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. [Round your answers to three decimal places, for example: 0.123]
Compute the probability that a randomly selected peanut M&M is not green.
Compute the probability that a randomly selected peanut M&M is blue or orange.
Compute the probability that two randomly selected peanut M&M’s are both orange.
If you randomly select three peanut M&M’s, compute that probability that none of them are orange.
If you randomly select three peanut M&M’s, compute that probability that at least one of them is orange.
I need answers
Compute the probability that two randomly selected peanut M&M’s are both orange.
Answer:
.23 x .23= .0529
To answer these questions, we can use basic probability principles. We will calculate the probabilities step by step, explaining the process along the way.
1. Probability that a randomly selected peanut M&M is not green:
To find this probability, we need to subtract the probability of selecting a green M&M from 1 (since the sum of probabilities of all possible outcomes should be 1).
P(not green) = 1 - P(green)
P(green) = 15% = 0.15 (as given)
P(not green) = 1 - 0.15 = 0.85
Therefore, the probability that a randomly selected peanut M&M is not green is 0.85.
2. Probability that a randomly selected peanut M&M is blue or orange:
To calculate this probability, we can simply add the individual probabilities of selecting a blue M&M and an orange M&M.
P(blue or orange) = P(blue) + P(orange)
P(blue) = 23% = 0.23 (as given)
P(orange) = 23% = 0.23 (as given)
P(blue or orange) = 0.23 + 0.23 = 0.46
Therefore, the probability that a randomly selected peanut M&M is blue or orange is 0.46.
3. Probability that two randomly selected peanut M&M's are both orange:
To calculate this probability, we need to multiply the probability of selecting an orange M&M in the first draw by the probability of selecting another orange M&M in the second draw.
P(both orange) = P(orange in 1st draw) * P(orange in 2nd draw)
P(orange in 1st draw) = 23% = 0.23 (as given)
P(orange in 2nd draw) = (total orange M&Ms - 1) / (total M&Ms - 1) [since we're drawing without replacement]
= (23 - 1) / (100 - 1) [reducing the fractions]
= 22 / 99 [simplifying]
P(both orange) = 0.23 * (22 / 99) ≈ 0.051
Therefore, the probability that two randomly selected peanut M&Ms are both orange is approximately 0.051.
4. Probability that none of the three randomly selected peanut M&M's are orange:
Since we're selecting M&Ms without replacement, we need to consider the probabilities for each draw, and multiply them together.
P(no orange in 3 draws) = P(not orange in 1st draw) * P(not orange in 2nd draw) * P(not orange in 3rd draw)
P(not orange in 1st draw) = 77 / 99 [remaining non-oranges after the first draw]
P(not orange in 2nd draw) = 76 / 98 [remaining non-oranges after the second draw]
P(not orange in 3rd draw) = 75 / 97 [remaining non-oranges after the third draw]
P(no orange in 3 draws) = (77 / 99) * (76 / 98) * (75 / 97) ≈ 0.455
Therefore, the probability that none of the three randomly selected peanut M&Ms are orange is approximately 0.455.
5. Probability that at least one of the three randomly selected peanut M&M's is orange:
To calculate this probability, we need to subtract the probability of selecting no orange M&M's from 1 (since the sum of probabilities of all possible outcomes should be 1).
P(at least one orange in 3 draws) = 1 - P(no orange in 3 draws)
P(no orange in 3 draws) ≈ 0.455 (as calculated in the previous question)
P(at least one orange in 3 draws) = 1 - 0.455 = 0.545
Therefore, the probability that at least one of the three randomly selected peanut M&Ms is orange is approximately 0.545.
If you randomly select three peanut M&M’s, compute that probability that none of them are orange.
Answer:
1-.23=.77
.77 x .77 x .77 = .4565
Examples
P(not green) = 1 - .15 = ?
Either-or probabilities are found by adding the individual probabilities.
P(blue or orange) = P(blue) + P(orange)
If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.
P(both orange) = P(orange)^2
P(none orange) = (1-.23)^3
"at least one" = 1, 2 or 3 are orange. Use methods above.
Online "^" is used to indicate an exponent, e.g., x^2 = x squared.
you can just as easily consider the % values as integers.
Then just count the ones you need, out of 100