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**Solution of Tim has a box with a set of 8 green, 5 blue, 6 yellow, and 7 black toys. what is the** **probability that a random selection would yield a blue or black toy?**

In this, they ask the probability of finding blue ball, Here the events are mutually exclusive events, i.e. one event does not influence other event, to be more clear, selection of green ball, does not affects the no. of red and blue balls.

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So, here P(**neither red nor green**)implies nothing but P(blue balls)

Let n(B)n(B) be no. of blue balls, so n(B)=7n(B)=7

Also, n(S)=n(R)+n(B)+n(G)=8+7+6=21n(S)=n(R)+n(B)+n(G)=8+7+6=21 where n(R),n(B),n(G)n(R),n(B),n(G)are no. of red, blue, green balls respectively.

Then P(B)=n(B)n(S)P(B)=n(B)n(S)

P(B)=721P(B)=721

P(B)=13P(B)=13

i.e. probability of getting neither red or green balls is 1313 or 0.3333…