I'm beginning some probability courses so please explain your reasoning as if I were stupid.

We have a 4-sided dice. Our experiment consists of rolling the dice twice:

  • Let event $A = \{$maximum of the two rolls is $2\}$.
  • Let event $B=\{$minumum of the two rolls is $2\}$.

The problem is asking to find whether or not the two events are independent; however, that's not the issue I'm having. I don't think finding out independence is too hard since you just apply the definition i.e. $P(A\cap B) = P(A)P(B)$ for independent events.

My issue is finding $P(A)$ and $P(B)$.

To reiterate, event $A$ is when the maximum of the two rolls is a 2. So, listing out all the possible outcomes we get:

  • (1,1), (2,1), (1,2), (2,2). Giving us 4 outcomes.

Event $B$ is when the minimum of the two rolls is a 2. Listing out all the possible outcomes we get:

  • (2,2), (3,2), (4,2), (4,3), (4,4), (3,4), (2,4), (2,3). Giving us 8 outcomes.

However, the book says event $A$ only has $3$ and that event $B$ only has $5$. Only possible logic I could think of why is that the author did away with "duplicates" such as $(2,1)$ and $(1,2)$ but I can't see why that is allowed since the two rolls are separate.

Can anyone explain this? Also, does anyone have a better title suggestion? LOL! Thank you.

  • 1
    $\begingroup$ For B you omitted (2,3) and (3,3) So that's 9 outcomes. I'd agree that for this kind of probability question it's more useful to treat (2,3) and (3,2) as distinct outcomes as then each outcome is equally likely. $\endgroup$ – IanF1 Jun 28 '15 at 23:20
  • $\begingroup$ Thank you! I have edited it in my post. $\endgroup$ – David South Jun 28 '15 at 23:23
  • $\begingroup$ Apologies, i then realised you missed out 3,3 as well while writing my answer $\endgroup$ – IanF1 Jun 28 '15 at 23:35

Each of the two rolls is independent. The best way of visualising the possibilities is to make a grid with the outcome of the first roll horizontally and the outcome of the second roll vertically. Then mark which of the squares in the grid correspond to A and which to B. \begin{array}{c c c c c} & 1 & 2 & 3 & 4\\ 1 & A & A & - & -\\ 2 & A & AB & B & B\\ 3 & - & B & B & B\\ 4 & - & B & B & B\\ \end{array}

So we can see there are 16 equally probable outcomes - each has probability $1/16$.

A is $4$ of the cases, giving a probability of $4/16 = 1/4$. B is $9$ of the cases, giving a prob of $9/16$.

It's fairly clear that A and B are not independent, as you correctly stated by the intersection formula.

I'm not sure why your book insisted on removing duplicates - if you treat $(2,3)$ as the same outcome as $(3,2)$, this outcome is twice as likely as $(2,2)$, making it not very useful for probability calculations.

  • $\begingroup$ Please see this tutorial on how to typeset mathematics on this site. I used an array to format the table you were having trouble aligning. $\endgroup$ – N. F. Taussig Jun 29 '15 at 8:55
  • $\begingroup$ Thanks NF. I'll have a practice. $\endgroup$ – IanF1 Jun 29 '15 at 16:35

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