Expert: Josh - 9/2/2004

Question
You have two quarters, three dimes and four nickels in your pocket .  You reach into your pocket and randomly select three of the coins.  What is the probability that their total value is exactly 25 cents?    Express your answer as a common fraction.

Thank you  we are stumped as to how to figure this out and need how the answer came to be. I have to show my answer and how I got it.  I have two answers and they are so far off that is why I need your help.  Thank you again for trying.  Mary Ann  

Answer
Hi Mary Ann,

Here is a summary of each coin's worth.

The Penny is equivalent to 1¢ or $0.01.
(One hundred pennies make a dollar.)

The nickel is equivalent to 5¢ or $0.05.
(Twenty nickels make a dollar.)

The Dime is equivalent to 10¢ or $0.10.
(Ten dimes make a dollar.)

The Quarter is equivalent to 25¢ or $0.25.
(Four quarters make a dollar.)

Let N=0.05, D=0.10, Q=0.25.
You have to select three items from {Q,Q,D,D,D,N,N,N,N} in random, so that they add up exactly to 0.25.

Part I - Initial reasoning
--------------------------
1. If we include a "Q" amongst the three coins, (which is worth 0.25 by itself), we will always exceed a total sum of 0.25. So, we can exclude any combination, which involves "Q","x","x" appearing in any order. Here, we don't care what "x" represents, it can be N,D or Q.

2. What other arrangement can possibly lead to a sum which is precisely 0.25?
Case 1: we can select at most two dimes (2D or 2xD) and this would take our subtotal to 2*0.10=0.20. Then, we must select a nickel (N) to bring the total to 0.25.

Case 2: observe that we can replace a dime with two nickels. If we select "N","N","D","N" in any order, this will also give us a sum of 3*0.05+0.10=0.25. However, this violate the condition that we can only pick three coins.

So, in fact, case 1 is the only condition which would give us a total of $0.25.

Part II - Considering the different combinations
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There are now two more things to think about.
Question 1: How many ways can we select three coins in random, given the set of coins {Q,Q,D,D,D,N,N,N,N}?

Question 2: How many ways can we select the three coins, in random, from the set of coins {Q,Q,D,D,D,N,N,N,N}, such that they satisfy the condition that their "total value is 0.25".

Important assumptions:
1) The order of appearance is not important.
2) The coins selected from the set will not be replaced (put back into the pocket before another selection).

Answer to 1: We work this out systematically.
The indistinguishable combinations are:
If Q is included,
{Q,D,N}
{Q,D,D}
{Q,N,N}
{Q,Q,D}
{Q,Q,N}
If Q is excluded,
{D,N,N}
{D,D,N}
{D,D,D}
If Q and D are both excluded,
{N,N,N}                       ...[#A]

Answer to 2: As we have analyzed in part I, amongst the 9 different configurations listed above, only combinations of {D,D,N} will give us a sum of 0.25.

Part III - Probability of occurrence
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Let us use the notation P({a,b,c}) to denote the probability of choosing the coins "a","b" and "c" from the set {Q,Q,D,D,D,N,N,N,N} in ANY given order.

Strictly speaking, we should use the notation P(a,b|a,c|(a,b)) to denote the probability of choosing the coins "a","b" and "c" strictly in this order. What this means is that we pick "a", then pick "b" given that we have already selected "a", and finally pick "c" given that we have already selected "a" and "b".
But to keep things simple (for your sake), we will use P("a,b,c") to represent the probability of choosing the coins "a","b" and "c" strictly in this order.
You must explain what you mean by these notations in your answer. Now, we are set.

Consider the first configuration from [#A] above,
P({Q,D,N})=P("Q,D,N")+P("Q,N,D")+P("D,N,Q")+P("D,Q,N")+P("N,Q,D")+P("N,D,Q") ......[#1]

I need your full attention to understand the following.

