At least one blue and one green

EasyProbability Combinatorics~4m

A box holds $6$ marbles of distinct colors, among them one blue and one green. You draw a marble, note its color, and replace it - doing this $4$ times. What is the probability that your four draws include at least one blue and at least one green?

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Inclusion–exclusion on $\bar B$ (no blue) and $\bar G$ (no green): $1-\mathbb{P}[\bar B]-\mathbb{P}[\bar G]+\mathbb{P}[\bar B\cap\bar G]$ .
$\mathbb{P}[\bar B]=\mathbb{P}[\bar G]=(5/6)^4$ , $\mathbb{P}[\bar B\cap\bar G]=(4/6)^4$ .

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At least one blue and one green

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