Question 1036434: please help me solve these questions , im really not sure how to to go about to solve them ive tried selection without repetition for q1 and ive tried multiplying the vertices by the degrees for q2 and but im still confused as to how to go about doing them
1) in a class of 12 students, 7 are good and 5 are evil. If one selects at random a team of 3 students from the class, what is the chance that the team contains some good student?
2) Let G be a simple graph with 7 vertices of degree 1, 6 vertices of degree 2, and 3 vertices of degree 3. Show that G is not connected.
Thanks
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! i don't know about number 2, but i think the solution to number 1 would be that the probability that the team contains some good students would be equal to 1 minus the probability that the team contains no good students.
the probability that the team contains no good students is the same as the probability that the team contains all evil students.
the probability that team contains no good students would be equal to c(5,3) / c(12,3).
that should be equal to 10 / 220 which is equal to 1/22.
i calculated it another way and got the same answer.
the probability that the first person is evil is 5/12, the second person evil is 4/11, the third person evil is 3/10.
multiply them together and you get 5/12 * 4/11 * 3/10 = 60 / 1320.
simplify to get 1/22.
that's the probability of them all being evil.
the probability that at least one of them would be good would therefore be 1 minus 1/22 = 21/22.
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