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anonymous

  • 5 years ago

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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  1. anonymous
    • 5 years ago
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    answer is \[\frac{3}{8}\] would you like a silly worked out method?

  2. anonymous
    • 5 years ago
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    yeah cuz i have nooo clue how to do it

  3. anonymous
    • 5 years ago
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    ok we can think like this: suppose the first card drawn is yellow. this happens with probability \[\frac{3}{8}\]

  4. anonymous
    • 5 years ago
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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)

  5. anonymous
    • 5 years ago
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    so the probability that the first card is yellow and the second card is yellow is their product \[\frac{3}{8}\times \frac{2}{7}=\frac{6}{56}\]

  6. anonymous
    • 5 years ago
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    now we consider what happens if the first card is green. this happens with probability \[\frac{5}{8}\]

  7. anonymous
    • 5 years ago
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    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

  8. anonymous
    • 5 years ago
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    therefore the probability that the first card is green and the second card is yellow is their product \[\frac{5}{8}\times \frac{3}{7}=\frac{15}{56}\]

  9. anonymous
    • 5 years ago
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    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

  10. anonymous
    • 5 years ago
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    i meant "either the first card was yellow OR it was green"

  11. anonymous
    • 5 years ago
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    \[\frac{6}{56}+\frac{15}{56}=\frac{21}{56}=\frac{3}{8}\]

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