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. Let us assume that all eight candidates are equally qualified for the positions. Let \(x\) represent the number of paid interns who are hired for these three positions. 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

a) The number of paid interns who got hired for these three positions follows a hypergeometric distribution. b) The probability that three paid interns are hired is 5/28. c) The probability that none of the three hired teachers was a paid intern is 1/56. d) The probability that x ≤ 1 is 31/56.

Step-by-step solution

Determine the distribution type for \(x\)

The number of paid interns who got hired for these three positions is based on a finite population with two types of individuals (paid interns and traditional student teachers). Therefore, \(x\) follows a hypergeometric distribution. Answer to part a: Hypergeometric distribution.

Probability of hiring 3 paid interns

To find the probability that three paid interns are hired, we use the hypergeometric probability formula: $$ P(X=k) = \frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}} $$ Here, \(N\) = Total candidates (interns + traditional) = 8 \(K\) = Number of paid interns = 5 \(n\) = Number of total positions = 3 \(k\) = Number of paid interns to be hired = 3 Now, we can plug these values into the formula: $$ P(X=3) = \frac{\binom{5}{3}\binom{8-5}{3-3}}{\binom{8}{3}} = \frac{\binom{5}{3}\binom{3}{0}}{\binom{8}{3}} $$ Using combination formula: \(\binom{n}{k}= \frac{n!}{k!(n-k)!}\), we can calculate these binomial coefficients. $$ P(X=3) = \frac{\frac{5!}{3!2!}\frac{3!}{0!3!}}{\frac{8!}{3!5!}} =\frac{10}{56} =\frac{5}{28} $$ Answer to part b: The probability that three paid interns are hired is \(\frac{5}{28}\).

Probability that none of the three hired was a paid intern

Now we want to find the probability that none of the hired teachers is a paid intern, which means all three hired teachers are traditional student teachers. We can apply the hypergeometric probability formula again with \(k = 0\): $$ P(X=0) = \frac{\binom{5}{0}\binom{8-5}{3-0}}{\binom{8}{3}} = \frac{\binom{5}{0}\binom{3}{3}}{\binom{8}{3}} $$ Using combination formula, calculate these binomial coefficients. $$ P(X=0) = \frac{\frac{5!}{0!5!}\frac{3!}{3!0!}}{\frac{8!}{3!5!}} =\frac{1}{56} $$ Answer to part c: The probability that none of the three hired teachers was a paid intern is \(\frac{1}{56}\).

Probability that \(x \leq 1\)

To find the probability that \(x \leq 1\), we need to calculate the probabilities for \(x = 0\) and \(x = 1\) and add them together. We already calculated the probability for \(x = 0\) in part c. Now, we'll calculate the probability for \(x = 1\): $$ P(X=1) = \frac{\binom{5}{1}\binom{8-5}{3-1}}{\binom{8}{3}} = \frac{\binom{5}{1}\binom{3}{2}}{\binom{8}{3}} $$ Using combination formula, calculate these binomial coefficients. $$ P(X=1) = \frac{\frac{5!}{1!4!}\frac{3!}{2!1!}}{\frac{8!}{3!5!}} =\frac{15}{28} $$ Now, we can add the probabilities for \(x=0\) and \(x=1\): $$ P(x \leq 1) = P(X=0) + P(X=1) = \frac{1}{56} + \frac{15}{28} = \frac{1}{56} + \frac{30}{56} = \frac{31}{56} $$ Answer to part d: The probability that \(x \leq 1\) is \(\frac{31}{56}\).

Key concepts

Probability Calculation

Probability calculation is an essential mathematical concept that helps us understand the likelihood of various outcomes in uncertain situations. In this exercise, we used the hypergeometric probability formula to calculate the chances of specific outcomes when selecting candidates for teaching positions.
When dealing with a finite population divided into two distinct groups, as in this scenario (paid interns and traditional student teachers), the hypergeometric distribution is suitable. This distribution allows us to find the probabilities without replacement, meaning once a candidate is chosen, they aren't put back in the pool for further selection.
The formula for the hypergeometric probability is:

• \(P(X=k) = \frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}\)

In the formula:

• \(N\) is the total number of individuals considered, 8 in our case.
• \(K\) is the number of individuals in the group of interest (paid interns), which is 5.
• \(n\) is the number of selections made (teaching positions), 3 in this instance.
• \(k\) is the number of individuals from the group of interest selected, varying as per the problem's requirement.

By replacing these values into the formula, we calculate the probabilities of selecting different combinations of paid interns for the positions.

Combinatorics

Combinatorics is the branch of mathematics that deals with counting, combination, and permutation. It is a crucial tool for calculating probabilities, especially for problems involving selection and arrangement, like the present exercise.
Using combinatorics, we can determine how many ways we can choose a subset of items from a larger set without consideration of the order in which they are chosen. This is done using combinations, denoted mathematically as \(\binom{n}{k}\). The combinations formula indicates how many ways \(k\) elements can be selected from a set of \(n\) elements:

• \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\), where \(!\) represents a factorial, meaning the product of all positive integers up to a specified number.

In the exercise, combinatorics was used to calculate how many ways certain numbers of paid interns could be selected out of the total 8 candidates. Each calculated combination feeds into the probability formula, giving us the necessary probabilities for hiring different numbers of paid interns.
Combinatorics makes it feasible to solve complex selection problems by breaking them down into simple counts of possible groupings.

Teaching Internships

Teaching internships are valuable experiences for aspiring educators, offering real-world classroom experiences and often remuneration compared to traditional student teaching. In Southern California, there's a noticeable trend of education candidates favoring paid internships over traditional routes.
This shift in preference is part of the exercise's context and impacts the probability calculations, as it affects the population and groups we consider, leading to a demonstration of the hypergeometric distribution in action.
By understanding both the theory (hypergeometric distribution) and the practical scenarios (real-life choice between internships or student teaching), students can appreciate how probability and combinatorics help assess such situations accurately. These calculations ensure fair and balanced selection of candidates when all are equally qualified, highlighting the vital role of internships in modern educational paths.
Internships help bridge academic knowledge with practical teaching skills, which potentially impacts the decisions educational authorities make regarding candidate selection, whether consciously or through set procedural means.