a die is thrown twice. what is the probability that the sum of the rolls is greater than 9 given that the first roll is a 4?
the 2nd throw must be a 6, with P=1/6
so, the answer would just be 1/6? or should it be 1/6 x
1/6= 1/36 since 2nd throw is 1/6 and fist is 1/6 too?
we already know that the 1st roll is a 4. So, its odds of being a 4 are 1
1 * 1/6 = 1/6
The odds of rolling a 4 and a 6 are indeed 1/6 * 1/6, but the 4 is not in question. You have a 4 on the table already.
To find the probability that the sum of the rolls is greater than 9 given that the first roll is a 4, we need to calculate the probability distribution of the sums of two dice rolls when the first roll is a 4.
First, let's consider the possible outcomes of the second roll. A standard die has 6 sides labeled 1 through 6. Since the first roll is fixed at 4, there are now 6 possible outcomes for the second roll: 1, 2, 3, 4, 5, or 6.
Next, we need to calculate the sums of the two rolls for each of these outcomes. We can create a table to list all the possible outcomes and their corresponding sums:
Second Roll | Sum
---------------------
1 | 5
2 | 6
3 | 7
4 | 8
5 | 9
6 | 10
Now let's count how many of these sums are greater than 9. In this case, there are only two sums that satisfy this condition: 10 and 12.
Therefore, out of the 6 possible outcomes for the second roll, only 2 have a sum greater than 9. Hence, the probability of the sum of the rolls being greater than 9 given that the first roll is a 4 is 2/6 or 1/3.