Question

Let \(x\) be a hypergeometric random variable with \(N=15, n=3,\) and \(M=4\). Use this information to answer the questions in Exercises 14-17. Construct a probability histogram for \(x\)

Short answer

Question: Construct a probability histogram for a hypergeometric random variable with parameters \(N=15\), \(n=3\), and \(M=4\). Answer: The probability histogram for the hypergeometric random variable with the given parameters will have the following probabilities for each value of \(x\): - \(P(X=0) \approx 0.3626\) - \(P(X=1) \approx 0.4835\) - \(P(X=2) \approx 0.1451\) - \(P(X=3) \approx 0.0088\) The histogram will have bars for each value of \(x\) (0, 1, 2, and 3) with heights corresponding to their respective probabilities, with the highest probability at \(x=1\).

Step-by-step solution

Determine the possible values of x

Based on the given parameters, the total population size is \(N=15\), the number of successes in the population is \(M=4\), and the size of the sample taken is \(n=3\). The random variable \(x\) represents the number of successes in the sample. The minimum value of \(x\) is 0 (no successes in the sample) and the maximum value is min\((n, M)\), which is equal to min\((3, 4) = 3\). So the possible values of \(x\) are 0, 1, 2, and 3.

Calculate the probabilities using the PMF

Now that we know the possible values of \(x\), we can calculate their corresponding probabilities using the hypergeometric PMF formula: \(P(X=x) = \frac{\binom{M}{x} \binom{N-M}{n-x}}{\binom{N}{n}}\) We will calculate the probabilities for each possible value of \(x\): \(P(X=0) = \frac{\binom{4}{0} \binom{15-4}{3-0}}{\binom{15}{3}}\) \(P(X=1) = \frac{\binom{4}{1} \binom{15-4}{3-1}}{\binom{15}{3}}\) \(P(X=2) = \frac{\binom{4}{2} \binom{15-4}{3-2}}{\binom{15}{3}}\) \(P(X=3) = \frac{\binom{4}{3} \binom{15-4}{3-3}}{\binom{15}{3}}\)

Calculate the probabilities for each x value

Now we will compute the probabilities using the previously determined combinations: \(P(X=0) = \frac{\binom{4}{0} \binom{11}{3}}{\binom{15}{3}} = \frac{1 \cdot 165}{455} = \frac{165}{455} \approx 0.3626\) \(P(X=1) = \frac{\binom{4}{1} \binom{11}{2}}{\binom{15}{3}} = \frac{4 \cdot 55}{455} = \frac{220}{455} \approx 0.4835\) \(P(X=2) = \frac{\binom{4}{2} \binom{11}{1}}{\binom{15}{3}} = \frac{6 \cdot 11}{455} = \frac{66}{455} \approx 0.1451\) \(P(X=3) = \frac{\binom{4}{3} \binom{11}{0}}{\binom{15}{3}} = \frac{4 \cdot 1}{455} = \frac{4}{455} \approx 0.0088\)

Construct the probability histogram

Finally, we can construct the probability histogram for the hypergeometric random variable \(x\) using the calculated probabilities: On the x-axis, we have the possible values of \(x\): 0, 1, 2, and 3. On the y-axis, we have the corresponding probabilities: - \(P(X=0) \approx 0.3626\) - \(P(X=1) \approx 0.4835\) - \(P(X=2) \approx 0.1451\) - \(P(X=3) \approx 0.0088\) For each value of \(x\), plot a bar with the corresponding probability as its height. The final probability histogram should resemble an (approximate) symmetric distribution with the highest probability at \(x=1\).

Key concepts

Hypergeometric Random Variable

A hypergeometric random variable is a type of discrete random variable that describes the number of successes in a series of draws from a finite population without replacement. To elaborate, imagine you have a bag with a mixture of colored balls, some red (successes) and some blue (failures). If you draw a handful of balls from this bag without putting any back, the number of red balls you get can be represented by a hypergeometric random variable.

For example, if you have a total of 15 balls (N=15), of which 4 are red (M=4), and you draw 3 balls at random (n=3), the number of red balls in your hand can be any number between 0 and the minimum of the number of draws or red balls available, in this case, 3. This range of numbers, 0 to 3, is the set of possible values for the hypergeometric random variable 'x' in the context of the exercise provided.

Probability Histogram

A probability histogram is a graphical representation that displays the probabilities associated with each possible value of a discrete random variable. It is a valuable tool for visualizing the distribution of probabilities in a dataset.

Each bar in the histogram corresponds to one possible value of the random variable and the height of the bar represents the probability of that value occurring. In the case of our hypergeometric random variable 'x', there would be bars for each value of x from 0 to 3, with heights equivalent to the likelihood of drawing exactly that number of red balls from the bag. The histogram helps to quickly identify the most probable outcomes and understand the shape of the distribution, which, in our case, tends to peak at the value with the highest probability.

Probability Mass Function (PMF)

The Probability Mass Function (PMF) is a mathematical formula that gives us the probabilities of discrete random variables, like the hypergeometric random variable we discussed. It's a function that maps the values of the random variable to their corresponding probabilities.

The PMF for the hypergeometric distribution is given by the formula:\[P(X=x) = \frac{{\binom{M}{x} \binom{N-M}{n-x}}}{{\binom{N}{n}}}\]
where \(\binom{M}{x}\) represents the number of ways to choose 'x' successes out of 'M' possible, \(\binom{N-M}{n-x}\) represents the ways of choosing the remaining 'n-x' failures from the 'N-M' failures in the population, and \(\binom{N}{n}\) is the total number of ways to draw 'n' samples from 'N' items. This formula helps determine the probabilities that are later used to construct the probability histogram.

Combinatorics in Probability

Combinatorics in probability refers to the use of combinatorial techniques to calculate the likelihood of various outcomes. Fundamental to these calculations are the concepts of permutations and combinations, which are methods to count the number of different ways to arrange a set of items.

For instance, in our hypergeometric example, the PMF includes terms like \(\binom{M}{x}\), which represents a combination—it's the number of ways to pick 'x' successes from 'M' without regard to order. Combinatorics is essential in calculating the probabilities of complex events where order does not matter or where there's no replacement, such as the situation presented in the hypergeometric distribution.