DrPizza
Administrator Elite Member Goat Whisperer
How sure are you that the students were completely randomly selected for which instructor they got? Perhaps the lazy, though smart enough to do so students requested the easy instructor.
Unless we know that the students were completely randomly chosen, we can't compute a probability. If students were assigned in order of registering/applying for the class, then we could perhaps surmise that there would be some clustering - many of the eager, organized students applying earlier than most people.
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To explain the probability, let's say I have a weighted coin that comes up heads 60% (3/5) of the time. The probability of getting HHTTT is .6 * .6 * .4 * .4 * .4
The probability of HTTTH is .6 * .4 * .4 * .4 * .6, which is the same value (just multiplying in a different order.) The probability of getting exactly 2 heads out of 5 flips, is the same for any possible order. There are exactly 10 different orders where 2 of the flips are heads, and 3 of the flips are tails. This can be calculated as a combination 5C2 So, the probability of getting heads exactly 2 times in 5 flips is 10 times larger if the order doesn't matter. 5C2 * .6^2 * .4^3
Now, with your students, the probability of each "flip" depends on the previous flip.
The probability that the first student chosen at random gets honors is 10/40. The 2nd is 9/39, the 3rd is 8/38, then 7/37, and 6/36, followed the 6th student who doesn't get honors - 30/35, 7th student 29/35,... 15th student 21/26.
Find the product. Or, honors (H), non-honors (N), the order could be HNHNHNHNHNNNNNN, which would be 10/40 * 30/39 * 9/38 * 29/37 *...
If you look at it, all the numerators are the same, just in a different order. The denominators are also the same (and same order.) There are 15C5 different orders, so 15C5 * 10*9*8*7*5 * 30*29*28*27*26*...*21 / 40*39*38*...*26.
And that's just the probability that instructor A has 5 honor students out of 15. If instructor B also had 15 students, then you can double the probability to find the probability for it to have occurred at all for either of the two instructors with 15 students (more or less; there's also the chance that both have 5, which is equal to the chance that 10 randomly chosen students were not honor students).
Unless we know that the students were completely randomly chosen, we can't compute a probability. If students were assigned in order of registering/applying for the class, then we could perhaps surmise that there would be some clustering - many of the eager, organized students applying earlier than most people.
----
To explain the probability, let's say I have a weighted coin that comes up heads 60% (3/5) of the time. The probability of getting HHTTT is .6 * .6 * .4 * .4 * .4
The probability of HTTTH is .6 * .4 * .4 * .4 * .6, which is the same value (just multiplying in a different order.) The probability of getting exactly 2 heads out of 5 flips, is the same for any possible order. There are exactly 10 different orders where 2 of the flips are heads, and 3 of the flips are tails. This can be calculated as a combination 5C2 So, the probability of getting heads exactly 2 times in 5 flips is 10 times larger if the order doesn't matter. 5C2 * .6^2 * .4^3
Now, with your students, the probability of each "flip" depends on the previous flip.
The probability that the first student chosen at random gets honors is 10/40. The 2nd is 9/39, the 3rd is 8/38, then 7/37, and 6/36, followed the 6th student who doesn't get honors - 30/35, 7th student 29/35,... 15th student 21/26.
Find the product. Or, honors (H), non-honors (N), the order could be HNHNHNHNHNNNNNN, which would be 10/40 * 30/39 * 9/38 * 29/37 *...
If you look at it, all the numerators are the same, just in a different order. The denominators are also the same (and same order.) There are 15C5 different orders, so 15C5 * 10*9*8*7*5 * 30*29*28*27*26*...*21 / 40*39*38*...*26.
And that's just the probability that instructor A has 5 honor students out of 15. If instructor B also had 15 students, then you can double the probability to find the probability for it to have occurred at all for either of the two instructors with 15 students (more or less; there's also the chance that both have 5, which is equal to the chance that 10 randomly chosen students were not honor students).
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