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there are 6 girls
6 choose 3 / 10 choose 3
there are 6 girls, you need to choose 3 of them, where order does not count. thats 6 choose 3 . that will be your numerator (the favorable event). the denominator is the total possible 3 positions from the set of 10 kids, so thats 10 choose 3
thats for part a)
part b) you want 1 boy and 2 girls. there are 4 boys in total, so you want to choose 1 boy. This is 4 choose 1. now for each one of these choices you want to pick 2 girls out of 6 ( 6 choose 2). so multiply them 4 choose 1 * 6 choose 2 . this is your numerator. the denominator is like before 10 choose 3
adam anthony? and agnes, not sure about that one
how about part c and e?
well i got you started
this is the proper question Ten students are applying for 3 positions on a team. the students include 4 boys (adam, alex, anthony, and arnold) and 6 girls (abbey, aurora, agnes, alice, amanda, and anna). all the students have an equal chance of being selected. find the probability that the students selected will include: a) 3 girls b) 1 boy and 2 girls c) at most 1 girl d) adam, anthony, and alice e) agnes and 2 other students
Hey, do you have the answers to this, in the book
so how do you know if its right
this is a worksheet
how do u do c,d,e
at most one girls is 0 girls or 1 girl
at most 1 is number of ways to count with 0 girls + number of ways to count with 1 girl
4c0 + 4c1 / 6c3???
if there are no girls, then the other 3 people chosen must be boys
ill be back in an hour
what is it then