A law firm has six senior and fourteen junior partners. A committee of 3 partners is selected at random to represent the firm at a conference. What is the probability that at least one of the junior partners is on the committee?

number of possible committees

= C(20,3) = 1140

number with NO junior partners = C(6,3) = 20
so the number with at least one junior = 1140-20 = 1120

prob of at least one junior = 1120/1140
= 56/57 = appr .98

Ah, the joys of probability! Let's see. To find the probability that at least one junior partner is on the committee, we need to calculate the opposite probability and then subtract it from one.

The total number of ways to select a committee of 3 partners from a group of 20 partners is 20 choose 3, denoted as C(20, 3). This is equal to (20!)/(3!(20-3)!), which simplifies to 1140.

Now, let's calculate the number of ways to select a committee with only senior partners. Since there are 6 senior partners and we need to choose 3, we have 6 choose 3, which is (6!)/(3!(6-3)!), and that is equal to 20.

Finally, the probability that there are only senior partners on the committee is 20 divided by 1140, which simplifies to 1/57.

Since we want the opposite probability, the probability that at least one junior partner is on the committee is 1 - 1/57, which is approximately 0.982. So, there's about a 98.2% chance that at least one junior partner will attend the conference.

To calculate the probability that at least one of the junior partners is on the committee, we need to calculate the probability of the complementary event (i.e., none of the junior partners is on the committee) and subtract it from 1.

First, let's calculate the total number of ways to select a committee of 3 partners from a group of 20 partners (6 senior partners + 14 junior partners). This can be done using the combination formula:

C(20, 3) = (20!)/(3!(20-3)!) = (20!)/(3!17!) = 1140.

Now, let's calculate the number of ways to select a committee of 3 partners from just the senior partners (i.e., no junior partners). This can be done using the combination formula again:

C(6, 3) = (6!)/(3!(6-3)!) = (6!)/(3!3!) = 20.

Therefore, the number of ways to select a committee where none of the junior partners is on the committee is 20.

Finally, we can calculate the probability of this complementary event:

P(complementary event) = number of ways to select committee with no junior partners / total number of ways to select a committee.
P(complementary event) = 20/1140.

Now, to find the probability that at least one of the junior partners is on the committee, we subtract the probability of the complementary event from 1:

P(at least one junior partner) = 1 - P(complementary event) = 1 - 20/1140.

Calculating this, we find:

P(at least one junior partner) ≈ 0.9825.

Therefore, the probability that at least one of the junior partners is on the committee is approximately 0.9825 or 98.25%.

To find the probability that at least one junior partner is on the committee, we need to calculate the probability of the complement event: the probability that none of the junior partners are on the committee.

First, let's determine the total number of ways to choose a committee of 3 partners out of the total of 20 partners (6 seniors + 14 juniors). This is a combination problem, so we can calculate it using the formula:

C(20, 3) = 20! / (3! * (20 - 3)!) = 20! / (3! * 17!) = (20 * 19 * 18) / (3 * 2 * 1) = 1140

Now, let's calculate the number of ways to choose a committee of 3 partners, where all of them are senior partners. The number of ways to choose 3 senior partners out of the total of 6 seniors is given by:

C(6, 3) = 6! / (3! * (6 - 3)!) = 6! / (3! * 3!) = (6 * 5 * 4) / (3 * 2 * 1) = 20

So, there are 20 ways to select a committee with all senior partners. Since we want to find the probability that none of the junior partners are on the committee, we need to subtract these 20 cases from the total number of possible committees.

Therefore, the number of committees with at least one junior partner is: 1140 - 20 = 1120.

Finally, we can calculate the probability by dividing the number of committees with at least one junior partner by the total number of possible committees:

P(at least one junior partner) = 1120 / 1140 = 28 / 57 ≈ 0.4912 or 49.12% (rounded to two decimal places).

Therefore, the probability that at least one of the junior partners is on the committee is approximately 49.12%.