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 $\displaystyle {\color{red}^{13}C_6}$ ..... (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.