# Combinatorics and leaving a choice out

Three actors are to be chosen out of five — Jack, Steve, Elad, Suzy, and Ali. What is the probability that Jack and Steve would be chosen, but Suzy would be left out?

The answer given was: Total Number of actors = $5$; Since Jack and Steve need to be in the selection and Suzy is to be left out, only one selection matters. Number of actors apart from Jack, Steve, and Suzy = $2$; Probability of choosing 3 actors including Jack and Steve, but not Suzy = $$\frac{C(2,1)}{C(3,5)} = \frac{1}{5}$$

I do not understand the answer. What do they mean by only one selection matters? It looks like they are choosing $1$ person from $2$ combinations? Why? Can anyone please explain this.

Thanks

• I guess C(2, 1)/C(5, 3) in your question? Oct 6 '17 at 10:37
• By "only one selection matters" they're making a meaningless statement but I think they intended to mean that there is only one selection of three from 5 which satisfies the criteria which you are asked to enumerate. Oct 6 '17 at 11:42

I must admit that I don't know either what they mean with "only one selection matters".

I would rather say that two selections matter: J|St|E and J|St|A.

There are $\binom53=10$ possible selections and this will be the denominator.

There are $\binom21=2$ choices for the "-" in J|St|- and this will be the numerator.

That gives: $$P(\text{J|St|E or J|St|A})=\frac2{10}=\frac15$$

• Thank you very much for all your explanations. I understand the denominator which is essentially C(5,3). It was the numerator that has me stumped. I had assumed initially it was C(5,2) x C(2,1) as you are choosing 2 out of the five first, and then multiplying by 1 out of the 2 remaining. Why is the initial calculation of the first 2 not in the numerator. First you select 2 out of the 5 do you not? Oct 7 '17 at 11:37
• Because we are dealing with choosing "including Jack and Steve and passing Suzy". This restricts the number of choices in such a way that only $2$ options stay open. The numerator corresponds with the number of options where Jack and Steven are chosen and Suzy is passed. Oct 7 '17 at 12:17

You could easily count this out yourself, but for the sake of practice, let us solve it this way:

The said probability is the number of different ways for Jack and Steve but not Suzy to be chosen divided by the number of ways to choose three actors from five.

Since our event space has predetermined fates for three of the actors, we’re really just looking at the fates of the remaining two. The event space is the number of ways to choose one actor (since there is only one spot) from the two actors whose fates are undecided.

Of course, the total number of ways to choose three actors from five is $\binom{5}{3}=C(5,3)$.

$$\frac{C(2,1)}{C(5,3)}$$
• @A.Lucas Yes, it sounds like you understand $\ddot\smile$ Oct 8 '17 at 18:04