A fair die is thrown 3 times. Find the probability that the largest outcome of the three throws is a 3.
I understand that the problem isn't as simple as $ \frac{3^3}{6^3}=1/8$, being the answer. That would mean getting any of the values from $1-3$ for all of the throws is favorable, which includes series of throws that do not even have $3$ as an outcome.
What's the solution to ensure that there is at least one throw with $3$ as the outcome, with the $2$ other throws being anywhere from $1-3$?
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