Problem 27 - Algebra and Trigonometry

Question

Pharmaceuticals \(\quad\) A drug used to prevent motion sickness is found to be effective about 75\(\%\) of the time. Six friends, prone to seasickness, go on a sailing cruise, and all take the drug. Find the probability of each event. (a) None of the friends gets seasick. (b) All of the friends get seasick. (c) Exactly 3 get seasick (d) At least 2 get seasick.

Step-by-Step Solution

Answer

(a) 0.178, (b) 0.000244, (c) 0.131836, (d) 0.525.

Step 1: Define the problem context

This is a binomial probability problem because each friend getting seasick or not can be considered as a trial with two possible outcomes: success (does not get seasick) with probability 0.75, and failure (gets seasick) with probability 0.25. We need to calculate the probability of different scenarios out of the six trials.

Step 2: Calculate probability that none get seasick

Let's calculate the probability that all 6 friends do not get seasick. The probability that one friend does not get seasick (success) is 0.75. The probability that none get seasick is the probability of success raised to the power of 6 trials:\[ P(X = 0) = (0.75)^6 \approx 0.178 \]

Step 3: Calculate probability that all get seasick

The probability that one friend gets seasick (failure) is 0.25. To find the probability that all 6 friends get seasick, raise the probability of failure to the power of 6:\[ P(X = 6) = (0.25)^6 \approx 0.000244 \]

Step 4: Calculate probability that exactly 3 get seasick

Use the binomial probability formula: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] For 3 friends getting seasick:\[ P(X = 3) = \binom{6}{3} (0.25)^3 (0.75)^3 \]Calculating the above gives:\[ P(X = 3) = 20 \times 0.015625 \times 0.421875 = 0.131836 \]

Step 5: Calculate probability of at least 2 getting seasick

"At least 2" means 2, 3, 4, 5, or 6 get seasick. This can be calculated by subtracting the probabilities of 0 or 1 getting seasick from 1.First calculate for 1 getting seasick:\[ P(X = 1) = \binom{6}{1}(0.25)^1(0.75)^5 \approx 0.297 \]Now calculate the probability:\[ P(X \geq 2) = 1 - (P(X = 0) + P(X = 1)) \]\[ P(X \geq 2) = 1 - (0.178 + 0.297) = 0.525 \]

Key Concepts

Probability Theory

Probability theory is a branch of mathematics that deals with the likelihood of events occurring. In our exercise, we determine the probability of various seasickness outcomes for six friends. The problem is a binomial probability problem, which means each friend is a trial that results in a success or failure.

• Success: A friend does not get seasick, which has a probability of 0.75.
• Failure: A friend gets seasick, which has a probability of 0.25.

Probability theory helps us calculate the chance of all, none, or some friends getting seasick. It requires understanding the nature of independent events, where one friend's condition does not affect another's. Each trial (a friend's reaction to the drug) is independent, making this a perfect illustration of probability theory in action.

Combinatorics

Combinatorics involves counting, arranging, and finding patterns within sets. In our problem, it helps in determining the number of ways events can occur. For example, when calculating the probability of exactly 3 out of 6 friends getting seasick, we first find how many ways 3 friends can be chosen from 6. This is where the binomial coefficient comes in, denoted as \(\binom{n}{k}\), calculated by:\[\binom{6}{3} = \frac{6!}{3!(6-3)!} = 20\]This means there are 20 different ways to select which 3 friends get seasick out of 6. By multiplying this with the probabilities from probability theory, we find the total probability of exactly 3 friends getting seasick. Combinatorics helps simplify these kinds of calculations by providing efficient techniques to count possible outcomes.

Probability Distribution

A probability distribution describes the likelihood of each possible outcome in an experiment. For our scenario, the distribution is a binomial distribution, because the outcomes (friends getting seasick or not) hold two possibilities and happen independently.In a binomial distribution, the probability \(P(X = k)\) for any outcome \(k\) is given by:\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]

• \(n\): Total number of trials (friends), here \(n = 6\).
• \(k\): Number of successes of interest (such as all being seasick, none, or 3).
• \(p\): Probability of success on a single trial, which in this context is 0.75 for not getting seasick.

Each probability provides insight into various real-world scenarios. For instance, the probability of at least 2 getting seasick is found by calculating and summing the probabilities for 2, 3, 4, 5, and 6, providing a comprehensive understanding of outcomes for this given situation. Probability distributions not only help predict likelihoods but also assist in decision making based on those predictions.