Probability Terminology

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 three decimal places, for example: 0.123]

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

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

Compute the probability that three randomly selected peanut M&M’s are all brown.

If you randomly select five peanut M&M’s, compute that probability that none of them are orange.

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

Answer 1

Hi Gladys,

Let's break this question down piece by piece:

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

You were given that 15% of M&Ms were green. Therefore:

P(Green)=0.15

P(Non-Green)=1-0.15

P(Non-Green)=0.850

Recall that in a discrete probability distribution like this, all probabilities must add to 1, so if you have "not" or "non" in the question, you can use the "1 minus trick," formally known as the Complement Rule.

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

Recall also that when "or" is in the question, we are looking at addition for a discrete probability distribution.

P(Orange or Yellow)=P(Orange) + P(Yellow)

P(Orange)=0.23

P(Yellow)=0.15

P(Orange or Yellow)= 0.23 + 0.15

P(Orange or Yellow)=0.380

Compute the probability that three randomly selected peanut M&M’s are all brown.

This requires the binomial probability equation. To recognize the binomial situation, determine if you have a binary outcome--success vs. failure--and if your trials are independent. In this case, you have a binary outcome--brown vs. non--and there is no relation between draws, so your trials are independent, so we can use the formula:

P(X=x)=C(n,x)p^xq^(n-x)

Let's break this down:

x=number of successes you are looking for, in this case number of brown M&Ms

C=combination, indicates that selection order does not matter

n=sample size

p=probability of success

q=1-p=probability of failure

For our problem:

x=3

n=3

p=0.12, from percentage given for brown

q=1-0.12=0.88

Thus:

P(X=3)=C(3,3)0.12³0.88^(3-3)

Recall:

C(3,3)=3!/(3!0!)

where:

!=Factorial=n*(n-1)*(n-2)....,

3*2*1 in this case.

Anyway,

C(3,3)=3!/3!=1

0.88⁰=1

Thus:

P(X=3)=1*0.12³*1

P(X=3)=0.12³

P(X=3)=0.002

If you randomly select five peanut M&M’s, compute that probability that none of them are orange.

This is essentially the same principle as our last part of the question.

P(X=x)=C(n,x)p^xq^(n-x)

x=0, no orange M&Ms

C still denotes combination, order does not matter

n=5

p=0.23, probability of orange given

q=1-p=0.77

Thus:

P(X=0)=C(5,0)0.23⁰0.77⁵

C(5,0)=5!/(0!5!)

C(5,0)=5!/5!

C(5,0)=1

0.23⁰=1

P(X=0)=1*1*0.77⁵

P(X=0)=0.77⁵

P(X=0)=0.271

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

We just computed the probability that no M&Ms were orange if we drew 5, so:

P(X>=1)=1-P(X=0)

P(X>=1)=1-0.271

P(X>=1)=0.729

The old 1 minus trick, or compliment rule, saves the day again. I hope this helps.