0
$\begingroup$

Suppose you have $4$ pieces of furniture, each with two drawers. Furniture pieces 1 and 2 have a gold coin in a drawer and a silver coin in the other drawer. Furniture piece 3 has a gold coin in each drawer and furniture piece 4 has a silver coin in each drawer. A random piece of furniture is chosen and one of the drawers opens. If the coin that appears is gold. What is the probability that the other drawer contains:
a) A silver coin?
b) A gold coin?

My try (for a)

I believe I have to use conditional probability:

Let $B = $ the event that the first coin is gold.

Let $A = $ given B, the event that the second coin is silver.

So I need to find $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$.

I believe that $P(B) = 1/2$ because there are the same chances to obtain a gold or silver coin.

My trouble is that I been thinking about how to calculate $P(A \cap B)$, which I believe is the same as the chances of getting one silver and then one gold coin, which could only happen if furniture piece 1 or 2 are chosen.

$\endgroup$
  • $\begingroup$ For part (a), you can just use $P(S|G) = \frac{P(F1).P(S|F1) + P(F2).P(S|F2) + P(F4).P(S|F4) }{P(G)} = \frac{\frac{1}{4}.\frac{1}{2} + \frac{1}{4}.\frac{1}{2} + \frac{1}{4}.0}{\frac{1}{4}.\frac{1}{2} + \frac{1}{4}.\frac{1}{2} + \frac{1}{4}.1}$ $\endgroup$ – ironX Sep 21 '19 at 0:22
  • 2
    $\begingroup$ Intuitively, there are $4$ silver coins, and you are equally likely to have chosen any of them. In $2$ cases, the other coin is gold, so $\Pr(G|S) = \frac24=\frac12$ $\endgroup$ – saulspatz Sep 21 '19 at 0:33
  • $\begingroup$ This tutorial explains how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Sep 21 '19 at 10:08
  • 3
    $\begingroup$ Possible duplicate of Probability problem $\endgroup$ – Matthew Daly Sep 21 '19 at 11:16
1
$\begingroup$

It seems very unusual and confusing to use "given $B,$ the event that the second coin is silver" as the description of an event. How do you interpret it? It should be a subset of the entire sample space; what is the sample space, and what members of it are in that subset?

There is some flexibility on how we describe the sample space, but as you have noted, certain things can happen only if we choose certain pieces of furniture, so I would make the choice of the piece of furniture part of the sample space. Perhaps like this: each member of the sample space consists of the choice of a piece of furniture and a sequence in which the coins are pulled out from that particular piece of furniture: $(n,c_1,c_2)$ where $n$ is the number of the piece of furniture and $c_1$ and $c_2$ describe the first and second coin pulled.

Whether you figure there are two possible sequences to pull coins from piece number $3$ (distinguishing $(3,G_1,G_2)$ from $(3,G_2,G_1)$) or just one sequence to pull the coins ($(3,G,G)$), the entire subset of the sample space in which piece number $3$ is selected should have a probability of $\frac14.$ I think you instinctively agree with that already. I also think sample spaces in which each member is equally probable are easiest to work with, so I'll distinguish $(3,G_1,G_2)$ from $(3,G_2,G_1)$. In particular, among other things,

$$ P(1,G,S) = P(2,G,S) = P(3,G_1,G_2) = P(3,G_2,G_1) = \frac18. $$

Those are the probabilities of all ways in which the first coin can be gold, and they are disjoint events, so $$ P(B) = 4\times\frac18 = \frac12. $$

I would simply define $A$ as the event in which the second coin is silver. It is the subset of the sample space containing all members in which $c_2$ is a silver coin. Then $P(A\mid B)$ is read as "the probability that the second coin is silver, given that the first coin is gold".

The event $A\cap B$ consists of all members of the sample space in which the second coin is silver and the first coin is gold -- not the events in which you get a silver coin and then a gold coin. It turns out (due to the particular setup of the probability space) that these two events have the same probability in this particular problem, but they are not the same event and you could easily make a problem in which they have different probabilities (for example by putting three drawers with three coins in each piece of furniture: one gold and two silver, all gold, or all silver).

To sum up the last paragraph in a nutshell, $A\cap B$ doesn't mean $A$ happens first.

So it turns out that the members of the sample space in $A\cap B$ are just the members of $B$ that have a silver coin in the third position (the second coin): that is, $A\cap B = \{(1,G,S),(2,G,S)\}.$ There is no other way to get a gold coin first and a silver coin second. So now you should easily be able to find $P(A\cap B).$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.