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Mathematics 19 Online
OpenStudy (anonymous):

Will fan and medal if you help!!!

OpenStudy (anonymous):

OpenStudy (rational):

How many "red cards" are there in a standard deck of cards ?

OpenStudy (anonymous):

52

OpenStudy (rational):

Total cards are 52, how many of them are "red" ?

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

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

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

OpenStudy (rational):

Yep!

OpenStudy (rational):

see if you can try the problem in 3rd box

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