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 4 decimal places where possible)
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 green or yellow.
c. Compute the probability that two randomly selected peanut
M&M’s are both orange.
d. If you randomly select four peanut M&M’s, compute that
probability that none of them are orange.
e. If you randomly select four peanut M&M’s, compute that
probability that at least one of them is orange.
Let total product = 1
a)
Probability of brown = total brown / total product = 0.12/1 = 0.12
Probability of not brown = 1- 0.12 = 0.88
b)
Probability of green or yellow = 0.15 +0.15 = 0.3
c)
Probability of orange = 0.23
Probability of both orange = 0.23 * 0.23 =
0.0529
d)
Probability of others (except orange) = 1 - 0.23 = 0.77
Probability of 4 others = 0.77 * 0.77 * 0.77 *
0.77=0.3515
e)
Probability of at least 1 orange = 1 - None orange
= 1 - 0.3515 = 0.6485
Please revert back in case of any doubt.
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