According to Masterfoods, the company that manufactures M&M’s, 12% of peanut M&M’s are brown, 15% are yellow, 12% are red, 23% are blue, 23% are orange and 15% are green. You randomly select peanut M&M’s from an extra-large bag looking for an orange candy. (Round all probabilities below to four decimal places; i.e. your answer should look like 0.1234, not 0.1234444 or 12.34%.)
Compute the probability that the first orange candy is the ninth M&M selected.
Compute the probability that the first orange candy is the ninth or tenth M&M selected.
Compute the probability that the first orange candy is among the first nine M&M’s selected.
If every student in a large Statistics class selects peanut M&M’s at random until they get a orange candy, on average how many M&M’s will the students need to select? (Round your answer to two decimal places.)
1. 0.0014
2. 0.0028
3. 0.4164
4. 8.46
To solve these probability questions, we first need to understand the given information. According to Masterfoods, the company that manufactures M&M's, we know the following probabilities:
- Brown: 12%
- Yellow: 15%
- Red: 12%
- Blue: 23%
- Orange: 23%
- Green: 15%
Now let's address each question one by one.
1. Compute the probability that the first orange candy is the ninth M&M selected.
To solve this, we need to consider the probability of selecting a non-orange candy for the first eight M&M's and then an orange candy for the ninth one.
Since the probability of selecting an orange candy is 23%, the probability of selecting a non-orange candy is 100% - 23% = 77%.
So, the probability of selecting a non-orange candy in the first eight M&M's is (77%/100%)^8.
Hence, the probability of the first orange candy being on the ninth M&M is (77%/100%)^8 * (23%/100%) ≈ 0.0242.
2. Compute the probability that the first orange candy is the ninth or tenth M&M selected.
To solve this, we need to consider the probability of selecting a non-orange candy for the first eight M&M's, then an orange candy for the ninth one, or an orange candy for the ninth one, and a non-orange candy for the tenth.
The probability of selecting a non-orange candy in the first eight M&M's is (77%/100%)^8.
The probability of selecting a non-orange candy in the ninth M&M is 77%/100% = 0.77.
The probability of selecting an orange candy for the ninth M&M is 23%/100% = 0.23.
The probability of selecting a non-orange candy for the tenth M&M is 77%/100% = 0.77.
So, the probability of the first orange candy being on the ninth or tenth M&M is (77%/100%)^8 * (0.23 + 0.77) + (0.77 * 0.23) ≈ 0.0394.
3. Compute the probability that the first orange candy is among the first nine M&M’s selected.
To solve this, we need to consider the probability of selecting an orange candy in the first nine M&M's, regardless of their order.
Since the probability of selecting an orange candy is 23%, the probability of selecting a non-orange candy is 77%.
So, the probability of selecting an orange candy in the first nine M&M's is 1 - (77%/100%)^9 ≈ 0.6869.
4. If every student in a large Statistics class selects peanut M&M’s at random until they get an orange candy, on average, how many M&M’s will the students need to select?
To solve this, we can calculate the expected value, which is the sum of each possible outcome multiplied by its probability.
Since the probabilities of selecting each color are given, we can calculate the expected value using the formula:
E(X) = 1/(probability of orange) = 1/0.23 ≈ 4.35
Hence, on average, students will need to select approximately 4.35 M&M's until they get an orange candy.
To solve these problems, we need to use the concept of conditional probability.
1. To compute the probability that the first orange candy is the ninth M&M selected, we need to calculate the probability of getting 8 non-orange M&M's followed by an orange one.
The probability of getting a non-orange M&M on each selection is 1 - 0.23 = 0.77.
So, the probability of getting 8 non-orange M&M's in a row is (0.77)^8 = 0.1210.
Since the probability of getting an orange M&M on the ninth selection is 0.23,
the probability that the first orange candy is the ninth M&M selected is 0.1210 * 0.23 = 0.0278.
Therefore, the probability is 0.0278.
2. To compute the probability that the first orange candy is the ninth or tenth M&M selected, we can calculate the probability of getting 8 non-orange M&M's followed by an orange one on the ninth selection, as well as the probability of getting 9 non-orange M&M's followed by an orange one on the tenth selection.
The probability of getting 8 non-orange M&M's followed by an orange one on the ninth selection is 0.0278 (calculated in step 1).
The probability of getting a non-orange M&M on the ninth selection is 0.77, and the probability of getting an orange one on the tenth selection is 0.23.
So, the probability of getting 9 non-orange M&M's followed by an orange one on the tenth selection is 0.77^9 * 0.23 = 0.0722.
The probability that the first orange candy is the ninth or tenth M&M selected is the sum of these two probabilities:
0.0278 + 0.0722 = 0.1000.
Therefore, the probability is 0.1000.
3. To compute the probability that the first orange candy is among the first nine M&M’s selected, we can calculate the probability of getting at least one orange M&M within the first nine selections.
The probability of getting a non-orange M&M on each selection is 1 - 0.23 = 0.77.
So, the probability of getting 9 non-orange M&M's in a row is (0.77)^9 = 0.1350.
The probability that the first orange candy is among the first nine M&M’s selected is 1 - 0.1350 = 0.8650.
Therefore, the probability is 0.8650.
4. To calculate the average number of M&M’s that the students need to select until they get an orange candy, we need to find the expected value.
The probability of getting a non-orange M&M on each selection is 1 - 0.23 = 0.77.
So, the expected number of M&M's needed to get an orange candy is calculated as follows:
(1 * 0.77) + (2 * 0.77^2) + (3 * 0.77^3) + ... = Σ (n * 0.77^n) from n=1 to infinity
By using the formula for the sum of an infinite geometric series, we get:
(1 * 0.77) / (1 - 0.77) = 0.77 / 0.23 ≈ 3.35
Therefore, on average, the students will need to select approximately 3.35 M&M’s to get an orange candy.