Question

Suppose you are going to form a committee of students and faculty. You have 10 total students and 11 total faculty to pick from. You want to have 9 total people on the committee. What is the probability that you select 3 students and 6 faculty, in that order? Enter your answer rounded to three decimals.

Answer 1

Answer:

0.189

Step-by-step explanation:

We are using combination to solve for this.

C(n,r) = nCr = n!/r! (n - r)!

We have 10 total students and 11 total faculty = 21 people

Step 1

Selecting 3 students out of 10 total students

= 10× 9× 8×7×6×5×4×3×2×1/ 3×2×1 × 7×6×5×4×3×2×1/

= 120

Step 2

Select 6 faculty out of 11 faculty

11C6 = 11!/6!(11 - 6)!

= 11!/6! × 5!

= 11×10× 9× 8×7×6×5×4×3×2×1/6×5×4×3×2×1 × 5×4×3×2×1/

= 462

Step 3

Selecting 9 people out of 21 people to form a committee

21C9 = 21!/9!(21 - 9)!

= 21!/9! × 12!

= 293,930

Step 4

The fourth and final step of having a committee of 9 people =

(120 × 462)/293,930

= 55440/293,930

= 0.1886163372

Rounded to 3 decimal places = 0.189

Therefore, the probability that you select 3 students and 6 faculty = 0.189

Answer 2

The probability that you select 3 students and 6 faculty, in that order is 0.189

The number (n) of ways of selecting 3 students from 10, and 6 faculty from 11 is:

$\mathbf{n = ^{10}C_3 \times ^{11}C_6}$

Using a calculator, we have:

$\mathbf{n = 120 \times 462}$

$\mathbf{n = 55440}$

The total number of people selected is 9 (i.e 3 students and 6 faculties), and the total number of people available is 21 (i.e 10 students and 11 faculties)

So, the total ways of selection is:

$\mathbf{Total = ^{21}C_9}$

Using a calculator, we have:

$\mathbf{Total = 293930}$

The probability is then calculated as:

$\mathbf{p = \frac{n}{Total}}$

$\mathbf{p = \frac{55440}{293930}}$

$\mathbf{p = 0.189}$

Hence, the probability that you select 3 students and 6 faculty, in that order is 0.189

Read more about probabilities at:

https://brainly.com/question/2264282