Question

There are 22 students, from which 6 people will be selected to be on a panel. If Allen and Joe are 2 of the 22 students, then what is the probability that neither will be selected to be on the panel?
Round your answer to three decimal places.

Answer 1

Answer:  0.519

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Explanation:

Let's start off counting the number of ways to form the panel that doesn't have Allen or Joe on it. In other words, both are not on the panel.

There are 22 students, but we want to exclude 2 of them (Allen and Joe) from the panel. So in reality, we only have 20 students to pick from. There are 6 slots to fill.

The first slot has 20 choices

The second slot has 19 choices

The third has 18 choices

And so on. We count our way down until all six slots are filled. We'll have this sequence of six values: 20, 19, 18, 17, 16, 15

They multiply out to 20*19*18*17*16*15 = 27,907,200

This is the number of permutations where we choose 6 from a pool of 20.

With permutations, order matters; however, order doesn't matter on this panel. All we care about is the group overall.

So we must divide by 6! = 6*5*4*3*2*1 = 720 and we get

(27,907,200)/(720) = 38,760

There are 38,760 different combinations, so there are 38,760 different panels where neither Allen nor Joe are selected.

We'll use this later later, so let A = 38,760

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Now onto the next part. We need to count the total number of possible panels, and this time we're including Allen and Joe.

We have our original total of 22 students. There are still 6 seats to fill.

We could follow the same steps as the prior section, but I'll use a different route.

Let n = 22 and r = 6. Plug these values into the nCr combination formula below

$_n C_r = \frac{n!}{r!*(n-r)!}\\\\_{22} C_6 = \frac{22!}{6!*(22-6)!}\\\\_{22} C_6 = \frac{22!}{6!*16!}\\\\_{22} C_6 = \frac{22*21*20*19*18*17*16!}{6!*16!}\\\\_{22} C_6 = \frac{22*21*20*19*18*17}{6!}\\\\_{22} C_6 = \frac{22*21*20*19*18*17}{6*5*4*3*2*1}\\\\_{22} C_6 = \frac{53721360}{720}\\\\_{22} C_6 = 74613$

There are exactly 74,613 ways to form a committee of 6 people from a pool of 22 total.

Let B = 74,613

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To find the probability of neither Allen or Joe are on the committee, we divide the values of A and B found earlier

A/B = (38,760)/(74,613) = 0.5194805

This value is approximate.

Rounding to three decimal places, or rounding to the nearest thousandth, leads to the final answer 0.519

There's roughly a 51.9% chance of forming a panel where neither Allen nor Joe are members.