A deck of cards has three yellow cards and five green cards. One green card is drawn from the deck and not replaced. One more card is randomly drawn. Find the probability that the second card is yellow.

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A deck of cards has three yellow cards and five green cards. One green card is drawn from the deck and not replaced. One more card is randomly drawn. Find the probability that the second card is yellow.

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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answer is \[\frac{3}{8}\] would you like a silly worked out method?
yeah cuz i have nooo clue how to do it
ok we can think like this: suppose the first card drawn is yellow. this happens with probability \[\frac{3}{8}\]

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then the probability the next card is yellow is \[\frac{2}{7}\] because there are two yellow cards left and 7 cards total (the other yellow was removed)
so the probability that the first card is yellow and the second card is yellow is their product \[\frac{3}{8}\times \frac{2}{7}=\frac{6}{56}\]
now we consider what happens if the first card is green. this happens with probability \[\frac{5}{8}\]
then the probability the second card chosen is yellow is \[\frac{3}{7}\] because there are still 3 yellow cards left and 7 to choose from
therefore the probability that the first card is green and the second card is yellow is their product \[\frac{5}{8}\times \frac{3}{7}=\frac{15}{56}\]
now either the first card was yellow it was green. those are the only possibilities. we have computed the probability that the first card is yellow and the second card is yellow, and the probability that the first card is green and the second card is yellow. since these events are 'disjoint" i.e. they cannot happen at the same time, we just add the probabilities
i meant "either the first card was yellow OR it was green"
\[\frac{6}{56}+\frac{15}{56}=\frac{21}{56}=\frac{3}{8}\]

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