Trigonometry Math Probability Problem

Question

Twenty people apply for seven jobs that are available at Fly-By-Night Aircraft Manufacturing Co. 

a. In how many different ways could such a group of seven people be composed?

b. Twelve of the applicants are men and the rest are women. How many of these seven-person groups have exactly three men? 

C. How many of these seven-person groups have exactly three men or exactly three women? 

d. What is the probability that a group of seven selected at random will have exactly three men or exactly three women?

Answer 1

Hi Emily!

To tackle this problem we will need to know an expression to calculate how many unique ways there are of choosing K out of N items (N>K) when the same items chosen in different orders does not count as unique solutions. The expression for this is called N choose K and is written in shorthand like a fraction inside parenthesis with N on top and K on the bottom and no fraction bar separating them. It is equal to

N!/(K!(N-K)!)

where N! is N factorial and is equal to N*(N-1)*(N-2)... (stop when you reach 1)

Let me use an elementary example to show the process.

6 choose 3

= 6!/(3!(6-3)!)

= 6!/(3!*3!)

= 6*5*4*3*2*1/((3*2(3*2))

=6*5*4/(3*2)

=6*5*4/6

=5*4

=20

In every case this sort of cancellation will occur N! between N! and K!. The procedure is to write out N! only up to and not including K or anything smaller, then write out (N-K)! in the denominator and see if you can combine some factors so the numerator and denominators share factors and you can cancel them. Finally, use a calculator to multiply.

a) 20 choose 7

= 20!/(7!(20-7)!)

= 20!/(7!*13!)

=20*19*18*17*16*15*14/(7*6*5*4*3*2)

=19*17*16*15

=77,520 unique ways of forming a group of 7 out of 20 people.

b) We begin by using the same trick to find how many ways there are of choosing 3 men out of the 12 applicants, call the result M. For each choice of the 3 men, there are a number of unique qualifying groups. They are made unique by their different choices of females for the remaining 4 openings and thus we multiply M by 8 choose 4.

c) Here you will follow the procedure in b) (except now M is 8 choose 3 and we multiply it by 12 choose 4) to find the number of unique groups with 3 women. Finally we sum our result with the result from b).

d) To find this probability, we need to know what proportion of the total groups satisfy the condition. We already calculated the total number in a) and the conditional number in c) and so the answer is simply our final result from c) divided by 77,520. I will let you work through b) and c) yourself but as a check of your work, I get Probability=0.56 for d).

As an aside:

Although the result of d) may seem high, it is a general characteristic of probabilities of this sort that the most likely outcomes are overwhelmingly likely (increasingly so as the group sizes get bigger).

Hope this helps!

Joss

Answer 2

_nC_r is the binomial coefficient of n things taken r at a time

The answer to part 1 is ₂₀C₇

Part 2 ₁₂C₃ * ₈C₄

Part 3 The answer to Part 2 + ₁₂C₄*₈C₃

Part 4 the answer to Part 3 divided by the answer to Part 1

Note: These are hypergeometric distribution questions.