# Elementary probability question involving a 4-sided dice rolled twice

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.

• 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. – IanF1 Jun 28 '15 at 23:20
• Thank you! I have edited it in my post. – David South Jun 28 '15 at 23:23
• Apologies, i then realised you missed out 3,3 as well while writing my answer – IanF1 Jun 28 '15 at 23:35

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 $9/16$.
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.