intro to stats and analytics

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. You randomly select six peanut M&M’s from an extra-large bag of the candies. (Round all probabilities below to four decimal places; i.e. your answer should look like 0.1234, not 0.1234444 or 12.34%.)

Compute the probability that exactly five of the six M&M’s are brown.

Compute the probability that four or five of the six M&M’s are brown.

Compute the probability that at most five of the six M&M’s are brown.

Compute the probability that at least five of the six M&M’s are brown.

If you repeatedly select random samples of six peanut M&M’s, on average how many do you expect to be brown? (Round your answer to two decimal places.)

With what standard deviation? (Round your answer to two decimal places.)

Answer 1

HI Lucy,

This is a binomial probability distribution, meaning essentially that you have two outcomes: "success" (getting the right color M&M) and "failure" (not getting the right color). Formula for probability of an exact number of successes:

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

n=sample size

x=number of "successes" you desire

p=probability of success

q=probability of failure=1-p

C(n,x) = n!/(x!(n-x)!

!=factorial i.e. 4*3*2*1, etc. You will likely need a calculator for these computations.

You will use these formulas frequently throughout this problem, so be prepared.

Question 1

You want exactly 5 brown M&Ms, so:

n=6

x=5

p=0.12--12% of all peanut M&Ms in box are brown

q=0.88=1-p

P(X=5) = C(6,5)*0.12⁵*0.88¹

C(6,5)=6!/(5!*1!)

C(6,5)=6

P(X=5)=6*0.12⁵*0.88¹

P=0.000131, to four decimals:

P=0.0001

It is very unlikely you get exactly five brown M&Ms.

Question 2

Now you are interested in 4 OR 5 brown M&Ms. We already know the probability for 5, so we need only to compute the probability for 4 and add the two together.

n=6

x=4

p=0.12

q=0.88

P(X=4) = C(6,4) * 0.12⁴*0.88²

C(6,4)=6!/(4!*2!) = 15

P(X=4) = 15* 0.12⁴ *0.88²

P(X=4)= 0.0024

Now, the question asked for 4 or 5 brown M&Ms, so we add:

P=0.0024 + 0.0001

P= 0.0025

Question 3

Now we want AT MOST 5 brown M&Ms. There are a couple ways we can approach this. We can compute the individual probabilities of 1, 2, and 3 brown M&Ms and add those to the probability for 4 or 5 above. That's a lot of work, though. The other option is to employ the Complement Rule aka the "One Minus Trick." Essentially, if we compute the probability that all six are brown, we can take one minus that probability and we will have our answer for the probability of 5 at most. So:

n=6

x=6

p=0.12

q=0.88

P(X=6)=C(6,6)*0.12⁶*0.88⁰

Remember, C(n,n) = 1 and a⁰=1, so really, we only need:

P(X=6) = 0.12⁶

P(X=6)=0.00000298

To four decimals, this simply rounds down to:

P=0.0000

Taking 1-0.0000

P=1.0000

It is a virtual, but not a true, certainty that you will not get six browns. The probability of that is so low that we cannot even round it to four decimals without getting all 0s.

Question 4

Now we want at least 5 brown M&Ms, so we add up the probabilities for x=5 and x=6. Recall P(X=6)=0.0000, essentially 0. So this is effectively equal to the probability in our first question

P(X>=5) = 0.0001

Question 5

Binomial mean formula is:

mu=np

n=sample size

p=probability of "success"

Here,

n=6

p=0.12

mu=6*0.12

mu= 0.72

Question 6

Standard deviation of a binomial distribution:

sigma=sqrt(n*p*q)

Variables are defined as above. Here,

n=6

p=0.12

q=0.88

sigma=sqrt(6*0.12*0.88)

sigma = 0.80

I hope this helps.