# Probability problem - binomial distibution

The keeper of a certain king’s treasure receives the task of filling each of 100 urns with 100 gold coins. While fulfilling this task, he substitutes one lead coin for one gold coin in each urn. The king suspects deceit on the part of the sentry and has two methods at his disposal of auditing the contents of the urns. The first method consists of randomly choosing one coin from each of the 100 urns. The second method consists of randomly choosing four coins from each one of 25 of the 100 urns. Which method provides the largest probability of uncovering the deceit?

The answers is "the second method" was solved by binomial distribution. The probabilities are 0.634 and 0.640 accordingly to the textbook "Understanding Probability"

Thanks!

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Hint: make it more formal: assign some names to events that can happen. – trutheality Jun 13 '11 at 5:22

Hint: call $p_1$ the probability that a given urn reveals the deceit when one draws one coin from it and $p_4$ when one draws four coins. Your first step would be to compute $p_1$ and $p_4$. The text of the exercise probably means that the urn reveals the deceit if and only if the lead coin is the coin drawn, respectively is among the four coins drawn. Then $p_1$ and $p_4$ are very simple functions of $n$ the number of coins in each urn. Your second step would be to write the global probabilities $P_1$ and $P_4$ of uncovering the deceit as functions of $p_1$ and $p_4$ respectively. Sub-hint: consider the probabilities of not uncovering the deceit.

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It is not actually necessary to calculate the numbers to reach the conclusion, though you may need to satisfy your teacher.

The expected number of lead coins found is $\frac{1}{100}$ times the number of coins inspected.

If you only looked at four coins, either drawn from four different urns or from the same urn, then you would be more likely to find two or more lead coins in the first case, and so also more likely to find zero lead coins in the first case (you might work out these probabilities of finding zero). So the probability of finding no lead coins when repeating these 25 times is higher in the first case (again you can work this out). So in the original problem the probability of finding at least one lead coin is higher in the second case (easily calculated from the previous numbers).

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Is the fact that it is more likely to find zero lead coins in the first case, a consequence of the fact that it is more likely to find two or more lead coins in the first case? – Did Jun 13 '11 at 7:10
@Didier Piau: Yes, since it is impossible to find two or more lead coins in the second case when looking at four coins from a single urn, and the expected number of lead coins found is the same in both cases. – Henry Jun 13 '11 at 8:19
Right, since $E(X)\ge1-P(X=0)$ for every nonnegative integer valued $X$, with an equality if and only if $X=0$ or $1$ with full probability. Thanks. – Did Jun 13 '11 at 9:10

thanks for help

regarding Didier Piau hint

$$p1=\frac{99}{100}$$ - uncovering deceit by the first method ,so

$$P1 = ( \frac{99}{100} ) ^ {100}$$

probability of uncovering the deceit for the second method might be looking like that

$$p2 = ( 1 - \binom{100}{4} ( \frac{1}{100} ) ^ 4 ( \frac {99}{100} ) ^ {96} )$$

and

$$P2 = p2 ^ {100}$$

I am not sure if this's right , but at least it makes sense for me :)

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Sorry but you are off the mark: the first step, using $n=100$, should give you $p_1=1/n$ and $p_4=4/n$. Let me suggest that now, either you signal that you understand why these formulas hold, or you explain what is unclear to you in them. Then, we shall proceed to the next step. – Did Jun 13 '11 at 17:46
I understand the way you are thinking and I just can't figure out how are you going to proceed the solution with your p1 and p4. Please show the continue at least with p1. Thanks! – com Jun 15 '11 at 4:10
First step: do you see why $p_1$ is $1/n$ and not $(n-1)/n$ and can you explain the reason, in formulas or in words? – Did Jun 15 '11 at 15:31
Of course p1=1/n - probability of covering the deceit for one urn. I used p1=(n-1)/n as probability of uncovering the deceit for one urn ,so the probability of uncovering for all runs will be P1=(99/100)^(100) – com Jun 15 '11 at 17:35
Right, $p_1$ is $1/n$. But forget this value for a moment and just call it $p_1$. Next question is: What could be $P_1$ as a function of $p_1$ and $n$? That is, you have $n$ trials, each with a probability $p_1$ of success, and you want the probability $P_1$ that at least one of these $n$ is successful (maybe one of them, maybe more, but not zero). Hint: $P_1$ is not $(1-p_1)^n$... – Did Jun 15 '11 at 17:42