At Ersatz University, the graduating class of 480 includes 96 guest students from Latvia. A sample of 10 students is selected at random to attend a dinner with the Board of Governors. Use the binomial model to obtain the approximate hypergeometric probability that the sample contains at least three Latvian students.
A).3222
B).1209
C).8791
D).6778

Answer:

A).3222

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they are from Latvia, or they are not. We are choosing students without replacement. However, since the sample size is large, we can use the binomial distribution to approximate the hypergeometric distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

480 students, of which 96 are from Latvia:

This means that [tex]p = \frac{96}{480} = 0.2[/tex]

Sample of 10 students:

This means that [tex]n = 10[/tex]

Probability that the sample contains at least three Latvian students.

Either there are less than three students from Latvia, or there are at least three. The sum of the probabilities of these events is 1. So

[tex]P(X < 3) + P(X \geq 3) = 1[/tex]

We want [tex]P(X \geq 3) = 1 - P(X < 3)[/tex], in which:

[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]

So

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{10,0}.(0.2)^{0}.(0.8)^{10} = 0.1074[/tex]

[tex]P(X = 1) = C_{10,1}.(0.2)^{1}.(0.8)^{9} = 0.2684[/tex]

[tex]P(X = 2) = C_{10,2}.(0.2)^{2}.(0.8)^{8} = 0.3020[/tex]

[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.1074 + 0.2648 + 0.3020 = 0.6742[/tex]

So

[tex]P(X \geq 3) = 1 - P(X < 3) = 1 - 0.6742 = 0.3258[/tex]

The closest option is A.

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