Question

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 red.

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 orange.

d. If you randomly select six peanut M&M’s, compute that probability that none of them are brown.

e. If you randomly select six peanut M&M’s, compute that probability that at least one of them is brown.

Answer 1

Given

P( Brown ) =12% = 0.12

P(Yellow) = 15% = 0.15

P( Red) =12% = 0.12

P( Blue) = 23% = 0.23

P( Orange ) = 23% = 0.23

P( Green) = 15% = 0.15

Part a. Compute the probability that a randomly selected peanut M&M is not red

Using Compliment rule

Using above rule :

probability that a randomly selected peanut M&M is not red is 0.88

Part b. Compute the probability that a randomly selected peanut M&M is yellow or orange.

P( Yellow ) = 0.15

P( Orange ) = 0.23

P( Yellow OR Orange) = P( Yellow ) + P( Orange ) = 0.15 + 0.23 = 0.38

P( Yellow OR Orange) = 0.38

Answer :

probability that a randomly selected peanut M&M is yellow or orange is 0.38

Part C. Compute the probability that three randomly selected peanut M&M’s are all orange.

P( Orange ) = 0.23

selected 3 peanuts

Answer :

probability that three randomly selected peanut M&M’s are all orange is 0.0122

Part d. If you randomly select six peanut M&M’s, compute that probability that none of them are brown.

Using compliment rule

P( Brown) = 0.12

P(Not Brown ) = 1- P( Brown) = 1- 0.12 = 0.88

Number of selected peanut is 6

Answer:

   Probability that randomly selected six peanut M&M’s, that none of them are brown is 0.4644.

Part e. If you randomly select six peanut M&M’s, compute that probability that at least one of them is brown

P( at least 1 brown ) = 1 - P( None of 6 brown)

Using part d . Probability that randomly selected six peanut M&M’s, that none of them are brown is 0.4644.

P( at least 1 brown ) = 1 - 0.4644 = 0.5356

Answer :

Probability that randomly selected six peanut M&M’s, that at least of them are brown is 0.5356