Help solving it.

Question

A jury pool has 21 people that are married and 23 people that are not married, from which 12 jurors will be selected. Assuming that each person is equally likely to be chosen and that the jury is selected at random without replacement, find the probability that the jury consists of the following. (Give answer as a fraction or a decimal out to at least 4 places. If your answer is very small use scientific notation out to 4 decimal places for example 3.3421E-6.)

(a) all married people.

  

(b) all not married people.

  

(c) 8 married and 4 that are not married.

  

(d) 6 married and 6 that are not married.

  

Answer 1

Assuming that order doesn't matter (ie. if we swap juror 1 and juror 2 we would have the same jury), then this is simply a combination problem. When finding probabilities, the numerator consists of the number of juries that fit the given condition and the denominator is the total possible number of juries. _nC_r will denote the number of total ways to choose a subset of r people from a group of n people.

(a) All married people: ₂₁C₁₂ / ₄₄C₁₂ (choose 12 people from the 21 married couples and divide that by choose 12 people from the total jury pool of 44)

(b) All unmarried people: ₂₃C₁₂ / ₄₄C₁₂

(c) and (d) are slightly more difficult. Think about it like this, you select 8 married people ₂₁C₈ (this gives us the number of unique groups of 8 married people out of 21 total married people). Now for each of those possible selections we can pair it with a unique group of 4 unmarried people selected from the 23 (₂₃C₄) unmarried people in the jury pool. Hence we multiply the two results

(c) (₂₁C₈)•(₂₃C₄) / ₄₄C₁₂

Similarly, we can find the jury with 6 married and 6 unmarried

(d) (₂₁C₆)•(₂₃C₆) / ₄₄C₁₂

I'll leave the actual computation to you.

Answer 2

Let us begin with a reminder of a combination.

The answer to the question

"How many different ways can we choose k objects out of n possible objects where order does not matter?"

is "n choose k" which can be represented by many different notations such as

C(n, k)

_nC_k

Cⁿ_k

For the given problem...

"A jury pool has 21 people that are married and 23 people that are not married"

so there is a total of 21 + 23 = 44 people

"from which 12 jurors will be selected"

How many different possible jury selections are there?

That is, how many different ways are there of selecting a group of 12 people out of 44 possible people when order does not matter?

That is precisely "44 choose 12" or C(44, 12).

C(44, 12) = 21,090,682,613

(a) Find the probability that the jury consists of all married people.

For the purposes of this part, we can only select the jurors out of the 21 married people.

How many different ways are there of selecting a group of 12 people out of 21 possible people when order does not matter?

That is precisely "21 choose 12" or C(21, 12).

C(21, 12) = 293,930

The desired probability can be found by simply dividing the number of outcomes with satisfy the given condition by the total number of outcomes. It is thus...

C(21, 12) / C(44, 12) = 293,930 / 21,090,682,613

(b) Find the probability that the jury consists of all not married people.

There are 23 not married people.

The logic is identical to that of part (a).

(c) Find the probability that the jury consists of 8 married people and 4 that are not married.

 

For the 8 married people

How many different ways are there of selecting a group of 8 people out of 21 possible people when order does not matter?

That is precisely "21 choose 8" or C(21, 8).

C(21, 8) = 203,490

For the 4 people who are not married

How many different ways are there of selecting a group of 4 people out of 23 possible people when order does not matter?

That is precisely "23 choose 4" or C(23, 4).

C(23, 4) = 8,855

Now, for each of the 203,490 groups of 8 married people, there are 8,855 possible groups of 4 not married people.

This gives us a total of 203,490 x 8,855 = 1,801,903,950 different groups of 8 married people and 4 not married people.

The desired probability can be found by simply dividing the number of outcomes with satisfy the given condition by the total number of outcomes. It is thus...

C(21, 8) x C(23, 4) / C(44, 12) = 1,801,903,950 / 21,090,682,613

(d) Find the probability that the jury consists of 6 married people and 6 that are not married.

The logic is identical to that of part (c).

Answer 3

This is a Combinatorics problem using Combinations, since the order in the jury doesn't matter.

The number of all possible outcomes is given by the ways to select 12 jurors out of 21 + 23 = 44 people i.e. by the combination coefficient: 44C12

You should calculate the combination coefficients with a calculator.

(a)

Ways to select 12 jurors out of 21 married: 21C12

probability = favorable / all outcomes = 21C12 / 44C12

(b)

Ways to select 12 jurors out of 23 not married: 23C12

probability = favorable / all outcomes = 23C12 / 44C12

(c)

Ways to select 8 out of 21 married: 21C8

Ways to select 4 out of 23 not married: 23C4

According to the counting principle, multiply the ways to select 8 out of 21 and 4 out of 23: 21C8 * 23C4

probability = favorable / all outcomes = 21C8 * 23C4 / 44C12

(d)

This is analogous to (c). I'll leave it to you as an exercise.