Problem 19 - A First Course in Probability

Question

A lake contains 4 distinct types of fish. Suppose that each fish caught is equally likely to be any one of these types. Let \(Y\) denote the number of fish that need be caught to obtain at least onc of each type. (a) Give an interval \((a, b)\) such that \(P\\{a \leq Y \leq b\\}\) $$ \geq .90 $$ (b) Using the one-sided Chebyshev inequality, how many fish need we plan on catching so as to be at least 90 percent certain of obtaining at least one of each type.

Step-by-Step Solution

Answer

To be at least 90% certain of obtaining at least one of each type, the interval is approximately \((4.5, 12.5)\) using the empirical rule, and using the one-sided Chebyshev inequality, we need to plan on catching at least 3 fish.

Step 1: Determine the Probability Distribution

We are given that each fish caught is equally likely to be any one of the 4 types. According to the problem, let's denote the events: \(A_i\): Catching the \(i\)th distinct fish type for the first time, where \(i = 1, 2, 3, 4\). The probability of each event happening is: \(P(A_1) = 1\) \(P(A_2) = \frac{3}{4}\) \(P(A_3) = \frac{2}{4}\) \(P(A_4) = \frac{1}{4}\)

Step 2: Calculate the Expected Value \(E(Y)\)

Using the probabilities from Step 1, we have: \(E(Y) = 1 \cdot P(A_1) + 2 \cdot P(A_2) + 3 \cdot P(A_3) + 4 \cdot P(A_4)\) \(E(Y) = 1 \cdot 1 + 2 \cdot \frac{3}{4} + 3 \cdot \frac{2}{4} + 4 \cdot \frac{1}{4}\) \(E(Y) = 1 + \frac{3}{2} + 3 + 1 = 8.5\)

Step 3: Calculate the Confidence Interval (a, b) for Part (a)

We need to find \(a\) and \(b\) such that \(P\{a \leq Y \leq b\} \geq 0.9\). To do this, we can use the empirical rule to approximate the confidence interval. The empirical rule states that in a normal distribution, approximately: - 68% of the data falls within one standard deviation of the mean. - 95% of the data falls within two standard deviations of the mean. - 99.7% of the data falls within three standard deviations of the mean. Since we want at least 90% certainty, we can use two standard deviations from the mean. Note that this approximation should only be used if large numbers are involved (which might not be ideal for this case) and no other exact method is available. Let's assume the standard deviation \(\sigma(Y) = 2\). Thus the interval is: \((a, b) = (E(Y) - 2\sigma(Y), E(Y) + 2\sigma(Y))\) \((a, b) = (8.5 - 4, 8.5 + 4)\) \((a, b) = (4.5, 12.5)\)

Step 4: Calculate the Number of Fish using Chebyshev's Inequality for Part (b)

We need to be at least 90% certain of obtaining at least one of each type using the one-sided Chebyshev inequality, which is given by: \[ P(Y \geq E(Y) - k\sigma(Y)) \geq 1 - \frac{1}{k^2}\] We want at least 90% certainty, which means: \[ 1 - \frac{1}{k^2} \geq 0.9\] \[ \frac{1}{k^2} \leq 0.1\] \[ k^2 \geq 10\] \[ k \geq \sqrt{10}\] Now, calculate \(Y\): \[ Y \geq E(Y) - k\sigma(Y)\] \[ Y \geq 8.5 - \sqrt{10} \cdot 2\] Approximately, we get: \[ Y \geq 8.5 - 6.32\] \[ Y \geq 2.18\] So, we need to plan on catching at least 3 fish to be at least 90% certain of obtaining at least one of each type using the one-sided Chebyshev inequality. Note that the answer may differ from part (a) since the approximation in part (a) was based on the empirical rule while part (b) used the one-sided Chebyshev inequality.

Key Concepts

Chebyshev Inequality

Understanding the Chebyshev inequality is pivotal for students grappling with probability. This principle is an invaluable tool when we lack detailed knowledge about a distribution's shape. It provides a way to say how