Counterfeit coin, conditional probability

Question

I am doing Problem AT9 [[ Harvard-MIT Math Tournament February 27, 1999 ]] here: https://hmmt-archive.s3.amazonaws.com/tournaments/1999/feb/adv/solutions.pdf

As part of his effort to take over the world, Edward starts producing his own currency. As part of an effort to stop Edward, Alex works in the mint and produces 1 counterfeit coin for every 99 real ones. Alex isn’t very good at this, so none of the counterfeit coins are the right weight. Since the mint is not perfect, each coin is weighed before leaving. If the coin is not the right weight, then it is sent to a lab for testing. The scale is accurate 95% of the time, 5% of all the coins minted are sent to the lab, and the lab’s test is accurate 90% of the time. If the lab says a coin is counterfeit, what is the probability that it really is?

I'm confused by the part that says:

The scale is accurate 95% of the time, 5% of all the coins minted are sent to the lab

If the scale is accurate 95% of the time, shouldn't the percentage of coins minted sent to the lab be$$(.01)(.95) + (.99)(.05) = .059?$$And not 5% as asserted?

Answer 1

The Organizers have given the Solution , which uses "Subjective Interpretation" of the Question. I do not agree with the Solution , because I think the Question itself is very ambiguous , & OP is rightly confused.
The Question & the Answers are not up to the Quality Standards.

Eg 1 : "and only .95% of the coins are sent to the lab and counterfeit" should be "and only 95% of the coins are sent to the lab are counterfeit"

Eg 2 : "5% of all the coins minted are sent to the lab" : Why ? because they failed the test ? or because of a quota ?

Eg 3 : "Since the mint is not perfect" : How imperfect is it ?

Eg 4 : "none of the counterfeit coins are the right weight" : Does this mean all real coins are the right weight ? Or is it Possible that real coins too have Deviations ?

Eg 5 : "Scale is accurate 95% of the time" : Provided the weight must be "$X$" , everything else is "not $X$" , this itself will Partially indicate the number of coins to be sent to the lab.

My Solution :

When we have $100000$ coins with $1000$ counterfeit coins , then $99000 \times 0.05 = 4950$ real coins will fail the Initial Check & $1000 \times 0.95 = 950$ counterfeit coins will fail that Initial Check.
We can not tell what happens to the other $50$ counterfeit coins whose weights are wrongly reported by the Scale : the "Subjective Interpretation" is that those counterfeit coins get magically reported with right weight.

Total coins sent to lab $4950+950=5900$

[[ It is matching what OP is rightly asking that "it must be .059 , not 5% as asserted" ]]

The lab then checks those coins. It is $90%$ accurate.
Hence , among real coins , $4950 \times 0.10 = 495$ are Wrongly reported.
While , among counterfeit coins , $950 \times 0.90 = 855$ are Correctly reported. Here too the "Subjective Interpretation" is that the other $95$ counterfeit coins magically get reported with Correct weight.

Thus , Probability that "when a coin is reported counterfeit , it really is counterfeit" is $855 / (855 + 495) = 855 / 1350 = 19/30$

SUMMARY :

Question given is ambiguous.
Answer given $19/28$ is wrong.
OP is right about it being confusing & wrong.
Answer must be $19/30$ , though it is necessary to make "Subjective Interpretations" to the Question.
Question should be reworded to make it rigorous & less ambiguous.
Answer should then be updated to match the Question.

ADDENDUM :

Normally , when Correct weight is $X$ & the coin has Wrong weight $Y$ , the Scale & the lab may wrongly report $Z$.
That is not the Case here !
The Scale & the lab are magically converting Wrong weight $Y$ to Correct weight $X$.

Answer 2

Question Maker View: Assume '5% of all the coins minted are sent to the lab'

It is given by the question and it means that 5% of coins will be sent to the lab, it is just a total amount and it is fixed by the qusetion. It is viewed as the estimation.

Why only .95% of the coins are sent to the lab and they are counterfeit: not the right weight: 1% and scale: 95% Multiply to get 0.95%

As 5% of all the coins minted are sent to the lab, so there is a 19% chance that a coin sent to the lab is counterfeit.

This value 19% chance comes from the calculation $\dfrac{0.95}{5}*$100% = 19%.

