Understanding the Problem: Probability of Ships Arriving Safely

The question asks for the probability that exactly 3 out of 5 ships would arrive safely during a voyage, given the average sinking rate.

This scenario involves a fixed number of trials (5 ships), each trial having only two possible outcomes (ship arrives safely or ship is sunk), and the probability of success (arriving safely) is constant for each ship. This is a classic case of a binomial probability distribution problem.

Defining Parameters for Binomial Probability

Let's define the parameters based on the problem statement:

• Total number of ships (trials), \(n = 5\).
• The probability of a ship being sunk is given as 1 out of 5 on average. This is the probability of "failure" in this context if we consider "arriving safely" as success. Let \(P(\text{sunk})\) be this probability.
• \(P(\text{sunk}) = \frac{1}{5}\).
• The probability of a ship arriving safely is the complement of being sunk. Let \(p\) be the probability of "success" (arriving safely).
• \(p = P(\text{safe}) = 1 - P(\text{sunk}) = 1 - \frac{1}{5} = \frac{5-1}{5} = \frac{4}{5}\).
• We are interested in the probability that exactly 3 out of the 5 ships arrive safely. So, the number of successful outcomes we are looking for is \(k = 3\).
• The probability of failure (ship being sunk) is \(1-p = \frac{1}{5}\).

Applying the Binomial Probability Formula

The probability of getting exactly \(k\) successes in \(n\) trials in a binomial distribution is given by the formula:

\(P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}\)

Where:

• \(\binom{n}{k}\) is the binomial coefficient, calculated as \(\frac{n!}{k!(n-k)!}\), representing the number of ways to choose \(k\) successes from \(n\) trials.
• \(p\) is the probability of success on a single trial.
• \((1-p)\) is the probability of failure on a single trial.
• \(k\) is the number of successes.
• \(n-k\) is the number of failures.

Calculating the Probability

Now, let's plug in our values: \(n=5\), \(k=3\), \(p=\frac{4}{5}\), and \(1-p=\frac{1}{5}\).

\(P(X=3) = \binom{5}{3} \left(\frac{4}{5}\right)^3 \left(\frac{1}{5}\right)^{5-3}\)

\(P(X=3) = \binom{5}{3} \left(\frac{4}{5}\right)^3 \left(\frac{1}{5}\right)^2\)

First, calculate the binomial coefficient \(\binom{5}{3}\):

\(\binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5!}{3!2!}\)

\(\binom{5}{3} = \frac{5 \times 4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1) \times (2 \times 1)} = \frac{5 \times 4}{2 \times 1} = \frac{20}{2} = 10\)

Next, calculate the powers of the probabilities:

\(\left(\frac{4}{5}\right)^3 = \frac{4^3}{5^3} = \frac{4 \times 4 \times 4}{5 \times 5 \times 5} = \frac{64}{125}\)

\(\left(\frac{1}{5}\right)^2 = \frac{1^2}{5^2} = \frac{1 \times 1}{5 \times 5} = \frac{1}{25}\)

Now, substitute these values back into the binomial probability formula:

\(P(X=3) = 10 \times \frac{64}{125} \times \frac{1}{25}\)

\(P(X=3) = \frac{10 \times 64 \times 1}{125 \times 25}\)

\(P(X=3) = \frac{640}{3125}\)

To simplify the fraction, we can divide both the numerator and the denominator by their greatest common divisor. Both numbers are divisible by 5.

Numerator: \(640 \div 5 = 128\)

Denominator: \(3125 \div 5 = 625\)

So, the simplified probability is \(\frac{128}{625}\).

The probability that exactly 3 out of 5 ships would arrive safely is \(\frac{128}{625}\).

Conclusion

Based on the binomial probability calculation, the probability of exactly 3 out of 5 ships arriving safely, given a 1/5 chance of being sunk, is \(\frac{128}{625}\).

Revision Table: Key Concepts

Additional Information: Binomial Distribution Conditions

For a probability problem to follow a binomial distribution, the following four conditions must be met:

1. Fixed Number of Trials (n): There must be a fixed number of times the experiment is repeated. In this problem, the fixed number of trials is the 5 ships making the voyage.
2. Each Trial is Independent: The outcome of one trial does not affect the outcome of other trials. We assume the fate of one ship does not influence the fate of another ship.
3. Each Trial has Only Two Outcomes: Each trial must result in either a "success" or a "failure". Here, the outcomes are "ship arrives safely" (success) or "ship is sunk" (failure).
4. Probability of Success (p) is Constant: The probability of success must be the same for each trial. We assume the probability of arriving safely is constant for every ship making the voyage.

Since all these conditions are met in the given problem, it is appropriate to use the binomial probability formula.