Question

In southern California, a growing number of persons pursuing a teaching credential are choosing paid internships over traditional student teaching programs. A group of eight candidates for three teaching positions consisted of five paid interns and three traditional student teachers. Assume that all eight candidates are equally qualified for the three positions, and that \(x\) represents the number of paid interns who are hired. a. Does \(x\) have a binomial distribution or a hypergeometric distribution? Support your answer. b. Find the probability that three paid interns are hired for these positions. c. What is the probability that none of the three hired was a paid intern? d. Find \(P(x \leq 1)\).

Short answer

Based on the problem, we identified the distribution as hypergeometric since selections are made without replacement from a finite pool of candidates. We then calculated the probabilities for various cases: a) Distribution: Hypergeometric Distribution b) Probability of hiring three paid interns: $\frac{5}{28}$ c) Probability of hiring no paid interns: $\frac{1}{56}$ d) Probability of hiring one or fewer paid interns: $\frac{2}{7}$

Step-by-step solution

Part a: Identifying the distribution

In this case, we have a finite population (8 candidates) with two types: paid interns (5) and traditional student teachers (3). We are interested in the probability of hiring a certain number of paid interns when selecting a sample (3 teaching positions). Since the selections are made without replacement, the distribution is a hypergeometric distribution, not a binomial distribution.

Part b: Probability of hiring three paid interns

To find the probability of hiring three paid interns, we can use the hypergeometric probability formula: $$ P(x=k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}} $$ Where \(N\) is the total number of candidates (8), \(n\) is the sample size (3), \(K\) is the total number of paid interns (5), and \(k\) is the desired number of paid interns (3). Calculating the probability: $$ P(x=3) = \frac{\binom{5}{3} \binom{3}{0}}{\binom{8}{3}} = \frac{10 \cdot 1}{56} = \frac{5}{28} $$ So the probability of hiring three paid interns is \(\frac{5}{28}\).

Part c: Probability of hiring no paid interns

Using the same process as above for part b, now we want to find the probability of hiring no paid interns (\(k = 0\)): $$ P(x=0) = \frac{\binom{5}{0} \binom{3}{3}}{\binom{8}{3}} = \frac{1 \cdot 1}{56} = \frac{1}{56} $$ The probability of hiring no paid interns is \(\frac{1}{56}\).

Part d: Probability of hiring one or fewer paid interns (\(x \leq 1\))

To find the probability of hiring one or fewer paid interns (\(x \leq 1\)), we need to calculate the probability of hiring 0 paid interns and 1 paid intern, and then sum the probabilities: First, we already calculated the probability of hiring 0 paid interns (\(P(x=0) = \frac{1}{56}\)). Now let's calculate the probability of hiring 1 paid intern (\(k = 1\)): $$ P(x=1) = \frac{\binom{5}{1} \binom{3}{2}}{\binom{8}{3}} = \frac{5 \cdot 3}{56} = \frac{15}{56} $$ Now we sum the probabilities for \(x=0\) and \(x=1\): $$ P(x \leq 1) = P(x=0) + P(x=1) = \frac{1}{56} + \frac{15}{56} = \frac{16}{56} = \frac{2}{7} $$ The probability of hiring one or fewer paid interns is \(\frac{2}{7}\).

Key concepts

probability

Probability is a fundamental concept in mathematics and everyday life, expressing the likelihood of a specific event occurring within a set of possible outcomes. To calculate the probability of an event, one needs to know the number of successful outcomes over the total number of possible outcomes. This calculation can often be symbolized as P(event) = number of favorable outcomes / number of total outcomes.

For instance, in our example of teaching positions, when we wished to determine the probability of hiring three paid interns, we looked at how many ways we could select three interns out of five (favorable outcomes) compared to the total number of ways to choose three candidates from eight (total outcomes). Understanding probability is essential to predict outcomes in real-life scenarios and solve mathematical problems involving uncertainty or risk.

combinatorics

Combinatorics is the branch of mathematics focused on counting, arranging, and grouping objects. It helps in solving problems that ask, 'How many different ways can a certain event occur?' by using tools like permutations and combinations. The combinations are particularly relevant to our example, which is represented by the notation \( \binom{n}{r} \), also known as 'n choose r.' It refers to the number of ways to select r items from a larger set of n items without considering the order.

In the exercise, the use of combinations \(\binom{K}{k} \binom{N-K}{n-k}\) to determine possible selections of interns and student teachers is a classic example of combinatorial logic. These calculations are crucial in predicting various scenarios and outcomes, especially when order does not matter.

probability distribution theory

Probability distribution theory is at the heart of understanding how probabilities are assigned to different events, particularly in situations where we deal with a large number of possible outcomes. It involves defining a probability distribution, which is a function that assigns probabilities to each possible outcome of a random experiment.

The hypergeometric distribution, shown in the educational example, is a discrete probability distribution that describes scenarios where one samples without replacement from a finite population comprised of two distinct classes. The formula \( P(x=k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}} \) is a textbook case of such a distribution. It’s used for situations where each pick affects the subsequent one—like choosing three qualified candidates from a group of eight without putting any back into the pool. This concept is essential not just in theoretical mathematics but also in various practical fields, including quality control, ecological studies, and business decisions, where you have to understand the likelihood of certain outcomes.