Two dice are rolled three times. What is the probability of getting a sum of 7 at least once?
<I would like to know how to work through this question please. I have an exam tomorrow asking the same question but changing the number of rolls and the sum. I appreciate it!!>
To find the probability of getting a sum of 7 at least once when two dice are rolled three times, we can use the concept of complementary probability.
First, let's calculate the probability of NOT getting a sum of 7 in a single roll.
The total number of outcomes when rolling two dice is 6 * 6 = 36 (each dice can land on one of the six numbers).
The number of favorable outcomes (sum of 7) is 6 (pairs: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)).
Therefore, the probability of getting a sum of 7 in a single roll is 6/36, which simplifies to 1/6.
Now, let's find the probability of not getting a sum of 7 in all three rolls. Since the rolls are independent events, we can multiply the probabilities together.
The probability of not getting a sum of 7 in a single roll is 1 - 1/6 = 5/6.
Therefore, the probability of not getting a sum of 7 in all three rolls is (5/6)^3 = 125/216.
Finally, to find the probability of getting a sum of 7 at least once, we subtract the probability of not getting a sum of 7 from 1.
1 - 125/216 = 91/216.
So, the probability of getting a sum of 7 at least once when two dice are rolled three times is 91/216.
Sure! I'll be happy to explain how to approach this question.
To find the probability of getting a sum of 7 at least once when rolling two dice three times, we can use the concept of complementary probability.
First, let's find the probability of NOT getting a sum of 7 in a single roll. We can do this by finding the favorable outcomes (not 7) divided by the total possible outcomes.
For a single roll, there are 6 x 6 = 36 possible outcomes since each die has six sides numbered from 1 to 6.
Out of these 36 outcomes, there are 6 favorable outcomes (not 7). Here are the possible outcomes that do not sum up to 7: (1, 1), (1, 2), (1, 4), (1, 5), (2, 1), (2, 3), (2, 6), (3, 2), (3, 4), (3, 5), (4, 1), (4, 3), (4, 6), (5, 1), (5, 3), (5, 4), (6, 2), (6, 3), (6, 5). So there are 18 favorable outcomes.
Therefore, the probability of NOT getting a sum of 7 on a single roll is 18/36 = 1/2.
Now, since we are interested in getting a sum of 7 at least once in three rolls, we can use the concept of complementary probability to find the probability of NOT getting a sum of 7 in any of the three rolls.
The probability of NOT getting a sum of 7 in any of the three rolls is (1/2) x (1/2) x (1/2) = (1/2)^3 = 1/8.
Finally, to find the probability of getting a sum of 7 at least once in three rolls, we subtract the probability of not getting a sum of 7 from 1:
P(getting a sum of 7 at least once in three rolls) = 1 - P(NOT getting a sum of 7 in any of the three rolls)
= 1 - 1/8
= 7/8.
So, the probability of getting a sum of 7 at least once in three rolls of two dice is 7/8.
I hope this explanation helps! Good luck on your exam.
this is a binary probability ... seven or not seven
for each roll ... seven can occur 6 ways out of 36 possible outcomes
... probability of seven (s) = 6/36 = 1/6
... probability of not seven (n) = 1 - 1/6 = 5/6
(n + s)^3 = n^3 + 3 n^2 s + 3 n s^2 + s^3
the solution is the sum of all terms containing s
...or, one minus the term not containing s ... 1 - (5/6)^3 = ?