In a certain liberal arts college with about 10,000 students, 40% are males. if two students from this college are selected at random, what is the probability that they are of the same gender?

Treat this almost as 2 separate problems.
P(male) = 0.40. P(two males) = (0.40)(0.40) = 0.1600.
P(female)= 0.60. P(two females) = (0.60)^2 = 0.3600

Now add these two probabilities together: P(students are of the same gender) = 0.52

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joaobezerra joaobezerra · Answer 2 · 2021-10-21T12:26:23+02:00

Using the binomial distribution, it is found that there is a 0.52 = 52% probability that they are of the same gender.

For each student selected, there are only two possible outcomes. Either it is a male student, or a female student. The gender of a student is independent of any other student, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, with p probability.

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

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

In this problem:

40% of the students are male, thus [tex]p = 0.4[/tex]
Sample of 2 students, thus [tex]n = 2[/tex].

Same gender is either two females, given by P(X = 0), or two males, given by P(X = 2), thus:

[tex]p = P(X = 0) + P(X = 2)[/tex]

In which:

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

[tex]P(X = 0) = C_{2,0}.(0.4)^{0}.(0.6)^{2} = 0.36[/tex]

[tex]P(X = 2) = C_{2,2}.(0.4)^{2}.(0.6)^{0} = 0.16[/tex]

Then

[tex]p = P(X = 0) + P(X = 2) = 0.36 + 0.16 = 0.52[/tex]

0.52 = 52% probability that they are of the same gender.

A similar problem is given at https://brainly.com/question/24863377