Pr of exactly one dice showing 2

[navi]

So I ran into this problem:

You roll two fair die. What is the probability of one of the dice showing 2 or the sum being at least 8?

I thought it should be: 15/36 for the sum being at least 8 and 10/36 for exactly one dice showing 2, but it tells me it is wrong. I got these numbers by constructing a chart that showed the outcomes of the 2 die rolls.

 

[Opalg]

So I ran into this problem:

You roll two fair die. What is the probability of one of the dice showing 2 or the sum being at least 8?

I thought it should be: 15/36 for the sum being at least 8 and 10/36 for exactly one dice showing 2, but it tells me it is wrong. I got these numbers by constructing a chart that showed the outcomes of the 2 die rolls.

You have given two answers, but the question wants a single answer.

Let $A$ stand for "the sum being at least 8", and let $B$ stand for "exactly one of the dice showing 2". You have correctly found the probabilities $\def\Prob{\operatorname{Prob}\,}\Prob(A) = \frac{15}{36}$ and $\Prob(B) = \frac{10}{36}.$ But the question asks for $\Prob(A\text{ or }B)$, in other words the probability of the dice showing either $A$ or $B$ (or possibly both). For that, you need to use the formula $\Prob(A\text{ or }B) = \Prob(A) + \Prob(B) - \Prob(A\text{ and }B)$.

 

[navi]

You have given two answers, but the question wants a single answer.

Let $A$ stand for "the sum being at least 8", and let $B$ stand for "exactly one of the dice showing 2". You have correctly found the probabilities $\def\Prob{\operatorname{Prob}\,}\Prob(A) = \frac{15}{36}$ and $\Prob(B) = \frac{10}{36}.$ But the question asks for $\Prob(A\text{ or }B)$, in other words the probability of the dice showing either $A$ or $B$ (or possibly both). For that, you need to use the formula $\Prob(A\text{ or }B) = \Prob(A) + \Prob(B) - \Prob(A\text{ and }B)$.

Thank you! I do not understand why you are subtracting the probability of A and B, though. Could you explain to me why?

 

[Opalg]

Thank you! I do not understand why you are subtracting the probability of A and B, though. Could you explain to me why?

It's in order to avoid double counting. Specifically, if the dice show 2 and 6, then they satisfy both conditions $A$ and $B$. If you just add the probabilities for $A$ and $B$ then you will have counted that case under both headings. So you need to subtract one of them off.

 

[navi]

It's in order to avoid double counting. Specifically, if the dice show 2 and 6, then they satisfy both conditions $A$ and $B$. If you just add the probabilities for $A$ and $B$ then you will have counted that case under both headings. So you need to subtract one of them off.

Okay, now I understand! I do have a follow up question, though: why would the probability of the union of A and B be 1/36 and not 2/36? Because it could be first dice 2, second dice 6 or first dice 6, second dice 2.

 

[Opalg]

Okay, now I understand! I do have a follow up question, though: why would the probability of the union of A and B be 1/36 and not 2/36? Because it could be first dice 2, second dice 6 or first dice 6, second dice 2.

It should indeed be 2/36.

 

[navi]

It should indeed be 2/36.

It is a WebWork problem and it accepted this as the correct answer: 10/36 + 15/36 - 1/36

 

[Opalg]

It is a WebWork problem and it accepted this as the correct answer: 10/36 + 15/36 - 1/36

Here is a possible explanation for that. You entitled this thread "Pr of exactly one dice showing 2", but the statent of the problem only refers to "one of the dice showing 2". I suspect that this may be intended to cover the case when both dice show 2. In that case, the probability of that event goes up from $\frac{10}{36}$ to $\frac{11}{36}$, and the probability of "$A$ or $B$" becomes $\frac{11}{36} + \frac{15}{36} - \frac{2}{36}$, which agrees with the answer accepted by WebWork.

 

[HOI]

By the way, your title, "Pr of one dice showing 2", should be "Pr of one die showing 2" and "you roll two fair die" should be "you roll two fair dice". "Dice" is the plural of the singular "die".

 

[Wilmer]

1st roll: must be 2 or 6: probability = 2/6 = 1/3
2nd roll: must be 2 if 1st was 6, or 6 if 1st was 2: probability = 1/6

1/3 * 1/6 = 1/18