Four marbles were drawn randomly from a box containing 5 red marbles, 4 blue marbles and 2 green marbles.

Question

(a) How many ways can the four marbles be drawn from the box?

(b) What is the probability of finding them all red?

(c) What is the probability of finding them all blue?

(d) What is the probability of finding all the colors in the marbles selected?

(e) What is the probability of finding at least one red?

(f) What is the probability of finding at least two greens from four marbles drawn?

(g) What is the probability of not finding a yellow marble?

Answer 1

(a) You are choosing 4 marbles from a box of 11. There are C(11, 4) = 11! / (4! * 7!) = 330 ways to do this.

(b) There are C(5, 4) = 5 ways of choosing all red marbles, so the probability is 5/330 = 1/66.

(c) There is C(4, 4) = 1 way of choosing all blue marbles, so the probability is 1/330.

(d) This can happen in three different ways: either (i) 2 red, 1 blue, 1 green; (ii) 1 red, 2 blue, 2 green; or (iii) 1 red, 1 blue, 2 green. Possibility (i) can happen in C(5, 2) * C(4, 1) * C(2, 1) = 80 ways; possibility (ii) can happen in C(5, 1) * C(4, 2) * C(2, 1) = 60 ways; and possibility (iii) can happen in C(5, 1) * C(4, 1) * C(2, 2) = 20 ways. They make a total of 160 ways, so the probability is 160/330 = 16/33.

(e) This is 1 - P(no red). There are 6 non-red marbles, so C(6, 4) = 15 ways to choose them. So the probability is 15/330 = 1/22.

(f) Since there are only 2 greens to begin with, the probability of finding at least 2 greens is the same as the probability of finding exactly 2 greens. You have to choose 2 greens, and the other 2 can be chosen from among the remaining 9 in any way. There are thus C(9, 2) * C(2, 2) = 36 ways to do this, so the probability is 36/330 = 6/55.

(g) Since there are no yellow marbles, the probability of not finding a yellow marble is 1.