M&Ms are blended in a ratio of 15 percent brown, 10 percent yellow, 11 percent r
ID: 3235630 • Letter: M
Question
M&Ms are blended in a ratio of 15 percent brown, 10 percent yellow, 11 percent red, 18 percent blue, 22 percent orange, and 24 percent green. Suppose you choose a sample of two M&Ms at random from a large bag.
How many elements are there in the sample space, assuming the order in which the M&Ms are drawn matters?
What is the probability that the first M&M is brown and the second is green? (Round your answer to 4 decimal places.)
M&Ms are blended in a ratio of 15 percent brown, 10 percent yellow, 11 percent red, 18 percent blue, 22 percent orange, and 24 percent green. Suppose you choose a sample of two M&Ms at random from a large bag.
Explanation / Answer
Given P(Brown) = 15% = 15/100 = 0.15 Similary, P(Yellow) = 0.1, P(Red) = 0.11, P(Blue) = 0.18, P(Orange) = 0.22 and P(Green) = 0.24. n = 2(no of picks)
There are 6 different colours (brown, yellow, red, blue, orange and green)
(a) The number of elements in the sample space(where the order matters i.e (BY) and (YB) are considered as 2 different sample spaces) = 2 * 6C2 = 2 * 15 = 30
(b) Probability of both being brown = 0.15 * 015 = 0.0225
(c) Probability of both being blue = 0.18 * 018 = 0.0324
(d) Probability of both being green = 0.24 * 024 = 0.0576
(e) Probability that the first is brown and the second is green = 0.15 * 0.24 = 0.0360
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