Probability 2 Dice Question

  • #1

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 = Ac ∩ 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(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?
 

Answers and Replies

  • #2
Ray Vickson
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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 = Ac ∩ 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(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?

Use inclusion-exclusion properly:
[tex] \text{P}(A \cup B) = \text{P}(A) + \text{P}(B) - \text{P}(A \cap B) = 6/36 + 3/36 - 1/36.[/tex]
 
  • #3
Thats a clean method.

Can I ask you why deMorgans doesn't work in this scenario?
 
  • #4
haruspex
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Let A = is a doublet
Let B = sum of dices is 10

so then what I am looking for is

P(Ac U Bc)
Not so.
neither a doublet nor the sum of 10
 
  • #5
Ray Vickson
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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:
[tex] \text{desired answer} = \text{P}(A^c \cap B^c) = 1 - \text{P}(A \cup B)[/tex]
If you evaluate ##\text{P}(A \cup B)## correctly you will get the right answer.
 
  • #6
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
haruspex
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I thought nor implied OR not AND.
"neither ... nor ..." is the same as "not ... and not ...".
 

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