Question

HELP!?! Please explain how to find the answer.
You roll 2 dice. What is the probability of getting a 3 or a 6 on the second die, given that you rolled a 1 on the first die?

Answer 1

2/5

Answer 2

What you get for the second has nothing to do with the first one, so the first die is just "noise". The two rolls (sequential or simultaneous) are independent.
Can you figure out what is the probability of rolling a 3 or a 6 on the second?

Answer 3

no 2/6

Answer 4

That is correct. Are you ok with that @britt1212 ?
We usually reduce it to 1/3.

Answer 5

I don't really get how you got the answer 2/3

Answer 6

@mathmate is it can be (1/36 )as the probability of getting (3,1) 3 from the second dice and 1 from the first dice (1/36 )as the probability of getting (6,1) 6 from the second dice and 1 from the first dice so total probability of getting 3 or 6 on the second when first is 1 =1/36 +1/36 =2/36 =1/18 (only explanation is it right ?)

Answer 7

The outcomes of throwing a fair die is {1,2,3,4,5,6} with equal probability.

The desired outcomes are either 3 or a 6, which is 2 out of 6 outcomes, so
P(3 or 6)=2/6=1/3

Answer 8

I thought when its 2 fair die its out of 36 @mathmate

Answer 9

yes it will 36 @britt1212

Answer 10

@Hanishala
The given problem is a conditional probability problem, i.e.
probability of getting a 3 or a 6 on the second die given the first is a one.
The problem can be simplified to just considering the second die, since the outcomes of the first and second are independent.
However, if we follow the motions, here's how to do it the "long" way in terms of conditional probability.

In general, for two given events,
P(A|B) (reads Probability of event A given B has happened)
=P(A intersect B) / P(B)
In this example, A={3, 6} (second die), B={1} (first die)
A intersect B = {13, 16} both dice
B={1} first die
So
P(A|B)=(2/36) / (1/6) = 2/6 = 1/3 as before.

Answer 11

how i cannot understand you @mathmate

Answer 12

(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)
this is all probabilities which will appear when you throw two dices
so probability of getting 3or 6 on second when first is 1 is 2/36=1/18 @mathmate

Answer 13

If you have not done conditional probability before, or it was a while back, you could read an article such as the following as a refresher:

http://www.mathsisfun.com/data/probability-events-conditional.html

The question asks what is the probability of getting a 3 or a 6 on the SECOND die, given the first is a one.

I.e. If you throw the first one, and get a one, then what is the probability of getting a 3 or a 6, which gives you 2/6 or 1/3.

If the first throw is a 2,3...6, it doesn't count (said given the first throw is a one), so one would throw the first one again until a one is obtained before throwing the second one. Then the result is still the same.

If you are still not convinced, then list all the 36 outcomes, and see how many have a 3 or 6 on the second throw OUT OF THOSE WHICH HAVE A "1" in the first throw.

You should get (1,3), (1,6) out of (1,1),(1,2),(1,3),(1,4),(1,5),(1,6), which gives 2/6=1/3.

Answer 14

The key phrase is "given that you rolled a 1 on the first die".

Answer 15

you count it from total probability not getting 1 from first die probabilities only look on probabilities again

Answer 16

The key phrase is "given that you rolled a 1 on the first die".
If we count those not having a one in the first roll, we are not following the rules.

Answer 17

When you see the phrase "given that" means that that event has already happened, and therefore the sample space has been narrowed.

Answer 18

not the rules but the probability of getting 3or 6 on 2d dice when you rolled a 1 on first dice when you throw 2 dices together so you will find that your probability will be (1,3) and (1,6) they are 2 probabilities of 36 you can search and try again if i were false tell me

Answer 19

I suggest you first read this article which may answer many of your questions.

http://www.mathsisfun.com/data/probability-events-conditional.html

Answer 20

i understand this but your first probability will not affect the other probability

Answer 21

That is exactly what I indicated at the very beginning.
The two events being independent, we do not have to consider the first die, and can come up with 2/6=1/3.