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 = ?
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!!>
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