1. ## Question about a Question: Inclusion-Exclusion theorm

QUESTION:

Okey.
So, i'm curious how
P(A) = 0.5
Because
Possible numbers (sum) = 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 = 11 Total
(because if two dice roll lowest number and we add them, you get two)

So why, is P(A) = 1/2?
Surely it should be = 6/11?

Thanks

2. Did you account for the various ways to throw 7 for example?

3. What do you mean?
I didn't think the 'way' in which the number's are aquired (aka 6 + 1, 3 + 3) would be required for this question?

4. Originally Posted by AshleyT
What do you mean?
I didn't think the 'way' in which the number's are aquired (aka 6 + 1, 3 + 3) would be required for this question?
Here are the probabilities.
$\displaystyle \begin{array}{*{20}c} {P(2)} & = & {\frac{1} {{36}}} \\ {P(4)} & = & {\frac{3} {{36}}} \\ {P(6)} & = & {\frac{5} {{36}}} \\ {P(8)} & = & {\frac{5} {{36}}} \\ {P(10)} & = & {\frac{3} {{36}}} \\ {P(12)} & = & {\frac{1} {{36}}} \\ \end{array}$

The ways to get 10 are $\displaystyle \{(5,5),~(4,6),~(6,4)\}$ that is 3 out of 36.

5. Originally Posted by Plato
Here are the probabilities.
$\displaystyle \begin{array}{*{20}c} {P(2)} & = & {\frac{1} {{36}}} \\ {P(4)} & = & {\frac{3} {{36}}} \\ {P(6)} & = & {\frac{5} {{36}}} \\ {P(8)} & = & {\frac{5} {{36}}} \\ {P(10)} & = & {\frac{3} {{36}}} \\ {P(12)} & = & {\frac{1} {{36}}} \\ \end{array}$

The ways to get 10 are $\displaystyle \{(5,5),~(4,6),~(6,4)\}$ that is 3 out of 36.
Ah, i see.
I was thinking: 11 outcomes, therefore P(2) would be 1/11.
Thankyou.

6. Hello, AshleyT!

You've never rolled a pair of dice before, have you?

You have two dice, each has a choice of six numbers.

There are 36 possible outcomes:

. . $\displaystyle \begin{array}{cccccc} (1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6) \\ (2,1) & (2,2) & (2,3) & (2,4) & (2,5) & (2,6) \\ (3,1) & (3,2) & (3,3) & (3,4) & (3,5) & (3,6) \\ (4,1) & (4,2) & (4,3) & (4,4) & (4,5) & (4,6) \\ (5,1) & (5,2) & (5,3) & (5,4) & (5,5) & (5,6) \\ (6,1) & (6,2) & (6,3) & (6,4) & (6,5) & (6,6) \end{array}$

Now count the cases . . .

Sum is even: .$\displaystyle \begin{Bmatrix}(1,1) & (1,3) & (1,5) \\\ (2,2) & (2,4) & (2,6) \\ (3,1) & (3,3) & (3,5) \\ (4,2) & (4,4) & (4,6) \\ (5,1) & (5,3) & (5,5) \\ (6,2) & (6,4) & (6,6) \end{Bmatrix} \quad \text{ 18 cases}$

$\displaystyle \displaystyle{\text{Therefore: }\;P(\text{Sum is even}) \;=\;\frac{18}{36} \;=\;\frac{1}{2}$