An urn contains 10 red balls, 10 green balls, and 5 white balls. 5 balls are selected. In how many ways can 5 balls be drawn if at least 3 are green?

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An urn contains 10 red balls, 10 green balls, and 5 white balls. 5 balls are selected. In how many ways can 5 balls be drawn if at least 3 are green?

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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Someone please help!!!!
@ganeshie8 I love the way you break the green as an independent variable. Please, do it again. :)
can someone just help me please?

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@JosephDeng what is the answer given in ur book
i think it is \(\dbinom{15}{2}+\dbinom{15}{1}+1=121\)
@JosephDeng Hint: We just need to know that there are 10 green and 15 non-green. The number of combinations of pulling 3 green (10 choose 3) and 2 non green (15 choose 2) =\(\dbinom{10}{3}\dbinom{15}{2}=12600\) Total number of combinations (25 choose 5) = \(\dbinom{25}{5}=53130\) Can you take it from here to find P(at least 3 green)?
the answer is 16002
Yes, the answer is correct. \(\dbinom{10}{3}\dbinom{15}{2}+\dbinom{10}{4}\dbinom{15}{1}+\dbinom{10}{5}\dbinom{15}{0}=16002\)

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