# Stats 213 Problem 2?

An instructor gives his class a set of 18 problems with the information that the next quiz will consist of a random selection of 9 of them. If a student has figured out how to do 16 of the problems, what is the probability the he or she will answer correctly

(a) all 9 problems?

(b) at least 8 problems?

### 6 Answers

- Anonymous2 months ago
So if the student has figured out how to do 16 out of 18 problems, the probability that he will answer 9 of them correct is 4/17, and the probability that he will answer at least 8 of them correct is 13/17.

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- AlanLv 72 months ago
You already have two correct answers, but

for more information

This can be treated as an instance of the

Hypergeometric distribution

See Wikipedia Article

https://en.wikipedia.org/wiki/Hypergeometric_distr...

= ( ( K over k ) ( (N-K) over (n-k) ) /

( N over n )

over mean combination

where is N is total number of original items = 18

K is the number of original which meet passing

requirement

K = 9

n is the number of item to be chosen from the

group of N = 9

k is how of n which meet passing requirement

(k =9 in one of your questions and k =8 in part of your

other question.)

This formula gives the two correct answers that you already have.

The formula in your 1st answer gives the same

answer as another answer you have if you multiplied it out.

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- 冷眼旁觀Lv 72 months ago
(a)

F: a certain problem is figured out (there are 16 F's)

N: a certain problem is not figured out (there are 2 N's)

Total number of ways to randomly choose 9 problems for 18

= ₁₈C₉

= 48620

To answer all 9 problems correctly, choose 9F (from 16) in the quiz.

Number of ways

= ₁₆C₉

= 11440

= P(all 9 problems correct)

= 11440/48620

= 4/17

====

(b)

To answer 8 problems correctly, choose 8F from (16) and 1N (from 2) in the quiz.

= ₁₆C₈ × ₂C₁

= 25740

P(8 problems correct)

= 25740/48620

= 9/17

P(at least 8 problems correct)

= P(all 9 problems correct) + P(8 problesm correct)

= (4/17) + (9/17)

= 13/17

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- 2 months ago
(16/18) * (15/17) * (14/16) * (13/15) * (12/14) * (11/13) * (10/12) * (9/11) * (8/10) =>

(1/18) * (1/17) * (16/16) * (15/15) * (14/14) * (13/13) * (12/12) * (11/11) * (10/10) * 9 * 10 =>

(9/18) * (10/17) * 1 * 1 * 1 * 1 * 1 * 1 * 1 =>

(1/2) * (10/17) =>

5/17

You have a 5/17 chance of getting all 9 correct

(16/18) * (15/17) * ... * (9/11) * (2/10) =>

(16 * 15 * 14 * 13 * 12 * 11 * 10 * 9 * 2) / (18 * 17 * 16 * 15 * ... * 10) =>

(9 * 2) / (18 * 17) =>

1/17

You have a 1/17 chance of getting 8 correct

1/17 + 5/17 = 6/17

You have a 6/17 chance of getting at least 8 correct

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- Lv 52 months agoReport
That is incorrect.

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- PopeLv 72 months ago
I am assuming that the student cannot possibly solve either of the remaining two problems.

P(all 9 correct) = C(16, 9) / C(18, 9)

P(8 or 9 correct) = [C(16, 8)C(2, 1) + C(16, 9)] / C(18, 9)

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