# Probability 2 Dice Question

1. Feb 12, 2014

### royalewithchz

1. The problem statement, all variables and given/known data

A pair of dice is rolled once, what is the probability that neither a doublet nor the sum of 10 will appear

2. Relevant equations

P(A) = 1 - P(A')

Demorgans law
(AUB)c = Ac ∩ B c

3. The attempt at a solution

I know how to do the solution through brute force and listing out all the possible scenarios:
(1,1), (1,2), (1,3) ...... (6,6) and doing it like that. But i wanted a more algebraic approach so I tried this:

Let A = is a doublet
Let B = sum of dices is 10

so then what I am looking for is

P(Ac U Bc)

and applying demorgans

P(Ac U Bc) = P(A∩B )c

using the law that Ac = 1 - A

I got the final easy equation

1 - P(A ∩ B )

solving for this

P(A) = 6/36 ...........(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)
P(B) = 3/36 ............(4,6), (5,5),(6,4)

plug in numbers

1 - (6/36)*(3/36) != the answer of 7/9

Apparently, math lies, j.k. Can someone let me know what i am doing wrong?

2. Feb 12, 2014

### Ray Vickson

Use inclusion-exclusion properly:
$$\text{P}(A \cup B) = \text{P}(A) + \text{P}(B) - \text{P}(A \cap B) = 6/36 + 3/36 - 1/36.$$

3. Feb 12, 2014

### royalewithchz

Thats a clean method.

Can I ask you why deMorgans doesn't work in this scenario?

4. Feb 12, 2014

### haruspex

Not so.

5. Feb 12, 2014

### Ray Vickson

It does work, and it is essentially what I used:
$$\text{desired answer} = \text{P}(A^c \cap B^c) = 1 - \text{P}(A \cup B)$$
If you evaluate $\text{P}(A \cup B)$ correctly you will get the right answer.

6. Feb 12, 2014

### royalewithchz

OHHHHH...my initial set up was wrong

this: P(Ac∩Bc)

and not this: P(AcUBc)

the phrase "neither a nor b"

confused me. I thought nor implied OR not AND.

7. Feb 12, 2014

### haruspex

"neither ... nor ..." is the same as "not ... and not ...".