there are 40 beads. 7/40 are yellow 5/40 are green 9/40 are blue 8/40 are brown 7/40 are orange 4/40 are red calculate WITH replacement and WITHOUT replacement for each: P(R1B2 OR B1R2) P(R1 AND G2) P(no yellows) P(doubles) P(no doubles)

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there are 40 beads. 7/40 are yellow 5/40 are green 9/40 are blue 8/40 are brown 7/40 are orange 4/40 are red calculate WITH replacement and WITHOUT replacement for each: P(R1B2 OR B1R2) P(R1 AND G2) P(no yellows) P(doubles) P(no doubles)

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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where is your attempt?
oh okay so for the or situation would it be like 2/1600 + 2/1600
B = blue or brown

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or B is Brown!!
*oh
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are we considering only exactly as written in that order 1 red then 2 green or do you need any combination of 1 red and 2 green ?
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can you clarify P(doubles) and P(no doubles) how many are you getting?
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with replacement means you are putting the chosen bead back in the "pool" before picking again. without replacement means you have one less each time you pull one bead out the group.
are you here?

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