According to Masterfoods, the company that manufactures M&M’s,
12% 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.
a) Compute the probability that a randomly selected peanut M&M is not brown.
b) Compute the probability that a randomly selected peanut M&M is yellow or orange.
c) Compute the probability that three randomly selected peanut M&M’s are all red.
d) If you randomly select four peanut M&M’s, compute that probability that none of them are brown.
e) If you randomly select four peanut M&M’s, compute that probability that at least one of them is brown.
a) The probability that a randomly selected peanut M&M is not brown can be calculated by subtracting the probability of selecting a brown M&M from 1:
Probability = 1 - 0.12
Probability = 0.88
So, the probability is 0.880.
b) The probability that a randomly selected peanut M&M is yellow or orange can be found by adding the probabilities of selecting a yellow M&M and an orange M&M:
Probability = 0.15 + 0.23
Probability = 0.38
So, the probability is 0.380.
c) The probability that three randomly selected peanut M&M's are all red can be calculated by multiplying the probability of selecting a red M&M three times:
Probability = 0.12 * 0.12 * 0.12
Probability = 0.001728
So, the probability is 0.001728.
d) The probability that none of the four peanut M&M's are brown can be found by multiplying the probabilities of selecting a non-brown M&M four times:
Probability = 0.88 * 0.88 * 0.88 * 0.88
Probability ≈ 0.599
So, the probability is approximately 0.599.
e) The probability that at least one of the four peanut M&M's is brown can be calculated by subtracting the probability of selecting none of them being brown from 1:
Probability = 1 - 0.599
Probability ≈ 0.401
So, the probability is approximately 0.401.
a) To compute the probability that a randomly selected peanut M&M is not brown, we need to subtract the probability that it is brown from 1.
Probability of not being brown = 1 - Probability of being brown
= 1 - 0.12
= 0.88
Therefore, the probability that a randomly selected peanut M&M is not brown is 0.880.
b) To compute the probability that a randomly selected peanut M&M is yellow or orange, we simply add the individual probabilities of being yellow and orange.
Probability of being yellow or orange = Probability of being yellow + Probability of being orange
= 0.15 + 0.23
= 0.38
Therefore, the probability that a randomly selected peanut M&M is yellow or orange is 0.380.
c) To compute the probability that three randomly selected peanut M&M's are all red, we multiply the individual probabilities of selecting a red M&M together.
Probability of selecting three red M&M's = Probability of selecting a red M&M * Probability of selecting another red M&M * Probability of selecting a third red M&M
= 0.12 * 0.12 * 0.12
= 0.001728
Therefore, the probability that three randomly selected peanut M&M's are all red is 0.001.
d) If you randomly select four peanut M&M's, the probability that none of them are brown can be calculated by multiplying the probability of selecting a non-brown M&M each time.
Probability of selecting four non-brown M&M's = Probability of selecting a non-brown M&M * Probability of selecting another non-brown M&M * Probability of selecting a third non-brown M&M * Probability of selecting a fourth non-brown M&M
= 0.88 * 0.88 * 0.88 * 0.88
= 0.59969536
Therefore, the probability that none of the randomly selected four peanut M&M's are brown is 0.600.
e) If you randomly select four peanut M&M's, the probability that at least one of them is brown can be calculated by subtracting the probability that none of them are brown from 1.
Probability of at least one of them being brown = 1 - Probability of none of them being brown
= 1 - (0.88 * 0.88 * 0.88 * 0.88)
= 0.39906464
Therefore, the probability that at least one of the randomly selected four peanut M&M's is brown is 0.399.
To compute the probabilities, we need to use the information provided about the percentages of each colored M&M.
a) The probability of selecting a peanut M&M that is not brown can be found by subtracting the percentage of brown M&Ms from 100%. So, the probability is 100% - 12% = 88%.
b) To compute the probability that a randomly selected peanut M&M is yellow or orange, we need to add the percentages of yellow M&Ms and orange M&Ms. So, the probability is 15% + 23% = 38%.
c) To compute the probability that three randomly selected peanut M&Ms are all red, we need to multiply the percentage of red M&Ms by itself three times, since we are selecting three M&Ms. So, the probability is (12%)^3 = 0.001728.
d) To compute the probability that none of the four randomly selected peanut M&Ms are brown, we need to multiply the probabilities of getting a non-brown M&M four times. So, the probability is (88%)^4 = 0.59969584.
e) To compute the probability that at least one of the four randomly selected peanut M&Ms is brown, we can subtract the probability of getting none of them being brown from 1. So, the probability is 1 - 0.59969584 = 0.40030416.
a), well it said prob(brown) = 12/100 = .12, so prob(not brown) = .88
b) prob(yellow or orange) = .15+.23 = .38 or 19/50
c) If they are not returned : prob(3 all red) = (12/100)(11/99)(10/98) = 1/735
or C(12,3)/C(100,3) = 220/161700 = 1/735
If the picked candy is returned: prob(3 red) = (12/100)^3 = 27/15625
d) number of non-browns = 88
so prob(4 non-browns) if they are not returned = (88/100)(87/99)(86/98)(85/97) = .59468..
or C(88,4)/C(100,4) = 2331890/3921225 = .59468... , same as above
e) well, that would be simply 1 - .59468.. = .....