# Thread: Probability help

1. ## Probability help

I'm having trouble with this probability question.

"Find the probability of rolling an even sum or a sum greater than nine when a pair of dice is rolled."

So P(A)= even sum when dice rolled
(1,1) (1,3) (1,5) (2,2) (2,4) (2,6) (3,3) (3,5) (4,4) (4,6) (5,5) (6,6)

$\displaystyle =\frac{12}{21}$

P(B)= sum greater than nine when dice rolled
(4,6) (5,5) (5,6) (6,6)

$\displaystyle =\frac4{21}$

P(A or B)$\displaystyle = \frac{12}{21} + \frac{4}{21} -P$(A and B)

..and now I'm stuck, I don't know if I'm even doing it right. help?

2. You're doing good. Now you just need to subtract off P(A and B).

The ones who are greater than 9 AND have an even sum. In other words, the pair who sum to 10 and 12.

3. so would I multiply $\displaystyle \frac{12}{21}$ and $\displaystyle \frac1{21}$? Which basically equals $\displaystyle \frac{12}{21}$. Then substitute that into P(A and B)?

4. You've got the right idea, but remember that you can roll both (6,4) and (4,6), whereas there is only one way to roll (5,5), for instance. The formula you wrote for calculating the probability of (A or B) is correct.

5. Hello, Hypertension!

You're using the wrong denominator.

When a pair of dice is rolled, there are: .$\displaystyle 6^2 \:=\:36$ outcomes.

Find the probability of rolling an even sum or a sum greater than nine
when a pair of dice is rolled.
$\displaystyle A$: Even sum is rolled: .(1,1) (1,3) (1,5) (2,2) (2,4) (2,6) (3,3) (3,5) (4,4) (4,6) (5,5) (6,6)

. . $\displaystyle P(A)\:=\:\frac{12}{36}$

$\displaystyle B$: Sum greater than nine is rolled: .(4,6) (5,5) (5,6) (6,6)

. . $\displaystyle P(B)\:=\:\frac4{21}$

$\displaystyle A \cap B$: Sum is even and greater than nine: .(4,6) (5,5) (6,6)

. . It's the overlap of $\displaystyle A$ and $\displaystyle B$: .$\displaystyle P(A \cap B) \:=\:\frac{3}{36}$

Therefore: .$\displaystyle P(A \cup B) \:=\: P(A) + O(B) - P(A \cap B) \:=\: \frac{12}{36} + \frac{4}{36} - \frac{3}{36} \;=\;\frac{13}{36}$

6. Originally Posted by Soroban Even sum is rolled: .(1,1) (1,3) (1,5) (2,2) (2,4) (2,6) (3,3) (3,5) (4,4) (4,6) (5,5) (6,6)

. . $\displaystyle P(A)\:=\:\frac{12}{36}$
Actually, the probability that an even sum is rolled is 18/36. Don't forget to include the outcomes (3, 1), (5, 1), (4, 2), (6, 2), (5, 3) and (6, 4).

7. ok, so I include the rolled numbers too?

so technically it would be,

P(A)= sum of even numbers: .(1,1) (1,3) (1,5) (2,2) (2,4) (2,6) (3,3) (3,5) (4,4) (4,6) (5,5) (6,6) (3,1) (5,1) (4,2) (6,2) (5,3) (6,4)

$\displaystyle =\frac{18}{36}$

P(B)= sum of numbers greater than nine: (4,6) (5,5) (5,6) (6,6) (6,4) (6,5)

$\displaystyle =\frac{6}{36}$

Then P(A and B) = $\displaystyle \frac{4}{36}$

So, P(A or B)=$\displaystyle \frac{18}{36} + \frac{6}{36} - \frac{4}{36}$
$\displaystyle =\frac{20}{36}$

$\displaystyle =\frac{5}{9}$

8. Originally Posted by Hypertension ok, so I include the rolled numbers too?

so technically it would be,

P(A)= sum of even numbers: .(1,1) (1,3) (1,5) (2,2) (2,4) (2,6) (3,3) (3,5) (4,4) (4,6) (5,5) (6,6) (3,1) (5,1) (4,2) (6,2) (5,3) (6,4)

$\displaystyle =\frac{18}{36}$

P(B)= sum of numbers greater than nine: (4,6) (5,5) (5,6) (6,6) (6,4) (6,5)

$\displaystyle =\frac{6}{36}$

Then P(A and B) = $\displaystyle \frac{4}{36}$

So, P(A or B)=$\displaystyle \frac{18}{36} + \frac{6}{36} - \frac{4}{36}$
$\displaystyle =\frac{20}{36}$

$\displaystyle =\frac{5}{9}$
Correct.

9. Thank you all for the help.

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