A bag of plain M&Ms; contains 60 candies. Twelve of the candies are yellow, 15 a

Question

A bag of plain M&Ms; contains 60 candies. Twelve of the candies are yellow, 15 are red, 6 are orange, 9 are blue, 11 are green and the rest are brown. 20. Suppose you select one M&M; from this bag at random and do not look at the color. A friend tells you that the candy you selected is a primary color (i.e. it is red, yellow, or blue). What is the probability it is yellow if you know it is a primary color? 21. If you select three M&Ms; in a row (without replacement), what is the probability they are all brown? If you select three M&Ms; in a row (without replacement), what is the probability at least one is green? 22.

Explanation / Answer

20) P(primary color) = P(red) + P(yellow) + P(blue) = 15/60 + 12/60 + 9/60 = 36/60

P(yellow | primary color) = P(yellow and primary color) / P(primary color) = P(yellow) / P(primary) = (12/60) / (36/60) = 12/36 = 1/3

21) number of brown candies = 60 - (12 + 15 + 6 + 9 + 11) = 7

P(all 3 are brown) = (7/60) * (6/59) * (5/58) = 7/6844 or 0.001

22) P(at least one is green) = 1 - P(none are green) = 1 - (49/60) * (48/59) * (47/58) = 0.4616

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.