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

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anonymous
  • anonymous
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)
anonymous
  • anonymous
so the probability that the first card is yellow and the second card is yellow is their product \[\frac{3}{8}\times \frac{2}{7}=\frac{6}{56}\]
anonymous
  • anonymous
now we consider what happens if the first card is green. this happens with probability \[\frac{5}{8}\]
anonymous
  • anonymous
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
anonymous
  • anonymous
therefore the probability that the first card is green and the second card is yellow is their product \[\frac{5}{8}\times \frac{3}{7}=\frac{15}{56}\]
anonymous
  • anonymous
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
anonymous
  • anonymous
i meant "either the first card was yellow OR it was green"
anonymous
  • anonymous
\[\frac{6}{56}+\frac{15}{56}=\frac{21}{56}=\frac{3}{8}\]

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