"At Least" Combination Problem

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Dec 30th 2008, 08:23 AM
skeptiko

"At Least" Combination Problem

I've been banging away at this... and I think I'm close, but would really appreciate someone looking over my shoulder... here goes:

- 13 students took part in a matching exercise
- the exercise asked them to match 4 questions to four answers
- 6 of the 13 students got them all right

What is the probability of achieving a result at least this good?

Here’s where I’m at so far:
- the probability of one student getting them all correct is 4! (1 in 24… 0.042)
- the probability of 6 students matching them all is 0.042 to the power of 6 (191 million to 1)
- the number of combinations of 6 out of 13 is 13! / 6! * 7! = 1,716
- so, the probability of getting 6 out of 13 correct is 191 million / 1,716 = 111,000 to 1

Right so far?

Since I want to know the probability of getting at least 6/13 do I have to go further? Do I have to calculate the probability of going 7/13, 8/13… and add all those to this one? Is there an easier way?

Many Thanks.
Dec 30th 2008, 02:13 PM
mr fantastic

Quote:

Originally Posted by skeptiko

I've been banging away at this... and I think I'm close, but would really appreciate someone looking over my shoulder... here goes:

- 13 students took part in a matching exercise
- the exercise asked them to match 4 questions to four answers
- 6 of the 13 students got them all right

What is the probability of achieving a result at least this good?

Here’s where I’m at so far:
- the probability of one student getting them all correct is 4! (1 in 24… 0.042)
- the probability of 6 students matching them all is 0.042 to the power of 6 (191 million to 1) Mr F says: This is not quite right. There's a factor of ${\color{red}^{13}C_6}$ ..... (The reason why is implied in my main post)

- the number of combinations of 6 out of 13 is 13! / 6! * 7! = 1,716
- so, the probability of getting 6 out of 13 correct is 191 million / 1,716 = 111,000 to 1

Right so far?

Since I want to know the probability of getting at least 6/13 do I have to go further? Do I have to calculate the probability of going 7/13, 8/13… and add all those to this one? Is there an easier way?

Many Thanks.

The number of ways of arranging the answers to the questions is 4! = 24. Only one of these arrangements gives a correct match to all 4 questions.

So the probability of getting a match in a single trial is 1/24.

Let X be the random variable number of students who answer all questions correctly.

X ~ Binomial(n = 13, p = 1/24).

You have to calculate $\Pr(X \geq 6)$ . The easy way to do this is to use technology. Using my TI-89 I get 0.0000069655.

Conclusion: It's highly improbable that 6 or more students would get a perfect match by chance.
Dec 30th 2008, 05:21 PM
skeptiko

Quote:

Originally Posted by mr fantastic

The number of ways of arranging the answers to the questions is 4! = 24. Only one of these arrangements gives a correct match to all 4 questions.

So the probability of getting a match in a single trial is 1/24.

Let X be the random variable number of students who answer all questions correctly.

X ~ Binomial(n = 13, p = 1/24).

You have to calculate $\Pr(X \geq 6)$ . The easy way to do this is to use technology. Using my TI-89 I get 0.0000069655.

Conclusion: It's highly improbable that 6 or more students would get a perfect match by chance.

Thx a bunch Mr. F...