Is below an example of Bayes theorem?

Question

I went to an institute for hiring with low gender diversity (1 female students out of 10 male students), and observed an application of bayes theorem. Can you please confirm if is TRUE/ my thinking is correct?

We hired 10 people, what is the probability that more than 5 girls are there, in the two below scenarios?

Scenario1 - Institute IIT A, with 10% girls

Scenario2 - Institute IIT C, with 50% girls Both IIT A & IIT C are equal in the academic standards

Bayes Theorem -

P(H|E) = (P(E|H)P(H))/(P(E))

H -> Hypothesis; E -> Evidence

We hired 10 people, what is the probability that more than 5 girls are there, for the two scenarios?

E -> Hired 10 people; H -> More than 5 girls in 10 hired

As institutes in IIT A & IIT C are equal in academic standards, I think that P(E|H) would be same for both of them. Please correct if I am wrong

Scenario 1 -> P(H) or P(more than 5 girls in 10 hired) = proportion of girls in Institute IIT A = 10%

Scenario 2 -> P(H) or P(more than 5 girls in 10 hired) = proportion of girls in Institute IIT C = 50%

As P(H) is higher for scenario 2, P(H|E) or O(more than 5 girls hired|10 hired) would be higher for scenario 2 than scenario 1.

Is this the correct application of Bayes theorem?

Answer 1

No, Bayes Theorem doesn't come into play here. Bayes Theorem typically is used for dependent events.

Here, it is reasonable to presume that the events are independent. For example, when you hire $10$ people, if there is a quota that only permits $10\%$ of the new hires from being female, then it would be impossible to hire as many as $5$ females, out of the $10$.

So, it is reasonable to conclude that each hiring is an independent event. To solve (for example) the first problem, see Binomial Distribution, specifically $\displaystyle \binom{n}{k}p^kq^{(n-k)}.$

In the specific case, you have $p = (0.1)$ and $q = (0.9)$.

Therefore, the answer is

$$\sum_{k=6}^{10} \binom{10}{k} (0.1)^k ~(0.9)^{(10-k)}.$$

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Bayes Theorem is discussed here.

As a simple example, suppose that you roll a (fair) 6-sided die, once.

Let $E_1$ be the event that the die shows 1,2,or 3.

Let $E_2$ be the event that the die shows an even number.

These two events are not independent of each other. That is, if event $E_1$ occurs, it becomes less likely than normal that event $E_2$ occurs.

More formally,

$$p(E_1) \times p(E_2|E_1) = p(E_1,E_2) \implies $$

$$p(E_2|E_1) = \frac{p(E_1,E_2)}{p(E_1)} = \frac{1/6}{3/6} = \frac{1}{3}.$$