Hello and thanks for looking at my questions.

I'm sure this is a very simple question but I just cant seem to understand it

A committee of three is selected from a pool of five individuals : two females (A and B) and three males (C,D,E)

If the committee is selected at random, what is the probability that contains both females?

(i figured for this one it would be P=(2/5)*(1/4) because the first one selected has a 2/5 chance of being a women, and the second one has a 1/4 chance but I was told that was not correct)


If instead, the committee is selected by randomly selecting one of the two females, one of the three males, and then one of the remaining three individuals, what is the probability it contains both females


I would suggest that you just count the instances for which the experiment yields the outcome that you are looking for and devide them by the total number of possible outcomes.

For f:=female and m:=male, we get:

(f,f,m) with probability $\frac{2}{5}\cdot\frac{1}{4}\cdot\frac{3}{3}$,

(f,m,f) with probability $\frac{2}{5}\cdot\frac{3}{4}\cdot\frac{1}{3}$, and

(m,f,f) with $\frac{3}{5}\cdot\frac{2}{4}\cdot\frac{1}{3}$.

Hence, adding all those probabilities up, we will get the solution $0.3$.

Using the hypergeometric distribution will yield the same result.

The question is similar to:

"What is the probability of having all the aces (4 in total) if you get dealt five cards in a row".

The second question just reduces the total number of possible outcomes in the denominator.

By knowing, that one female and one male is already drawn the, remaining "pool" of possible outcomes comes from looking at what the problabiltiy of drawing the remaining female is. I.e., we look at

(f,m,f),(f,f,m) and (m,f,f)

and will have all possible outcomes as before, but, it is not possible to have (m,m,m).


The probability that a randomly selected committee of three includes both females must equal the probability that the two individuals not selected for the committee are both male. The latter can be calculated as $\frac{3}{5}\cdot\frac{2}{4} = 0.3$.

For the second question, after selecting one female and one male there remain three unselected individuals, one female and two males. To obtain a committee with two females, the female must be selected from those three, which has probability $\frac{1}{3}$.


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