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OpenStudy (anonymous):
26
OpenStudy (rational):
Good, how many of those "red cards" are "Aces" ?
OpenStudy (anonymous):
2
OpenStudy (rational):
Very good, so the probability of drawing a "red Ace" first time is :
\[\frac{2}{52}\]
After that drawing an Ace is
\[\frac{3}{51}\]
Multiply them to get the total probability
\[\frac{2}{52}\times \frac{3}{51}\]
simplify
OpenStudy (anonymous):
so 1/442
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OpenStudy (rational):
Yes drag the first box and place it before 1/442
OpenStudy (anonymous):
i did
OpenStudy (rational):
Lets look at second box
OpenStudy (rational):
look at the deck of cards again
how many of them are either "3" or "5" ?
OpenStudy (anonymous):
3 = 2
5= 2
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OpenStudy (rational):
count them again
OpenStudy (rational):
count both red and black
OpenStudy (anonymous):
ok so 3=4
5=4
OpenStudy (rational):
So, the cards in favor are 8
therefore the probability of drawing "3" or "5" is
\[\frac{8}{52}\]
Similarly, after putting the drawn card back in the deck, the probability of drawing "4" or "6" is
\[\frac{8}{52}\]
Multiply them to get the total probability
OpenStudy (anonymous):
so 4/169
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