In a mathematics class of 20 students, 5 are bilingual. If theclass is randomly divided into 4 project teams,
a) what is the probability that a team has fewer than 2 bilingualstudents?
b) what is the expected number of bilingual students on a team?

In this problem, we are dealing with a class of 20 students, 5 of whom are bilingual, and the class is randomly divided into 4 project teams. Let's break down the two parts of the problem.

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### Part a) Probability that a team has fewer than 2 bilingual students

We first calculate the probability that a randomly selected team has fewer than 2 bilingual students. Let’s assume that each team has the same size. Since there are 4 teams and 20 students, each team will have \( \frac{20}{4} = 5 \) students.

Let’s define the following:
- \( X \) is a random variable representing the number of bilingual students on a randomly chosen team.
- The total number of bilingual students in the class is 5.

We are tasked with calculating the probability that \( X < 2 \), i.e., that a team has fewer than 2 bilingual students.

This problem follows a **hypergeometric distribution** because we are choosing teams without replacement from a group where some students are bilingual and some are not.

- Population size \( N = 20 \)
- Number of bilingual students \( K = 5 \)
- Team size \( n = 5 \)
- Success (bilingual student) is \( X \).

The probability mass function of the hypergeometric distribution is:

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

We want to compute \( P(X = 0) \) and \( P(X = 1) \), and sum them for the final probability.

#### Case 1: \( P(X = 0) \) (no bilingual students)
\[
P(X = 0) = \frac{\binom{5}{0} \binom{15}{5}}{\binom{20}{5}} = \frac{1 \times 3003}{15504} \approx 0.1937
\]

#### Case 2: \( P(X = 1) \) (one bilingual student)
\[
P(X = 1) = \frac{\binom{5}{1} \binom{15}{4}}{\binom{20}{5}} = \frac{5 \times 1365}{15504} \approx 0.4401
\]

So, the probability that a team has fewer than 2 bilingual students is:

\[
P(X < 2) = P(X = 0) + P(X = 1) = 0.1937 + 0.4401 = 0.6338
\]

Thus, the probability that a randomly chosen team has fewer than 2 bilingual students is approximately **0.6338** (or 63.38%).

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### Part b) Expected number of bilingual students on a team

The expected number of bilingual students on a team can be calculated using the formula for the expected value of the hypergeometric distribution:

\[
E(X) = \frac{nK}{N}
\]

where:
- \( n = 5 \) (team size),
- \( K = 5 \) (total bilingual students),
- \( N = 20 \) (total students).

Substitute the values into the formula:

\[
E(X) = \frac{5 \times 5}{20} = \frac{25}{20} = 1.25
\]

So, the expected number of bilingual students on a randomly chosen team is **1.25**.

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### Summary:
- a) The probability that a randomly chosen team has fewer than 2 bilingual students is approximately **0.6338**.
- b) The expected number of bilingual students on a randomly chosen team is **1.25**.

Do you want more details or have any questions?

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Here are 5 related questions for you to explore:
1. What is the variance of the number of bilingual students on a team?
2. How would the probability change if the class had 30 students, but still only 5 bilingual?
3. What is the probability that a team has exactly 2 bilingual students?
4. How can the problem be modified if teams are not of equal size?
5. Can the binomial distribution approximate the probability in part a)?

**Tip:** The hypergeometric distribution models probabilities when sampling without replacement. This is different from the binomial distribution, where sampling is with replacement.

In a mathematics class of 20 students, 5 are bilingual. If the class is randomly divided into 4 project teams, a) what is the probability that a team has fewer than 2 bilingual students? b) what is the expected number of bilingual students on a team?

Explore the probability that a randomly selected team of students from a class of 20 has fewer than 2 bilingual students, as well as the expected number of bilingual students on a team. This solution involves hypergeometric distribution and expected value calculations, ideal for students in Grades 11-12.

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