Peter is giving marbles to some children at a carnival. He has 5 red marbles, 4 blue marbles, and 3 yellow marbles. If Peter selects a marble randomly without looking, what is the probability that he will give a blue marble to the first child and then a yellow marble to the second child? 4 over 12 plus 3 over 12 is equal to 7 over 12 4 over 12 plus 3 over 11 is equal to 80 over 132 4 over 12 multiplied by 3 over 12 is equal to 12 over 144 4 over 12 multiplied by 3 over 11 is equal to 12 over 132

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Peter is giving marbles to some children at a carnival. He has 5 red marbles, 4 blue marbles, and 3 yellow marbles. If Peter selects a marble randomly without looking, what is the probability that he will give a blue marble to the first child and then a yellow marble to the second child? 4 over 12 plus 3 over 12 is equal to 7 over 12 4 over 12 plus 3 over 11 is equal to 80 over 132 4 over 12 multiplied by 3 over 12 is equal to 12 over 144 4 over 12 multiplied by 3 over 11 is equal to 12 over 132

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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The initial number of marbles is 12, of which 4 are blue. Therefore the probability of selecting a blue marble first is 4/12. Having selected a blue marble, there is a total of 11 marbles remaining of which 3 are yellow. Therefore the probability of a yellow marble being selected second is 3/11. The required probability is therefore given by \[\large \frac{4}{12}\times\frac{3}{11}\]
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