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Mathematics 8 Online
OpenStudy (anonymous):

I cannot find the answer about playing cards for statistics!!!

OpenStudy (anonymous):

Whenever I try to find how many of each type of card there is, I find that there is 13 of spades, hearts, diamonds, and clover. But what about ace, kings, queens, and the other types?

OpenStudy (anonymous):

@OOOPS

OpenStudy (anonymous):

@franzmller682

OpenStudy (anonymous):

@alyssa_fields50

OpenStudy (anonymous):

@veronica.montanez

OpenStudy (anonymous):

i dont understand

OpenStudy (anonymous):

I cannot find how many of each type of playing cards there are. I need it for statistics

OpenStudy (anonymous):

could u help me on a question?

OpenStudy (anonymous):

post the original problem, please

OpenStudy (anonymous):

Jim picked a card from a standard deck, looked at it, and then put it back. He then picked a second card. What is the probability that Jim picked a heart or an ace on either pick?

OpenStudy (anonymous):

I would know how to solve these, except I do not know how many of each type of cards there.

OpenStudy (anonymous):

I know that the formula is p(a U b) = p(a) + p(b) - p(a n b)

OpenStudy (anonymous):

The picks are independently, right?

OpenStudy (anonymous):

Yes

OpenStudy (anonymous):

for the first time: for a heart, we have 13 hearts on the deck, right? so the probability of the heart is 13/52, ok?

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

and your next problem is OR, that is the probability of an Ace is 4/52, ok?

OpenStudy (anonymous):

right

OpenStudy (anonymous):

so, for the first time, we have the probability of heart OR ace is (13/52)*(4/52) ,ok?

OpenStudy (anonymous):

Wouldn't that be the probability for heart AND ace

OpenStudy (anonymous):

It's another topic!! AND is both happen at the same time, right? and you have only 1 Ace which is heart, right?

OpenStudy (anonymous):

and the probability to pick that card is 1/52, right?

OpenStudy (anonymous):

I thought that there were four aces, so it should be 4/52

OpenStudy (anonymous):

you have 4 aces but only 1 is heart ace, right?

OpenStudy (anonymous):

Oh, now I understand

OpenStudy (anonymous):

Back to your problem, it is just for the first time. the second pick is the same since you put the card back. Hence the probability is just *2

OpenStudy (anonymous):

I figured it out, the answer is 16/52 because 13/52 + 4/52 - 1/52 = 16/52 is this correct?

OpenStudy (anonymous):

The real question I had was the format of playing cards, and now I understand. I always thought that ace and heart were separate, but once I understood the format of cards I figured out how to do the problem.

OpenStudy (anonymous):

you have to multiply to 2, because you have 2 picks

OpenStudy (anonymous):

multiply what by 2?

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