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.
Compute the probability that a randomly selected peanut M&M is not green.
Compute the probability that a randomly selected peanut M&M is red or orange.
Compute the probability that two randomly selected peanut M&M’s are both green.
If you randomly select six peanut M&M’s, compute that probability that none of them are green.
If you randomly select six peanut M&M’s, compute that probability that at least one of them is green.
I ONLY need the formula for the last one. I already got the first four questions, but the last one I can't seem to get.
Thanks in advance!
that will be 1 - P(none green) = 1 - 0.85^6
To calculate the probability that at least one of the six randomly selected peanut M&M's is green, you can use the concept of complementary events. The probability of the complementary event (i.e., none of them being green) is easier to calculate, and then we can subtract it from 1 to find the probability of the desired event (i.e., at least one being green).
The formula to calculate the probability of the complementary event is:
P(complementary event) = (1 - P(green))^6
In this formula, P(green) represents the probability of selecting a green M&M in a single trial.
Since the given data states that 15% of peanut M&M's are green, we can convert this to decimal form: P(green) = 0.15.
Now, we can substitute this value into the formula and calculate the probability of the complementary event:
P(complementary event) = (1 - 0.15)^6
P(complementary event) ≈ 0.3105
Finally, to find the probability of at least one M&M being green, we subtract the probability of the complementary event from 1:
P(at least one green) = 1 - P(complementary event)
P(at least one green) = 1 - 0.3105
P(at least one green) ≈ 0.6895
Therefore, the probability that at least one of the six randomly selected peanut M&M's is green is approximately 0.6895.