A single card is drawn from a standard deck of cards. Find the probability if the given information is known about the chosen card. (Enter the probability as a fraction.) P(diamond|not a heart)

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in this case, we're being asked that if we have a full deck of cards, what are the chances that if we pull a card, we would get in a CE or heart? Okay, so it can be either or doesn't have to be the ace of hearts. That can be an ace, or it can be a heart. All right, First off, if you're unfamiliar with a full deck of cards, I've written all the options out here. Toe left. You've got a skin queen, Jack. 10 98765432 for all four of the suits. Okay, uh, that is 13 cards pursuit. And there are four suits, which means we have 52 cards total. Any time you're doing probability, you're gonna be basically creating yourself a fraction where you're gonna have the total options on the bottom in your denominator. And then in your numerator on top, you're gonna have all the ways that you can get a success all the ways that you can get what you actually want. Meeting for a probability of an ace or heart. It's going to be over 52. That much is certain. Now our final answer might not, because we might be able to simplify the fraction. But, you know, only time will tell with that. But for now, we know our total is 52. So the question is for our in numerator, how many different cards would be a success would be either an ace or heart. Well, to begin with, it says it can be any heart right? Does not specify it just says an ace or a heart so any heart will do that alone is 13 options, right? This whole row or this whole column, I mean, is 13 different cards. So there's 13 different ones that we could pull that would work. But it does say along with the heart, it could also be an ace. That's all of these. Now this ace right here of hearts was already counted in my column of 13 so I don't want to count it a second time. You can't double count because you can't pull. Um, there's into a sea of hearts, right? There's only one card, so there's only these three new aces that we need to add. So 13 plus, these three aces would give me a total of 16 possible cards that would work for this problem. So that's my probability. Now, in this case, 16/52 is able to simplify. 16 can divide by four and 52 can also divide by 4 16 divided by four is 4 52 Divided by four is 13. That is what we want for a final answer. Because I mean, any time you guys are dealing with fractions, if you can simplify, it should.

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Algebra 2

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Holliday, Luchin, Cuevas, Carter Marks, Day, Casey, Hayek

Algebra 2

Fuzail S.

Foster W.

Andrew K.

Dading C.

Robin C.

Lucas F.

Chris T.

Ryan M.

Samuel C.

Erik R.

Algebra 2

Holliday, Luchin, Cuevas, Carter Marks, Day, Casey, HayekAlgebra 2

Fuzail S.

Foster W.

Andrew K.

Dading C.

Robin C.

Lucas F.

Chris T.

Ryan M.

Samuel C.

Erik R.

Holliday, Luchin, Cuevas, Carter Marks, Day, Casey, Hayek

Algebra 2