Question 1190172: Hello, I would appreciate it if you could help me out!
Donna reaches into her purse and randomly draws two coins from those that have accumulated at the bottom of her purse. The coins are five nickels, three dimes, and four quarters.
A. Determine the probability for each of the outcomes.
B. What is the expected amount that Donna withdraws from the two coins?
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Part A
5 nickels + 3 dimes + 4 quarters = 12 coins total
N = nickel
D = dime
Q = quarter
Notation like NN refers to selecting two nickels.
Something like ND means a nickel and a dime in either order, NQ means a nickel and a quarter, and so on.
Here is the sample space of all possible outcomes
NN, DD, QQ
ND, NQ
DQ
There are 6 items in this sample space.
You can make a 3 by 3 table to help list out the possibilities.
Assume that Donna is not replacing the coins when selecting them
I'll be using this formula
P(A and B) = P(A)*P(B given A)
P(NN) = probability of selecting two nickels
P(NN) = P(N first)*P(N second, given N is first)
P(NN) = (5/12)*(4/11)
P(NN) = 20/132
Let's keep this fraction unreduced. I'll do the same for the other results as well.
You'll see why in a moment.
P(DD) = probability of selecting two dimes
P(DD) = P(D first)*P(D second, given D is first)
P(DD) = (3/12)*(2/11)
P(DD) = 6/132
P(QQ) = probability of selecting two quarters
P(QQ) = P(Q first)*P(Q second, given Q is first)
P(QQ) = (4/12)*(3/11)
P(QQ) = 12/132
P(ND) = probability of getting a nickel and a dime (either order)
P(ND) = P(N first)*P(D given N is first) + P(D first)*P(N given D is first)
P(ND) = (5/12)*(3/11) + (3/12)*(5/11)
P(ND) = 15/132 + 15/132
P(ND) = 30/132
P(NQ) = probability of getting a nickel and a quarter (either order)
P(NQ) = P(N first)*P(Q given N is first) + P(Q first)*P(N given Q is first)
P(NQ) = (5/12)*(4/11) + (4/12)*(5/11)
P(NQ) = 20/132 + 20/132
P(NQ) = 40/132
P(DQ) = probability of getting a dime and a quarter (either order)
P(DQ) = P(D first)*P(Q given D is first) + P(Q first)*P(D given Q is first)
P(DQ) = (3/12)*(4/11) + (4/12)*(3/11)
P(DQ) = 12/132 + 12/132
P(DQ) = 24/132
Summary of the probabilities
P(NN) = 20/132
P(DD) = 6/132
P(QQ) = 12/132
P(ND) = 30/132
P(NQ) = 40/132
P(DQ) = 24/132
Note that the fractions add to 1 to help confirm we have a probability distribution
You could choose to reduce these fractions, but I'll keep them as is so that the denominators are all the same.
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Part B
X = combined value of the two coins in cents
For the scenario of selecting NN, Donna has 5+5 = 10 cents in her hand.
For DD she has 10+10 = 20 cents, and so on.
Make a table for all the possible X values and their corresponding probabilities
| Coins | X | P(X) | | NN | 10 | 20/132 | | DD | 20 | 6/132 | | QQ | 50 | 12/132 | | ND | 15 | 30/132 | | NQ | 30 | 40/132 | | DQ | 35 | 24/132 |
Now multiply each X value with its corresponding P(X) value. Write the result in a third column labeled X*P(X)
For example, 10*(20/120) = 200/132 for the first row.
| Coins | X | P(X) | X*P(X) | | NN | 10 | 20/132 | 200/132 | | DD | 20 | 6/132 | 120/132 | | QQ | 50 | 12/132 | 600/132 | | ND | 15 | 30/132 | 450/132 | | NQ | 30 | 40/132 | 1200/132 | | DQ | 35 | 24/132 | 840/132 |
The last step is to add up everything in the X*P(X) column. Like earlier, I didn't reduce any of the fractions so I could keep the denominator of 132.
Since the denominator is the same for each fraction, we can ignore it for now and just add the numerators:
200+120+600+450+1200+840 = 3410
Then we divide over the denominator we left out earlier
3410/132 = 25.833
The value is approximate
Let's say Donna conducts 10,000 trials where she selects two coins at random without replacement.
Let's say she records the total value of the two coins in a spreadsheet.
Each new trial will have the original coins (meaning she puts the two coins back after the particular trial is over).
If she were to add up those results in the spreadsheet, and divide by the number of trials, then she should get somewhere close to 25.833
Answer: The expected value from the two coins is roughly 25.833 cents.
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