Probability of no 6 or no 5 when dice is rolled n times

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Can anyone guide me in the general direction of the answer to the following:

A die is rolled $n$ times

$$A = \text{no $6$s}$$

$$B = \text{no $5$s}$$

$$P(A\cup B) = \;?$$

I am first finding $P(A)$ and $P(B)$ as (where $n$ = 10):

$$P(A)=\left(\frac 56\right)^{10}$$ $$P(B)=\left(\frac 56\right)^{10}$$

$$P(A\cup B)=P(A)+P(B)$$

Is this correct or there is some intersection I have to worry about?

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Yes, you have to worry about the intersection. Use the formula $$P(A \cup B)=P(A)+P(B)-P(A \cap B).$$ In your case $$P(A\cap B)=(4/6)^{10}.$$