# Rolling a die twice in either order with not mutual exclusive events

I am reading the following problem:

If a single die is rolled twice, find the probability of rolling an odd number and a number greater than $$4$$ in any either order

My solution:
Probability of rolling an odd number = $$\frac{3}{6}$$
Probability of rolling a number greater than $$4$$ = $$\frac{2}{6}$$
Since the events are not mutually exclusive ($$5$$ is counted twice) the probabilty of throwing an odd number followed by a number greater than $$4$$:
$$\frac{3}{6}\cdot \frac{2}{6} - \frac{1}{36} = \frac{5}{36}$$

The probablity of throwing a number greater than $$4$$ followed by an odd number is: $$\frac{2}{6}\cdot \frac{3}{6} - \frac{1}{36} = \frac{5}{36}$$

Therefore the my final solution is $$\frac{5}{36} + \frac{5}{36} = \frac{10}{36}$$

But it is wrong, as the solution states that it should be $$\frac{11}{36}$$

What am I doing wrong here?

• What is 1/36, and why do you think it corrects for double counting? Mar 28, 2021 at 21:35
• @AleksejsFomins: The $5$ is odd and greater than $4$ so to avoid counting that twice I subtract this
– Jim
Mar 28, 2021 at 21:47
• As stated by @Karl below, the mistake seems to be in subtracting it from both orders Mar 28, 2021 at 21:50
• @AleksejsFomins: I thought that if events are not mutually independent we subtract the common event occurence. E.g. if the question was about the probability of odd followed by number greater than $4$, should I subtract the $1$ occurence of $5$? I apply this logic for the reverse order too
– Jim
Mar 28, 2021 at 21:54

The outcome potentially counted twice is $$(5,5)$$, not $$5$$.

The number of pairs $$(a,b)$$ where $$a$$ is odd and $$b>4$$ is $$3\times2=6$$. This includes $$(5,5)$$.

The number of pairs $$(a,b)$$ where $$a>4$$ and $$b$$ is odd is $$2\times3=6$$. This includes $$(5,5)$$.

The sum of these is $$6+6=12$$, but $$(5,5)$$ was counted twice; there are actually only $$11$$ pairs in the union of these two sets.

• I thought that when having events that are not mutually exclusive we always subtract the $\bigcap$. So for the number of pairs (𝑎,𝑏) where 𝑎 is odd and 𝑏>4 is 3×2=6. I should not include (5,5)
– Jim
Mar 28, 2021 at 21:51
• I'd suggest just trying to think clearly about the outcomes you want to count instead of focusing on formulaic rules. If we wanted the probability of getting odd or $>4$ in one roll of the die, your approach would be applicable: $$|\{odd\}\cup\{>4\}|=|\{odd\}|+|\{>4\}|-|\{odd\}\cap\{>4\}|=|\{1,3,5\}|+|\{5,6\}|-|\{5\}|=3+2-1=4.$$ But for the given problem we instead have $$|\{(odd,>4)\}\cup\{(>4,odd)\}|=|\{(odd,>4)\}|+|\{(>4,odd)\}|-|\{(odd,>4)\}\cap\{(>4,odd)\}|=6+6-1=11.$$
– Karl
Mar 28, 2021 at 22:20

In saying that "5 is counted twice", meaning (presumably) you're removing the duplicate event $$(5, 5)$$, what you should be doing is saying "The event $$(5, 5)$$ is part of both of the cases I've considered, so I need to only count it once, so I will remove it once from my calculation."

Instead, what it looks like you've done is removed it from both of your cases, each of which assumes the other case has already counted it.

• Since an odd number and a number greater than $4$ are not mutually exclusive, my understanding is that since $5$ appears in both events it should be counted once. Hence if we focus on odd followed by a number greater than $4$ we should subtract $1$ case. Similar logic for the reverse order.
– Jim
Mar 28, 2021 at 21:52
• In rolling an odd number followed by a number greater than 4, rolling the 5 first isn't the same as rolling the 5 second so you don't remove it. Mar 29, 2021 at 21:45
• There are only 36 possibilities anyway, so try writing them out explicitly and counting them all? That might help you see what you missed. Mar 29, 2021 at 21:46