Question

A box contains 5 red balls, 3 green balls, and 4 blue balls. You pull 2 balls from the box (one at a time) WITHOUT replacement.

a.) Draw a tree diagram and list out all the color combination possibilities for the two selections (Please draw your diagram on paper and list color combinations on paper, submit with your work via email, you DO NOT need to compute the probabilities for all the outcomes).

b.) What is the probability of drawing 2 red balls in your 2 pulls?

c.) After selecting two balls, what is the probability of ending up with 1 red ball AND 1 green ball (can be selected in either order)?

d.) What is the probability of selecting a green ball on your second pull, GIVEN a green ball was already selected on the first pull? Hint: P(G ∣ G) not P(G AND G)

e.) What is the probability of selecting two balls of the same color?

Answer 1

The tree diagram for the color combinations is as follows: RR, RG, RB, GR, GG, GB, BR, BG, BB. The probability of drawing 2 red balls is 4/33. The probability of ending up with 1 red ball and 1 green ball is 1/11. The probability of selecting a green ball on the second pull, given a green ball was already selected on the first pull, is 2/11. The probability of selecting two balls of the same color is 21/66.

Step-by-step explanation:

a.) The tree diagram for the color combinations is as follows:

Red (R) - Green (G) - Blue (B)

- First draw: R, G, B

- Second draw: R, G, B

- Combination possibilities for two selections: RR, RG, RB, GR, GG, GB, BR, BG, BB

b.) The probability of drawing 2 red balls is 4/33, because there are 4 red balls out of the 33 total balls remaining after the first draw.

P(RR) = (5/12) * (4/11) = 20/132 = 5/33

c.) The probability of ending up with 1 red ball and 1 green ball is 12/132 = 1/11. This can be calculated by adding the probabilities of drawing a red ball followed by a green ball (P(RG)) and drawing a green ball followed by a red ball (P(GR)).

P(RG) + P(GR) = (5/12) * (3/11) + (3/12) * (5/11) = 15/132 + 15/132 = 30/132 = 5/22 = 1/11

d.) The probability of selecting a green ball on the second pull, given a green ball was already selected on the first pull, is 2/11. This can be calculated by dividing the number of green balls remaining after the first draw (3) by the total number of balls remaining after the first draw (11).

e.) The probability of selecting two balls of the same color is 21/66, because there are 21 combinations where two balls have the same color out of the 66 total possible combinations. This can be calculated by adding the probabilities of drawing two red balls (P(RR)), two green balls (P(GG)), and two blue balls (P(BB)).