two-cards-are-drawn-from-a-standard-deck-of-cards-find-each-probability-if-no-replacement-occurs-pte

Question

Two cards are drawn from a standard deck of cards. Find each probability if no replacement occurs.
$P(	ext { two hearts })$

Charles Carter · Accepted Answer

Yeah. This problem wants us to find the probability of drawing two hearts out of a standard, uh, playing card deck with no replacements. That means these air depended events. We have to evaluate them in order. So the first heart of a standard A deck would just be a one in four chance. The second heart. We&#x27;re going to take one from the total amount of cards as 51 and the amount of hearts left in the deck would be 12 two. Then we can reduce a little bit. So, uh, we have one ever four. So the four will cancel with the 12. And so that&#x27;s just gonna leave us with 3/51. That which is just gonna be equal to 1/17. Okay, thank you very much.

Trinity Steen · Answer

So for Number 46 were trying to figure out the probability of during, ah heart than a club in a center deck of cards, but without replacement. So we recall that there are 52 cards in a deck. There are 13 hearts in a deck, so there&#x27;s a 13 out of 52 chance that I would be able to draw a heart. Okay, now then, there&#x27;s also 13 clubs now, since the heart and a club can&#x27;t co exist on the same card, they&#x27;re separate from each other. But we&#x27;re not replacing the first part. We draw. So now we&#x27;ll aggressively to. We have 51. Sorry to do. 13 times 13 which is 169 and then 52 times 51 is 2652. Now. This can be reduced. That can be reduced by 13. And so we ended up getting ah 13 out of 20 four

Trinity Steen · Answer

so we&#x27;re trying to find the probability of during a heart than a club with replacement. So we have to remember that a Senator cards was 52 them and that there are 13 hearts in a deck. So there&#x27;s a 13 out of 15 to sue chance that I&#x27;m going to be able Teoh, get ah heart. And then because we are replacing, that means we still have 52 cards in the deck. And then there&#x27;s a 13 out of the chance I could get clubs, So I&#x27;m going to get 1 69 on the top when we get 52 on the bottom. That&#x27;s 2704 and it turns out that we can divide the top and bottom both by 1 69 So because of their being replacement, this probability is one out of 16

Charles Carter · Answer

wants us to find the probability probability of kidding three hearts out of a standard deck of cards if there is no replacement. So it means these in these events are depended on each other, so we seem to find the probability of the first event. So they&#x27;re probably getting a heart when there&#x27;s no cards taken out. Is just 14 for a standard deck of playing cards. Case after we take one of the cards out, the next probability will have one less card in the total. So the total goes down to 51 instead of 52 in the hearts instead of being at 13. Will now be at 12 because we pulled one of the hearts out. Okay, now our third choice, we&#x27;re dropping again. So, um, said 51 cards and I will be at 50 cards and instead of having 12 hearts were now only gonna have 11 hearts. Okay, so we can reduce this a little bit just for mental math. Say so that 1/4 can cancel with the 12 s. Oh, that&#x27;s gonna leave us with 3/50 times. 11/50 and 3/51. Don&#x27;t go up there. Times 11/50. Okay. And then 3/51 3 goes into 51 17 times, so be won over 17 times 11/50. Okay, that&#x27;s just gonna give us 11 over 850. King is our final probability of getting three hearts without replacement. Thank you.

Charles Carter · Answer

Yes. So this probably wants us to find the problem of getting three hearts, and that&#x27;s with replacements of these air independent events. So we just need to find the probability of getting one heart. So probably getting, getting a heart and a standard that complain cards. It&#x27;s 1/4 because there are four suits. And then we&#x27;re gonna do this three times. And since we&#x27;re replacing, these are independent events so we can repeat this probability. And so it&#x27;s gonna be 1/4 time&#x27;s 1/4 time&#x27;s 1/4 which is just 1/64. That would be the probability of drawing three hearts with replacement. Yeah, thank you very much.

James Macpherson · Answer

Okay, so we won the probability of heart. Second heart given first heart comes the probability of one heart. So we&#x27;ll have one out of 52 er sorry. You&#x27;re 10. 52. The probability of one heart no have 12 out of 51 probability that our second is a heart given our first was a heart. And so then this is also could be simplified toe 1/4 times four set at 4/17 which can then be simplified to 1/17 which is the same as 0.5 uh, or should be an eight, Your 0.588

Problem

Two cards are drawn from a standard deck of cards…

Problem 10 Easy Difficulty

Two cards are drawn from a standard deck of cards. Find each probability if no replacement occurs.
$P(\text { two hearts })$

Answer

$\frac{1}{17}$

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