In a classroom of 12 students, only 4 passed the physics test. Unhappy with the students’ performance, the teachers forms a group of 6 of the students to provide feedback on the test. If she randomly chooses the 6 students from the class of 12, what is the probability that the group consists of 3 students who passed the test and 3 who did not? But it does not provide correct answer.
My reasoning was
\frac{4}{12}*\frac{3}{11}*\frac{2}{10}*\frac{8}{9}*\frac{7}{8}*\frac{6}{7}
This is incorrect. A lot of extraneous information in this question. You want to filter them out.
I basically thought we need to select 6 students out of 12. 6 students should consist of 3 pass students and 3 fail students
How do you actually solve the problem? There doesn’t seem to be extraneous information
I may be wrong but I think its something like this (4c3*8c3)/12c6
basically what i did was select 3 students who passed the test AND out of remaining 8 i selected 3 more divided by the number of ways i can select 6 students out of 12 Hope this helps
Your method is not correct in this situation, because the order of the students is not important. However, the method you’ve used differentiates between, say, the order of the three students being A, B and C, or B, A and C (where the failed students were named A, B, C and D)
Can be solved in 2 ways
The way you have mentioned:
But because we do not care about the order - you need to multiply it with 6!/(3! x 3!) - as there are 6 students in total (3 of each type)
As Greg would explain: ways that make you happy/Total no of possibilities
(4C3 x 8C3)/ 12C3
Both lead you to 8/33 which is ~ 0.25
solved in the same way.
Do you mean (4C3 x 8C3)/ 12C6 for way 2? Because (4C3 x 8C3)/ 12C3 gets me 1.01 as an answer.
My bad, it is 12C6
is it 12C6 right??
can it be done like 3/4 *3/8??