Consider each of the terms in [#1] independently.
Recall that P("Q,D,N") represents the probability of selecting "Q", followed by "D", then "N" strictly in this order.
From the set of coins {Q,Q,D,D,D,N,N,N,N}
.There is a 2/9 chance of picking a "Q" in the first selection. We write P(Q)=2/9 for this.
Now, we are left with {Q,D,D,D,N,N,N,N}
.Given that we have selected a "Q", there is now only a 3/8 chance of picking up a "D" in the second selection. We write P(D|Q)=3/8 for this.
Now, we are left with {Q,D,D,N,N,N,N}
.Given that we have already selected a "Q", followed by a "D", there is now a chance of 4/7 of picking a "N" in the third and final selection. We write P(N|D,Q)=4/7 for this.
.So, the probability of selecting the string "Q,D,N" is given by P("Q,D,N") = (2/9)*(3/8)*(4/7).
Repeating this argument for the other terms in [#1],
We find that
P("Q,N,D")
=P("D,N,Q")
=P("D,Q,N")
=P("N,Q,D")
=P("N,D,Q")
=(2/9)*(3/8)*(4/7).

So, the probability for P({Q,N,D})=6*(2/9)*(3/8)*(4/7).

...
You need to repeat this procedure for the other eight configurations given in [#A].
If we look at just one more example, say, {Q,D,D}.

P({Q,D,D})=P("Q,D,D")+P("D,Q,D")+P("D,D,Q")

From the set of coins {Q,Q,D,D,D,N,N,N,N}
.There is a 2/9 chance of picking a "Q" in the first selection. We write P(Q)=2/9 for this.
Now, we are left with {Q,D,D,D,N,N,N,N}
.Given that we have selected a "Q", there is now only a 3/8 chance of picking up a "D" in the second selection. We write P(D|Q)=3/8 for this.
Now, we are left with {Q,D,D,N,N,N,N}
.Given that we have already selected a "Q", followed by a "D", there is now a chance of 2/7 of picking another "D" in the third and final selection. We write P(D|D,Q)=2/7 for this.
.So, the probability of selecting the string "Q,D,D" is given by P("Q,D,D") = (2/9)*(3/8)*(2/7).
Repeating this argument for the other terms in [#1],
We find that
P("Q,D,D")
=P("D,Q,D")
=P("D,D,Q")
=(2/9)*(3/8)*(2/7).

So, the probability for P({Q,D,D})=3*(2/9)*(3/8)*(2/7).

Exercise: You now have to work out the probability for the relevant (or remaining 7) configurations. You should find, if I have not mistaken, that
P({Q,D,N})=6*(2/9)*(3/8)*(4/7)
P({Q,D,D})=3*(2/9)*(3/8)*(2/7)
P({Q,N,N})=3*(2/9)*(4/8)*(3/7)
P({Q,Q,D})=3*(2/9)*(1/8)*(3/7)
P({Q,Q,N})=3*(2/9)*(1/8)*(4/7)
P({D,N,N})=3*(3/9)*(4/8)*(3/7)
P({D,D,N})=3*(3/9)*(2/8)*(4/7)
P({D,D,D})=(3/9)*(2/8)*(1/7)
P({N,N,N})=(4/9)*(3/8)*(2/7) by inspection.

Please, after all this work, check that the answers make sense. What do I mean by that? Well, for a start all probabilities must have a value between 0 and 1. So, each of the terms P({.,.,.}) must be bounded between 0 and 1 for a start. Otherwise, it makes no sense. What you really should make sure of is that the sum of the probabilities for these 9 configuration indeed add up to 1. This means that we have covered all possible outcomes and haven't left anything out.

Part IV - Final answer to your question

All you need to compute for your answer is P({D,D,N}). It has a value of 72/504 (approx. 14.28%)

But now having seen this, you could have answered a whole lot more, right?

Cheers,
Josh

P.S. It is only fair to other students that you acknowledge the help that you have got from an anonymous source and demonstrate that you have at least fully understood the procedure and the theory behind it. I don't want to take any personal credit, but I think you should also help other students if they approach you for help. A candle doesn't lose its flame from litting others.