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OpenStudy (anonymous):
Suppose that you have 10 cards. 6 are green and 4 are yellow. The 6 green cards are numbered 1, 2, 3, 4, 5, and 6. The 4 yellow cards are numbered 1, 2, 3, and 4. The cards are well shuffled. You randomly draw one card. • G = card drawn is green • Y = card drawn is yellow • E = card drawn is even-numbered How do we find: P(G | E) P(G or E)
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OpenStudy (dan815):
P(G | E) means prob of GREEN given that the card is EVEN
OpenStudy (dan815):
there are a total of 5 even cards, of which 3 are green
OpenStudy (anonymous):
ohhh so it would be 3/5?
OpenStudy (dan815):
yes
OpenStudy (anonymous):
oh okay I got it and how about approaching an "OR"
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OpenStudy (anonymous):
would it be 8/10?
OpenStudy (dan815):
yes
OpenStudy (loser66):
dan?
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