# A box contains 22 red apples and 3 green apples. Three apples are selected at random, one after the

Discussion in 'Calculator Requests' started by math_celebrity, Jan 8, 2017.

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A box contains 22 red apples and 3 green apples. Three apples are selected at random, one after the other, without replacement. please show the steps.

(a) The first two apples are green. What is the probability that the third apple is red?
(b) What is the probability that exactly two of the three apples are red?

a) You have 22 red apples left and 1 green left leaving 23 total apples left. Therefore, probability of red is
P(R) = 22/23

b) Determine our sample space to select exactly two red apples in three picks.
1. RRG
2. RGR
3. GRR

Now determine the probabilities of each event in the sample space
P(RRG) = 22/25 * 21/24 * 3/23 = 0.1004
P(RGR) = 22/25 * 3/24 * 21/23 = 0.1004
P(GRR) = 3/25 * 22/24 * 21/23 = 0.1004

We want the sum of the three probabilities
P(RRG) + P(RGR) + P(GRR) = 0.1004 + 0.1004 + 0.1004
P(RRG) + P(RGR) + P(GRR) = 3(0.1004)
P(RRG) + P(RGR) + P(GRR) = 0.3012