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Assume the weight of real coin is x. And there exists only two types of coins, real and counterfeit coins.

real:99%, counterfeit:1%

Since the weighing scale is accurate 95% of the time, divide the weighing parts into 4 cases. Case 1. real and correct weighing: 99% * 95% -> not sent to lab

Case 2. real and incorrect weighing: 99% * 5% -> sent to lab

Case 3. counterfeit and correct weighing: 1% * 95% -> sent to lab

Case 4. counterfeit and incorrect weighing: 1% * 5% -> sent to lab is unknown (unknown part)

Unknown part: if weight of counterfeit coin is $y\neq x$, it is possible that the counterfeit coin is weighted x incorrectly. Then it will not be sent to lab. If we combine part 2,3, it can been seen that $99\%*5\%+1\%*95\%>5\%$.

From Question Maker View, he assumes that '5% of all the coins minted are sent to the lab' and it is an estimation of case 2,3 and 4.

So I agree with the solution of Prem more than the solution of Question Maker.

Answer 3

Suppose that the mint produces a million coins. Of these, 10,000 (1%) are counterfeit, and 990,000 are real.

We are given that the scale used to determine whether to send a coin to the lab is 95% accurate. Assume that the same inaccuracy applies equally to false positives and false negatives. Then, 9,500 counterfeit coins will be sent to the lab.

Since there are a total of 50,000 (5%) coins sent to the lab, then there must be 40,500 real coins sent to the lab alongside the counterfeits. (Some will be incorrect-weight coins flagged correctly, and some will be correct-weight coins flagged incorrectly, but it doesn't matter what the breakdown of these is.)

Assume that the lab's 90% accurate rate applies independently to false positives and false negatives. Then:

• $(0.90)(9,500) = 8,550$ counterfeit coins will be correctly detected as counterfeit.
• $(0.10)(9,500) = 950$ counterfeit coins will incorrectly pass as real
• $(0.90)(40,500) = 36,450$ real coins will correctly pass as real.
• $(0.10)(40,500) = 4,050$ real coins will incorrectly be detected as counterfeit.

So the lab will detect 12,600 coins as counterfeit, of which 8,550 actually will be. The conditional probability in question is thus $\frac{8550}{12600} = \frac{19}{28} \approx 0.678571$.

Answer 4

The scale is accurate 95% of the time, 5% of all the coins minted are sent to the lab, and the lab’s test is accurate 90% of the time.

The use of the term 'accurate' is not very accurate here. The errors and inaccuracies can be of different kinds. We can have

• false positives: for a coin that is not a counterfeit, the scale/test detects a counterfeit.
• false negative: for a coin that is a counterfeit, the scale/test does not detect a counterfeit.

In these type of problems it is often assumed that both cases occur with the same rate (5% in this case).

If the scale is accurate 95% of the time, shouldn't the percentage of coins minted sent to the lab be$$(.01)(.95) + (.99)(.05) = .059?$$And not 5% as asserted?

I imagine that the problem was supposed to be computed as you did and consider false positive and false negative rates of 5%.

Then the 5% coins that are sent to the test is a mistake and should indeed be 5.9%.

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However, you can have a situation with 5% inaccuracy and 5% of the coins being sent to the test.

Let's look at it in a contingency table:

$$\begin{array}{c|cc|c} &\text{actually positive} & \text{actually negative}& \text{total} \\ \hline \text{weighted positive} &x&4+x&4+2x\\ \text{weighted negative} &1-x&95-2x& 96-2x\\\hline \text{total}&1&99&100 \end{array}$$

If we set the true positives cell (actually positive and weighted positive) equal to $x$ then the other 3 cells are all based on $x$ because of the following three conditions

• fixed marginals for total truly positives
• fixed marginals for total truly negatives
• the diagonal, sun of false positives and false negatives, equals 5%

And with a fourth condition we can also fix $x$

• the total number of positive weighted cases is 5%. ($4+2x = 5$)

Then $x = 0.5$.

This means that with a false negative rate of $\frac{1}{2} = 0.5$ and a false positive rate of $\frac{1}{22} \approx 0.045$ your total rate of error will be

$$0.01 \frac{1}{2} + 0.99 \frac{1}{22} = 0.05$$