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n coins are given, among which exactly 3 are bad and heavier than the good ones. A balance is used to identify the bad coins. Assume k coins are picked in both sides of the balance at a time. What is the probability of

  1. left side being heavier
  2. right side being heavier
  3. both sides being equal in weight

Thanks.

  • 1
    2kCn is the number of ways to choose$n$coins from 2k coins. Since both sides must have an equal (k) number of coins, it's slightly more complicated. It might help to think about first choosing $k$ coins for the left side, and then choosing $k$ coins for the right side, taking into account that for the right side, you have $n-k$ coins to choose from.2012-05-11

4 Answers 4

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I'm not great at statistics (which is why I try answering these questions sometimes), but here it goes and I'll count on the community to tell me if I'm wrong.

Because I'm not great, I have to break these problems down into small quite elementary chunks.

First, I started with the probability that each side is equal. I believe that that could be expressed as $\mathbb{P}(\text{Both Sides Equal}) = \mathbb{P}(\text{0 Heavy}) + \mathbb{P}(\text{2 Heavy}) \cdot \mathbb{P}( \text{Even Split})$

I decided that that's the probability that you don't choose any heavy coins (thus they must be equal) or that you choose 2 heavies and one is on the left.

I started out by saying that: $\mathbb{P}(\text{0 Heavy}) = \frac{\binom{n - 3}{k}}{\binom{n}{k}}$

After trying to figure out $\mathbb{P}(\text{2 Heavy}) $ I think that I stumbled on this: $\mathbb{P}(i\text{ Heavy}) = \frac{\binom{3}{i}\binom{n - 3}{k - i}}{\binom{n}{k}}$

I also decided that most of the time I wanted to look at selecting both sides at once and then splitting them evenly so I started thinking of $k$ as $2k$.

OK, so that covered $\mathbb{P}(\text{0 Heavy})$ and $\mathbb{P}(\text{2 Heavy}) $, but I had to figure out the probability that given 2 heavies

Thanks to Gerry Myerson, I'm convinced this section was incorrect.

I'd randomly split them evenly between the sides of the scale. I decided I'd flip the heavy coins and if they were heads I'd put them on the left and tails I'd put on the right. That's metaphorically of course. But, basically, there are $2^2$ outcomes of flipping 2 coins and 2 of them result in an even split so $\mathbb{P}(\text{Even Split}) = 2 / 2^2 = 2 / 4 = .5$.

My new thought is that there will always be only 2 ways to have the 2 heavy coins together: both on the left or both on the right. There are $\binom{k}{\frac{k}{2}}$ ways to split the coins. Thus, I think this is correct: $\mathbb{P}(\text{Even Split}) = 1 - \frac{2}{\binom{k}{\frac{k}{2}}}$

If that's correct, I think, the probability of both sides weighing the same is: $\mathbb{P}(\text{Both Sides Equal}) =\frac{\binom{n - 3}{k}}{\binom{n}{k}}+\mathbb{P}(\text{Even Split})\cdot\frac{\binom{3}{2}\binom{n - 3}{k - 2}}{\binom{n}{k}}$

OK. So, the conditions for the left side being heavier than the right side are the same as the conditions for the right side being heaver than the left so $\mathbb{P}(\text{Left Heavier}) = \mathbb{P}(\text{Right Heavier})$.

Thus, I only tried to calculate $\mathbb{P}(\text{1 Side Heavier})$. I spent more time than I'd like to admit before I realized that $\mathbb{P}(\text{1 Side Heavier}) = \mathbb{P}(\neg\text{Both Sides Equal})$.

Nervously, I submit this answer as I run some tests to make sure I'm at least on the right track.

  • 0
    I see what you're saying. I was thinking about randomly assigning the coins to either side, but that obviously won't work. So, in your 4 coin scenario, you're saying there are 4 choose 2 ways to distribute the coins evenly. That makes sense as I'm picking half of the coins to put on the left. But then there are 2 choose 2 ways to put both coins on the same side and there are 2 sides. So really there are 2 / (k choose k/2) ways to be off balance (which would be a lot easier if I didn't redefine k. :)2012-05-12
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Hint: one side is heavier if it has more bad ones than the other. That can happen if one side has one bad coin and the other doesn't have any, or if one side has two bad ones and the other doesn't have any or ... There aren't too many possibilities. Can you see a relation between part 1 and part 2?

The number of unordered ways to pick left side coins that include one bad one are $3\binom {n-3} {k}$ You might see Wikipedia on combination if this is the question.

1

We first assume that $n \geqslant 2k$, and $k \geqslant 3$

Let's first sample $2k$ coins from the heap on $n$. The number of bad coins $X$ in the sample follows the hypergeometric distribution $\operatorname{Hyp}(2k,3,n)$. Then we sample $k$ coins to put on the left side from these $2k$ coins. The number of bad coins in this sample $Y$, conditioned on $X=x$ also follows the hypergeometric distribution, $Y|X=x \sim \operatorname{Hyp}(k,x,2k)$. The left side is heavier if $Y > X-Y$, i.e. $2Y > X$. $ \begin{eqnarray} \mathbb{P}(2Y > X) &=& \sum_{x=0}^3 \mathbb{P}(2(Y|X=x) > x) \mathbb{P}(X=x) = \sum_{y=0}^3 \sum_{x=0}^3 I(2y > x) \mathbb{P}(X=x) \mathbb{P}(Y|(X=x)=y) \\ &=& \frac{k \left(10 k^2-9 k n+6 k+3 n^2-6 n+2\right)}{(n-2) (n-1) n} \end{eqnarray} $ Similarly, the probability of right side being heavier is $\begin{eqnarray} \mathbb{P}(2Y < X) &=& \sum_{y=0}^3 \sum_{x=0}^3 I(2y < x) \mathbb{P}(X=x) \mathbb{P}(Y|(X=x)=y) \\ &=& \frac{k \left(10 k^2-9 k n+6 k+3 n^2-6 n+2\right)}{(n-2) (n-1) n} \end{eqnarray} $ Then, the probability of both sides being equal is $ \mathbb{P}(Y=X-Y) = 1 - \mathbb{P}(2Y>X) -\mathbb{P}(2Y

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Probability of either left or right side of balance being heavier is:

Probability of left/right side getting 2 heavy coins + probability of left / right side getting one heavy coin * probability of other side getting 0 heavy coins.

(k)C2 + {K(n-2k)/(n)C2 } / (n)C(2)

  • 0
    i wasn't, thank you. so it should have been (k)C3 + (k)C2 + {K(n-2k)/(n)C2 } / (n)C(2)2012-10-12