the number of parasitized specimens in the collected sample.
There are two main types of random variables: discrete and continuous. Here, \( X \) is a discrete random variable because it can only take specific values, such as 0, 1, 2, ..., up to 10 specimens.

• The value of \( X \) depends on the outcome of a random experiment, which in this case is picking random specimens.
• For a binomial distribution, \( X \) can be thought of as counting the number of 'successes' in repeated, independent trials.
• Its range is limited by the number of trials: it can't exceed the total number of specimens sampled.

The random variable helps translate real-world scenarios into mathematical terms that can be analyzed using probability and statistical formulas.

Binomial Theorem

The binomial theorem is a powerful mathematical tool used to expand expressions raised to a power. However, in probability, the binomial distribution relies heavily on a related concept for modeling scenarios with binary outcomes.
In this problem, the binomial distribution models the number of parasitized insects among a fixed number of specimens. It uses parameters:

• \( n \): number of trials or specimens (here, 10)
• \( p \): probability of success on each trial (here, 0.20 or 20%)

The probability mass function for a binomial distribution is given by: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}, \] where \( \binom{n}{k} \) is the binomial coefficient, calculating how many combinations of \( k \) successes can occur out of \( n \) trials.
Using this theorem, we calculated probabilities such as \( P(X = 0), P(X = 1), \) and \( P(X = 2) \), which then helped determine the overall likelihood that no more than 2 specimens are parasitized.