Yahoo Answers: Answers and Comments for Stats 213 Problem 2? [Mathematics]
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From max
enNZ
Mon, 17 Feb 2020 03:25:18 +0000
3
Yahoo Answers: Answers and Comments for Stats 213 Problem 2? [Mathematics]
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From 冷眼旁觀: (a)
F: a certain problem is figured out (there...
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Mon, 17 Feb 2020 04:36:07 +0000
(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

From Anonymous: So if the student has figured out how to do 16...
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Tue, 18 Feb 2020 22:31:26 +0000
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.

From max: What? I dont get it plese ezplain
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Mon, 17 Feb 2020 03:48:42 +0000
What? I dont get it plese ezplain

From Alan: You already have two correct answers, but
for...
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Tue, 18 Feb 2020 10:23:07 +0000
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_distribution
= ( ( K over k ) ( (NK) over (nk) ) /
( 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.

From Pope: I am assuming that the student cannot possibly...
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Mon, 17 Feb 2020 03:49:35 +0000
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)

From Captain Matticus, LandPiratesInc: (16/18) * (15/17) * (14/16) * (13/15) * (12/14...
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Mon, 17 Feb 2020 03:49:54 +0000
(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