Question

You roll two dice. What is the probability of rolling a 2 or a 4 on the second die, given that you rolled an even number on the first die

Answer 1

The sample space when you roll two dice:

$$ \begin{array}{c|c|c|C|}
&1&2&3&4&5&6 \\ \hline
1 &(1,1)&(1,2)&(1,3)&(1,4)&(1,5)&(1,6)\\ \hline
2 &(2,1)&(2,2)&(2,3)&(2,4)&(2,5)&(2,6)\\ \hline
3 &(3,1)&(3,2)&(3,3)&(3,4)&(3,5)&(3,6)\\ \hline
4 &(4,1)&(4,2)&(4,3)&(4,4)&(4,5)&(4,6)\\ \hline
5 &(5,1)&(5,2)&(5,3)&(5,4)&(5,5)&(5,6)\\ \hline
6 &(6,1)&(6,2)&(6,3)&(6,4)&(6,5)&(6,6)\\ \hline
\end{array}$$

Answer 2

Well, first of all, does the result of the first die affect what you'll get on the second die?

Answer 3

Can you spot your desired ones?

Answer 4

Check out the column 2 and 4.

Answer 5

that's some heck of latex

Answer 6

well the full question is this:
You roll two dice. What is the probability of rolling a 2 or a 4 on the second die, given that you rolled an even number on the first die? A 6X6 table of dice outcomes will help you to answer this question.

Answer 7

I just posted that \( 6\times 6\) table for you.

Answer 8

the thing is i don't know how to figure it out, even with the chart

Answer 9

help please!

Answer 10

Do you understand conditional probability?

Answer 11

i don't understand conditional probability.

Answer 12

there's 6 pairs? so 6 out of 36? is 1/6?

Answer 13

Let A is an event where you get an even number on the first die.
Let B is the event of rolling 2 or a 4 on the second die.

Can you spot the the pairs satisfying the event A and B respectively?

Answer 14

A=\(\{(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)\cdots (6,6)\} \)

There would be 18 such cases.

B= all the elements under the column 2 and 4 respectively.

Answer 15

There will be 12 such cases.

Answer 16

so 18/36 which is 1/2 ?

Answer 17

wait 12? i thought you said 18?

Answer 18

\(A\cap B = ?\)

Answer 19

ok so there are 2 dice and the possibilites are 36. but out of those only 12 are effective. so 12/36= is 1/3 ?

Answer 20

The probability of an even number on A is 3/6 and the probability of a 2 or a 4 on B is 2/6. The compound event is made up out of 2 separate and independent events. Therefore the probability of the compound event is the product of the separate probabilities:
\[P=\frac{3}{6}\times \frac{2}{6}=\frac{6}{36}\]