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What is the probability that on 8 rolls of dice, there will be at least one '3' or one '1' showing up?

My attempt was:

$\frac{{8 \choose 1} * 6^7 + {8 \choose 1} * 6^7}{6^8}$

Because 7 dice have 6 possible results, there are $6^7$ variations. The last one has to have the result 3. There are ${8 \choose 1} = 8$ ways to combine it with the other results. Becuase the question asks for a '3' or a '1', I added the same calculation for the result '1'.

To get the probability, I divided everything by the total number of variations, which is $6^8$.

Indeed, my result is bigger than $1$.

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    You are repeatedly counting scenarios in which you get both $1$ and $3$. – 2017-02-07
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    Hint: it's easier to compute the probability that you get neither a $1$ nor a $3$. – 2017-02-07
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    u calculated P(Atleast one 1)+P(Atleast one 3).U need to find P(Atleast one 1 or atleast 3).By booles' inequality your calculated probability is bigger(and hence wrong) – 2017-02-07

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You want the probability for "at least" not that for "exactly".

Your calculation also over counts the intersections; where you have more than one result of $3$ or $1$ among the eight results.

Let $X_n$ be the count for showing a result of $n$ among eight rolls of a die.   This is easiest evaluated by way of the complement.

$$\mathsf P(X_3\geq 1\cup X_1\geq 1) = 1-\mathsf P(X_3=0\cap X_1=0) \\ = 1-\frac{4^8}{6^8}$$