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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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A bag has 4 green marbles, 3 red marbles, and 3 yellow marbles. What is the probability that you pick a green marble, do not replace it, and pick a red marble?
Well, I believe these are two scenarios. First calculate the probability you have to pick a green marble, you count all the marbles and and the green marbles: \[\frac{ 4 }{ 10 }.100\]
why the 100?

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it's there by mere definition, but it's a formula that gives you the probability.
You have to first perform the division, remember that.
I got 40
Good, now, that gives the probability you would have to get a green marble, now, you won't put it back in the bag, so now you have a total of 9 marbles inside the bag. What would be the probability to find a red marble in the new quatity?: \[\frac{ 3 }{ 9 }.100\]
got 33.33
Better not multiply by a 100. Only if you want to convert the final answer to a percentage $$ \Large \sf Probability = \frac {4}{10}\cdot \frac{3}{9}$$
2/15??
yes
awwwh okay thank you! i'll look at this next time i have trouble with this
Haha, I was doing it by parts, but thanks, perl.
@perl @Owlcoffee so for the problem A bag has 5 red marbles, 6 blue marbles and 4 black marbles. What is the probability of picking a blue marble, replacing it, and then picking another blue marble? the answer would be 4/25 ?

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