# Thread: A Pair of Dice

1. ## A Pair of Dice

A pair of dice was thrown. What is the probability of getting 2 on the first die with any number on the second die OR getting any number on the first die with number 2 on the second die?

My Attempt was Like this

(1/6)(6/6) + (6/6)(1/6) = 0.3333

The book answer is 0.31. Where did I go wrong?

2. ## Re: A Pair of Dice

You have not done anything wrong. 1/3 is the correct answer.

3. ## Re: A Pair of Dice

Thanks a Lot.

4. ## Re: A Pair of Dice

Actually, you are double counting the case when you get 2 and 2. It is counted in both probabilities. You need to subtract it off once. The correct probability is:

$\dfrac{1}{6}\dfrac{6}{6}+\dfrac{6}{6}\dfrac{1}{6} - \dfrac{1}{6}\dfrac{1}{6} \approx 0.31$

5. ## Re: A Pair of Dice

Originally Posted by SlipEternal
Actually, you are double counting the case when you get 2 and 2. It is counted in both probabilities. You need to subtract it off once. The correct probability is:

$\dfrac{1}{6}\dfrac{6}{6}+\dfrac{6}{6}\dfrac{1}{6} - \dfrac{1}{6}\dfrac{1}{6} \approx 0.31$
Originally Posted by joshuaa
A pair of dice was thrown. What is the probability of getting 2 on the first die with any number on the second die OR getting any number on the first die with number 2 on the second die?
$\begin{array}{*{20}{c}}{(1,1)}&{(1,2)}&{(1,3)}&{(1 ,4)}&{(1,5)}&{(1,6)}\\{(2,1)}&{(2,2)}&{(2,3)}&{(2, 4)}&{(2,5)}&{(2,6)}\\{(3,1)}&{(3,2)}&{(3,3)}&{(3,4 )}&{(3,5)}&{(3,6)}\\{(4,1)}&{(4,2)}&{(4,3)}&{(4,4) }&{(4,5)}&{(4,6)}\\{(5,1)}&{(5,2)}&{(5,3)}&{(5,4)} &{(5,5)}&{(5,6)}\\{(6,1)}&{(6,2)}&{(6,3)}&{(6,4)}& {(6,5)}&{(6,6)}\end{array}$

@joshuaa, To see what SlipEterneral's point study the outcome table. There are only eleven pairs that contain a $2$.
So the answer is $\dfrac{11}{36}$

I do not like the wording of this question. If one throws a pair of dice onto a gaming table, which one is the first and which the second? Now even if one die is red and the other blue, would not change the answer to the question as stated.

6. ## Re: A Pair of Dice

SlipEternal & Plato

What you did was awsome!

Imagine that this problem came in the exam which means I would not know the correct answer is 0.31. Also, I would not have enough time to make the outcome table. If I used the rules I have been taught the answer will be 0.33 and I would be confident it is correct since I used the correct way. But of course the answer would be wrong. Is there a way that can alert me that I have to do extra calculations Like subtracting (1/6)(1/6) from the answer?

7. ## Re: A Pair of Dice

The idea is to fully understand what the numbers mean rather than just plugging things into formulas without any understanding. The best way I found is to get lots of experience with different types of problems.

8. ## Re: A Pair of Dice

You are right. I have to try diverse problems. Thanks a lot.