For this problem, assume 7 males audition, one of them being Miles, 5 females audition, one of them being Margaret, and 6 children audition.

Question

The casting director has 3 male roles available, 1 female role available, and 2 child roles available.

 

A)How many different ways can these roles be filled if exactly one of Miles and Margaret gets a part? 

B)What is the probability (if the roles are filled at random) of both Miles and Margaret getting a part? 

Answer 1

Hi Mauro

 

The basics: if you are picking, for example, the male part, then you have to pick 3 out of the 7 men available. The formula for the number of ways to do this is "7 choose 3" or ₇C₃. It is equal to 7! / ((3!)(7-3!))  or 7! / (3!4!), which comes out to 35.

 

This is pretty hard. For part A, if Miles is picked and Margaret is not, then that leaves 4 choices for the female role (Margaret is excluded) and 6 children for the two roles. So there are ₄C₁·₆C₂ choices. On the other hand if Margaret is picked and Miles is not, then that leaves 6 men for the three roles, and again the 6 children for the two roles. So there are ₆C₃·₆C₂ choices. To get the total you just add them:

 

₄C₁·₆C₂ + ₆C₃·₆C₂ = 4·15 + 20·15 = 360.

 

For part B, first think of the probability of Miles getting picked. This is equal to:

 

(# of male casts including Miles) / (total # of possible male casts)

 

If Miles is picked, there are two other slots to fill from 6 remaining choices. So there are ₆C₂ possible male casts including Miles. The total number of possible male casts is ₇C₃. So Miles's chance of getting picked is ₆C₂ / ₇C₃.

 

As for Margaret, if she's picked there are zero other slots to fill so there are ₄C₀ possible female casts including her. So her chance of getting picked is ₄C₀ / ₅C₁, or 1/5 (duh. But at least the formula works).

 

For the chance of both getting picked, just multiply:

 

(₆C₂ / ₇C₃)·(₄C₀ / ₅C₁)

 

This works out to (15/35)·(1/5), or 3/35.