Not related to a specific exam question, but i figure someone here will know the answer.

You have 31 people. 18 are in group 1, and 13 are in group 2.

25 are chosen randomly. What is the probably that all 18 people in group 1 are chosen?

Not related to a specific exam question, but i figure someone here will know the answer.

You have 31 people. 18 are in group 1, and 13 are in group 2.

25 are chosen randomly. What is the probably that all 18 people in group 1 are chosen?

I think i have an answer but would like to get feedback.

There are 736,281 different combinations for choosing 25 people out of a group of 31.

We want 18 choose 18 of group 1 (there is only 1 way) and 13 choose 7 from group 2 (=1716).

1716/736,281 = .00233. So if you did this drawing 1000 times you would expect it to happen about twice.

The odds of it happening are 1716/(736281-1716) = one in 428

The number of ways that you can choose 25 from 31 – 31C25 – is 736,821.

The number of ways that you can choose all 18 from group 1 and 7 from group 2 is: 18C18 × 13C7 = 1,716

Thus, the probability of getting all 18 people in group 1 when you choose 25 people from the larger group is 1,716 ÷ 736,821 = 0.002331, or 0.2331%.

Not very likely.

Why do you ask?

so the probability is .002331 like i came up with? Just wanted a second opinion.

THis is a real life problem. A lottery for kids at a school was done by computer. They were in two groups but the computer was said to have chosen randomly without regard for which group they were in. But all 18 kids in the first group were chosen. The person that asked me thought it was fishy and is going to question whether the computer was choosing randomly.

Fun!

Good luck on your exam.

Thanks, Cheeks, but I don’t need any luck on my exam: if you check my status you’ll see that I’m a charterholder.

However, best of luck on yours!

I saw that after i posted, sorry about that. But Congrats!

I’m jealous–taking level II… again.