SOLUTION: The Web site for M&M™ candies claimed that 24% of plain M&M candies are blue, 20% are orange, 16% green, 14% yellow, and 13% each red and brown. (a) Pick one M&M at random fro

Question 1171071: The Web site for M&M™ candies claimed that 24% of plain M&M candies are blue, 20% are orange, 16% green, 14% yellow, and 13% each red and brown.
(a) Pick one M&M at random from a package.
1.Describe the sample space.
2.What is the probability that the one you pick is blue or red?
3.What is the probability that the one you pick is not green?
(b) You pick three M&M’s in a row randomly from three separate packages.
1.Describe the sample space for the outcomes of your three choices.
2.What is the probability that every M&M is blue?
3.What is the probability that the third M&M is red?4.What is the probability that at least one is blue?
Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
Let's break down this M&M problem.
**(a) Picking One M&M**
1. **Describe the Sample Space:**
* The sample space consists of all possible colors of an M&M.
* Sample Space = {Blue, Orange, Green, Yellow, Red, Brown}
2. **Probability of Blue or Red:**
* P(Blue) = 24% = 0.24
* P(Red) = 13% = 0.13
* Since these are mutually exclusive events (an M&M can't be both blue and red), we add the probabilities:
* P(Blue or Red) = P(Blue) + P(Red) = 0.24 + 0.13 = 0.37
3. **Probability of Not Green:**
* P(Green) = 16% = 0.16
* P(Not Green) = 1 - P(Green) = 1 - 0.16 = 0.84
**(b) Picking Three M&M's**
1. **Describe the Sample Space:**
* Each M&M can be one of six colors.
* The sample space consists of all possible combinations of three colors.
* We can represent each outcome as a sequence of three colors (e.g., Blue, Orange, Red).
* The sample space would be a list of all possible combinations of three colors, such as (Blue, Blue, Blue), (Blue, Blue, Orange), (Blue, Orange, Blue), etc.
* The number of possible outcomes is 6 * 6 * 6 = 216.
2. **Probability of Every M&M Being Blue:**
* P(Blue) = 0.24
* Since the picks are independent, we multiply the probabilities:
* P(Blue, Blue, Blue) = P(Blue) * P(Blue) * P(Blue) = (0.24)^3 = 0.013824
3. **Probability of the Third M&M Being Red:**
* The probability of the third M&M being red is independent of the first two picks.
* P(Third M&M is Red) = P(Red) = 0.13
4. **Probability of At Least One Blue:**
* It's easier to find the probability of no blue M&M's and subtract it from 1.
* P(Not Blue) = 1 - P(Blue) = 1 - 0.24 = 0.76
* P(No Blue in 3 picks) = (0.76)^3 = 0.438976
* P(At Least One Blue) = 1 - P(No Blue) = 1 - 0.438976 = 0.561024

RELATED QUESTIONS
Please help me solve this question. The Web site for M&M™ candies claimed that 24% (answered by Boreal)

Mars claims that 15% of its plain M&M candies are green. Determine the probability of... (answered by richard1234)

M&M’s As noted in an earlier chapter, the Master-foods Company says that until very... (answered by ikleyn)

In a bag containing 143 M&M candies, there are 52 brown, 39 orange, 20 yellow, and 17 (answered by Fombitz,macston)

A snack-size bag of M&Ms candies is opened. Inside, there are 12 red candies, 12 blue, 7... (answered by Boreal)

A snack-size bag of M&Ms candies is opened. Inside, there are 12 red candies, 12 blue, 7... (answered by ikleyn)

Mars Inc, claims 24% of M&M plain candies are blue. Suppose that a sample of 100 M&Ms are (answered by edjones)

The color of M&M candies can be brown,yellow, red, blue, orange and green. The following... (answered by Fombitz)

A jar contains 12 M&M candies, of the 12: 2 are red, 3 are green, 3 are blue, 4 are... (answered by ewatrrr)