So I know how to find the classic probability of at least 1 6, when rolling 10 die. Its just $$1-P(\text{no } 6) = \left(1-\frac{5}{6}\right)^{10}.$$ How would I do this for both 5 and 6? Right now I am thinking to do $$1-P(\text{no } 5)-P(\text{no } 6)-P(\text{no } 5 \text{ and no } 6).$$ Is this correct?
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Related: http://math.stackexchange.com/questions/2140363/a-fair-die-is-rolled-10-times-define-n-to-be-the-number-of-distinct-outcomes-f/2141307#2141307 as this was the remaining missing information that I left to the user at the time to complete. – JMoravitz Feb 13 '17 at 22:11
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You’re on the right track, but you have to add the probability of “no 5 and no 6” back in because you’re counted it twice in the previous two terms. – amd Feb 13 '17 at 22:11
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This is almost correct. Remember your Inclusion-Exclusion principle. $\Pr(A\cup B)=\Pr(A)+\Pr(B)-\Pr(A\cap B)$, or rearranged we have $\Pr(A\cap B)=\Pr(A)+\Pr(B)-\Pr(A\cup B)$
Let $A$ be the event that a $5$ was among those that were rolled. Let $B$ be the event that a $6$ was among those that were rolled.
Then, $A\cap B$ is the event that at least one five and at least one six were rolled.
$\Pr(A\cap B)=\Pr(A)+\Pr(B)-\Pr(A\cup B)$
Now, notice $\Pr(A)=1-\Pr(A^c)$, $\Pr(B)=1-\Pr(B^c)$, and $\Pr(A\cup B) = 1-\Pr((A\cup B)^c)=1-\Pr(A^c\cap B^c)$
$A^c$ represents the event that no fives were rolled, which you correctly calculate the probability of above as $\Pr(A^c)=(\frac{5}{6})^{10}$ implying $\Pr(A)=1-(\frac{5}{6})^{10}$
What does $A^c\cap B^c$ represent as an event?
The event that simultaneously no fives and no sixes were rolled
What is $\Pr(A^c\cap B^c)$?
$(\frac{4}{6})^{10}$
What then is $\Pr(A\cup B)$? What then is the probability $\Pr(A\cap B)$?
$\Pr(A\cup B)=1-\Pr(A^c\cap B^c) = 1 - (\frac{4}{6})^{10}$
$~$
$\Pr(A\cap B) = \Pr(A)+\Pr(B)-\Pr(A\cup B) = \dots$
JMoravitz
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I see. Essentially what amd commented above, I was double counting one of the conditions. Thank you so much! – Bill Feb 13 '17 at 22:30