SOLUTION: You have 2 dimes, 6 nickels, and 7 pennies in your pocket. You randomly remove two coins without replacement. A. What is the probability that both of them are nickels? B. What is

Hi, there--

THE PROBLEM:
You have 2 dimes, 6 nickels, and 7 pennies in your pocket. You randomly remove two coins 
without replacement.

A. What is the probability that both of them are nickels?
B. What is the probability that the first is a nickel and the second is a dime?

A SOLUTION:
Before you use your calculator you need to set up the solution. I assume you want to use the
combinatoric method since you are using a calculator. 

A. What is the probability that both coins are nickels?

This is a random sampling without replacement from a finite population (15 coins) whose 
elements can be classified into two mutually exclusive categories (nickels, not nickels)

X is the random variable = the result of drawing a coin from your pocket
N is the population size = 15 (the total number of coins in your pocket)
K is the number of "successful" coins = 6 nickels (these are the ones you want to draw)
n is the number of draws = 2
k is the number of successes = 2 (because you want to draw 2 nickels)
(aCb) binomial coefficient, or counting combinations (the [nCr] button on your calculator)

The general formula for this type of probability distribution is

P(X=k) = [(KCk) * ((N-K)C(n-k))] / (NCn)

Don't freak out at all these letters. I'll explain each one in the context of your problem:

P(X=k) is P(X=2), the probability that you will draw successfully twice (two nickels.)

KCk is 6C2, or "6 choose 2," because there are 6 successful coins (nickels) and you want to 
draw two of them. "6 choose 2" is the number of ways to select 2 things from a set of 6 when 
order doesn't matter.

(N-K)C(n-k) is 9C0, or "9 choose 0," because there are 15-6=9 unsuccessful coins (not 
nickels) and you don't want to choose any of them.

NCn is 15C2, or "15 choose 2," because there are 15 coins all together and you are choosing 
2 of them.

So, the probability of drawing two nickels is
P(X=2) = [(6C2)*(9C0)] / (15C2) 

Now we are ready for the calculator. Do this in three steps.

For 6C2, press these buttons:
[6] [MATH] [<] [3] [2] [ENTER]

(You should get 15.) The [MATH] button takes you to the Math commands, and pressing the 
left arrow key once takes you to the PRB menu. Choosing [3] brings up the nCr function. Finish 
by typing [2] [ENTER] to get 15.

For 9C0, press:
[9] [MATH] [<] [3] [0] [ENTER]. (You should get 1.)


For 15C2, press:
[1] [5] [MATH] [<] [3] [2] [ENTER] (You should get 105.)

p(X=2) = (15)(1)/105 = 1/7


B. What is the probability that the first is a nickel and the second is a dime.

I personally would not use a calculator this this problem.

The probability of drawing a nickel, then a dime is

[probability of drawing the nickel]

P(draw nickel) = 6/15 because there are 15 coins altogether and 6 chances to pick on of them.

P(draw a dime after a nickel) = 2/14 because there are 14 coins left in your pocket and 2 of 
them are dimes.

P(draw nickel, then a dime = (6/15)(2/14) = 2/35

In the calculator method you have

[(6C1)*(9C0)/(15C1)] = 2/35

Hope this helps! If you have questions about any of this, email me and I'll explain further.

Mrs. Figgy
math.in.the.vortex@gmail.com