What is the sample space that you use for rolling two dice, a first one and then a second one? Using this sample space, explain why the event "i dots are on the top of the first die" and the event "j dots are on the top of the second die" are independent if you roll two dice.
Please Help!!!! Please explain. Thanks so much.
Please Help!!!! Please explain. Thanks so much.
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Sample space is pretty much all of the possible outcomes you can have for the experiment.
So if you roll two dice, you can have the rolls be:
1 and 1 or 2 or 3 or 4 or 5 or 6
2 and 1 or 2 or 3 or 4 or 5 or 6
3 and 1 or 2 or 3 or 4 or 5 or 6
4 and 1 or 2 or 3 or 4 or 5 or 6
5 and 1 or 2 or 3 or 4 or 5 or 6
6 and 1 or 2 or 3 or 4 or 5 or 6
this is your sample space
the second question is asking why the two events are independent of each other.
This is because whatever number the first dice lands on does not effect what number the second dice will land on.
proof of this:
if you count up all of the possible outcomes, you should see that there are 36 in the sample space.
the probability of anyone of these happening is 1/36
two events are independent if P(AB) = P(A)*P(B)
1/36 = 1/6*1/6
1/6 refers to the probability of landing on any single number of the dice.
hope this helped
So if you roll two dice, you can have the rolls be:
1 and 1 or 2 or 3 or 4 or 5 or 6
2 and 1 or 2 or 3 or 4 or 5 or 6
3 and 1 or 2 or 3 or 4 or 5 or 6
4 and 1 or 2 or 3 or 4 or 5 or 6
5 and 1 or 2 or 3 or 4 or 5 or 6
6 and 1 or 2 or 3 or 4 or 5 or 6
this is your sample space
the second question is asking why the two events are independent of each other.
This is because whatever number the first dice lands on does not effect what number the second dice will land on.
proof of this:
if you count up all of the possible outcomes, you should see that there are 36 in the sample space.
the probability of anyone of these happening is 1/36
two events are independent if P(AB) = P(A)*P(B)
1/36 = 1/6*1/6
1/6 refers to the probability of landing on any single number of the dice.
hope this helped
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36. Think of each dice as a seperate set any number has a (1/6) chance of being rolled. now you introduce a second dice that has the same probability. The probability of rolling and combination of numbers is (1/6)*(1/6) or (1/36" so 36 seperate cases. They are independent because the roll and number of one dice does not affect the outcome of the other.
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36. because they each happen in six cases. In other words you have to make a table and show that choosing a column and a row does not affect the probability of the other role.