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 blue or brown.
c. compute the probability that two randomly selected peanut m& m’s are both red.
d. if you randomly select six peanut m& m’s, compute that probability that none of them are blue.
e. if you randomly select six peanut m& m’s, compute that probability that at least one of them is blue.

a) 0.88

b) 0.35

c) 0.0144

d) 0.2084

e) 0.7916

Step-by-step explanation:

a) The probability of a peanut being brown is 12/100 = 0.12. Hence the probability of it not being brown is 1-0.12 = 0.88

b) 12% of peanuts are brown, 23% are blue. So 35% are either blue or brown. The probability of a peanut being blue or brown is, therefore 35/100 = 0.35.

c) 12% of peanuts are red, so the probability of a peanut being red is 12/100 = 0.12. In order to calculate the probability of 2 peanuts being both red, we can assume that the proportion doesnt change dramatically after removing one peanut (because the number of peanuts is absurdly high. We can assume that we are replenishing the peanuts). To calculate the probability of 2 peanuts being both red, we need to power 0.12 by 2, hence the probability is 0.12² = 0.0144.

d) Again, we will assume that the probability doesnt change, because we replenish. The probability of a peanut being blue is 0.23. The probability of it not being blue is 0.77, so the probability of 6 peanuts not being blue is obtained from powering 0.77 by 6, hence it is 0.77⁶ = 0.2084

e) The event 'at least one peanut is blue' is te complementary event of 'none peanuts are blue', so the probability of this event is 1- 0.2084 = 0.7916

853.3

step-by-step explanation:

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