Economic Statistics I

Does anyone have a formula for these type of problems?

## 30% of all Girl Scout cookies sold are Thin Mints. If we select 9 boxes of cookies at random, what is the probability at most 5 of them are Thin Mints?

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# 4 Answers

Short answer: you will need the binomial theorem. I'm typing out a response, but it will take a few minutes.

Long answer:

The precise answer requires that we use the binomial theorem. The binomial theorem states that, given a probability p of an event happening, the probability of a certain number of successes is equal to n_C_k (p^k)(1-p)^(n-k). n_C_k is the number of ways that you can choose k things from a total of n objects. p is the probability of the success, and, because probabilities have to total to 1, (1-p) is the probability of failure.

Now, how do we apply this in this case? You want to find out the probability of getting at most 5 of them with thin mints.

A success in this case means that the box is thin mints. A failure in this case means a box is not thin mints. So p = probability of a box being thin mints = 0.3, and 1- p = probability of a box not being thin mints = 0.7.

Long answer:

The precise answer requires that we use the binomial theorem. The binomial theorem states that, given a probability p of an event happening, the probability of a certain number of successes is equal to n_C_k (p^k)(1-p)^(n-k). n_C_k is the number of ways that you can choose k things from a total of n objects. p is the probability of the success, and, because probabilities have to total to 1, (1-p) is the probability of failure.

Now, how do we apply this in this case? You want to find out the probability of getting at most 5 of them with thin mints.

A success in this case means that the box is thin mints. A failure in this case means a box is not thin mints. So p = probability of a box being thin mints = 0.3, and 1- p = probability of a box not being thin mints = 0.7.

The probability of having at most 5 boxes of thin mints out of 9 total boxes can be calculated one of two ways:

P(<=5) = P(0) + P(1)+ P(2) + P(3) + P(4) + P(5) (formula 1)

OR

P(<=5) = P(0) + P(1)+ P(2) + P(3) + P(4) + P(5) (formula 1)

OR

P(<=5) = 1 - P(9) - P(8) - P(7) - P(6) (formula 2)

where P(number) is the probability that you have exactly "number" boxes of thin mints in the sample.

Notice that the second formula takes advantage of the fact that the probability of all possibilites is 1. You can always calculate the probability of a certain condition by adding up the probability of all outcomes that satisfy that condition (formula 1) or start from a total probability of 1 and subtract the probability of each outcome that does

*not*satisfy the condition.

(Simply put: you can either build a sculpture piece by piece, or start from the whole block and chisel out what you don't want -- the result is basically the same, at least mathematically.)

Because the second formula involves fewer terms, let's go with that one. We sitll have to calculate P(9), P(8), P(7), P(6).

You still with me?

Ok.

Here's P(9):

P(9) = 9_C_9 *(0.3)^9*(0.7)^0

= 1*(0.3)^9*(0.7)^0

= 1*(0.3)^9*(1)

= 0.3^9

= 0.000019683

(By the way, you can calculate n_C_k using the COMBIN function in Excel, or using your calculator.)

P(8)= 9_C_8*(0.3)^8*(0.7)^1

= 9*(0.3)^8*(0.7)^1

= 9*(0.3)^8*(0.7)

= 0.00413343

Similarly, for P(7) and P(6), I get

P(7) = 0.003857868

P(6) = 0.021003948

Plugging it into the formula 2 for P(<=5), I get

P(<=5) = 1 - 0.000019683 - 0.00413343 - 0.003857868 - 0.021003948

= 0.974705158

Obviously, you should probably round to whatever your teacher recommends (usually about 4 decimal places).

P(<=5) = 0.9747

Now, this may have seemed "hard", but the key fundamental conceptual principle is the application of the binomial theorem. Take some time to struggle with it and convince yourself that the probability of precisely a certain number of thin mint boxes

*k*appearing in a sample

*n*follows the binomial formula. Once that's done, then convince yourself that the probability for the given condition is equal to the formulas listed above (formula 1 and formula 2) by the rules of probability.

Is this the exact wording of the problem?

It's not clear if the selection of each box is with or without replacement-are you drawing 9 boxes all at once, or are you choosing one at a time and then putting the box back before drawing again?

Also, it's unclear what the probability is of *choosing* a box of Thin Mints. Knowing 30% of boxes *sold* are Thin Mints doesn't tell you anything about how many Thin Mints boxes are present in the "population" of cookie boxes you are drawing from!

Ryan Y. gave a great answer! I just wanted to add that you can use the Poisson Formula. (Yes, I know poisson=fish...) If you need to look up how he did his data, that's the name to use.

total mints boxes is 3