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Answer choices P(Grade 9 | opposed) = P(opposed | Grade 9) P(Grade 9 | opposed) > P(opposed | Grade 9) P(Grade 9 | opposed) < P(opposed | Grade 9) There is not enough information.
to find P(Grade 9 | opposed) that says Given the opposed (so circle that whole column) how many are Grade 9. write down the fraction # in Grade 9 in the opposed column divided by the total number in the opposed column (all Grades)
to find P(opposed | Grade 9). which means Given Grade 9 (i.e. circle the Grade 9 row) how many in that row are opposed? make the fraction # in Grade 9 who are opposed divided by total number in Grade 9 (add up all the numbers in that row)
so it B @phi
are you able to find P(Grade 9 | opposed) ? do you see the opposed column? what is the sum of that column?
and 2 out of the 33 are Grade 9 so the probability of Grade 9 given opposed is 2/33
okay i understand that
now do P(opposed | Grade 9) we are "given Grade 9" what is the sum of numbers in Grade 9 ?
and how many of that 14 are "opposed" ? what is the fraction you should make?
so now you have 2/33 2/14 which one is bigger ?
so its c
so you want the relation that has the "big side" next to the 2/14 2/33 < 2/14
yes, choice C
alright thank you so much