Probability 2 Dice Question

[royalewithchz]

Homework Statement

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

Homework Equations

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

Demorgans law
(AUB)^c = A^c ∩ B ^c

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(A^c U B^c)

and applying demorgans

P(A^c U B^c) = P(A∩B )^c

using the law that A^c = 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?

 

[Ray Vickson]

Homework Statement

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

Homework Equations

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

Demorgans law
(AUB)^c = A^c ∩ B ^c

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(A^c U B^c)

and applying demorgans

P(A^c U B^c) = P(A∩B )^c

using the law that A^c = 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?

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

 

[royalewithchz]

Thats a clean method.

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

 

[haruspex]

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

so then what I am looking for is

P(A^c U B^c)

Not so.

neither a doublet nor the sum of 10

 

[Ray Vickson]

Thats a clean method.

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

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.

 

[royalewithchz]

OHHHHH...my initial set up was wrong

this: P(A^c∩B^c)

and not this: P(A^cUB^c)

the phrase "neither a nor b"

confused me. I thought nor implied OR not AND.

 

[haruspex]

I thought nor implied OR not AND.

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