two-cards-are-drawn-from-a-standard-deck-of-cards-find-each-probability-ptext-both-either-black-or-a

Question

Two cards are drawn from a standard deck of cards. Find each probability. 
$P(	ext { both either black or an ace) }$

Garrett Bess · Accepted Answer

So for this problem, if we draw two cards from the top of the deck, what are the chances of them being both either black or an ace? So this problem is actually very simple. We just were only given two terms here, so both have to be either Black Rene. So we&#x27;re gonna actually multiply our two fractions together. So our first fraction are our first fraction is going to be. Well, what is the chance of drawing the first card? It&#x27;s a black or an ace. Well, there&#x27;s 26 black cards, and then there&#x27;s too aces that are not black. So 26 plus two is going to 28/52. That&#x27;s how many cards in the deck. If you&#x27;ve already drawn one of these cards, then there should only be 27 out of 51 cards left in the deck. So if we multiply that together, then we end up with a fraction of 756 over 2652. You write that 2652 and if you simplify that, then you should end up with the fraction of 63 over 221. That is the probability of drawing either a black or nace for both of them.

Garrett Bess · Answer

So for this problem, we want to look for the probability. If you draw out two of two cars off the top of the deck of finding both face cards or both black, in order to solve this problem, we need to recognize went to multiply your fractions and went to add. So when you see an or you want to add, if you if you intend on using and so black and black we want to multiply face and face, we wanna multiply. That&#x27;s what that means. So let&#x27;s get started. So the probability of finding both face cards well face cards make up 12 cards out of 52 and if you&#x27;ve already drawn one card, then you only have 11 out of 51. So is our first set. Plus, because there&#x27;s an or both black, so they&#x27;re black. Cars make up half the deck, so that&#x27;ll be 26 out of 52. You&#x27;ve already drawn a black card, then you&#x27;ll be at 25/51. But since these fund these probabilities are not mutually exclusive there inclusive because there could be a face card that is black, we have to subtract the probability of finding the same thing of the blackface card. Well, there&#x27;s 12 face cards and black cards make up half the deck, so we&#x27;re gonna end up with six out of 52 multiplied by Well, if I already have one than five. So in orderto combine these terms using plus and minus, we have to find a common denominator. And luckily for us, the common denominator is already there. It&#x27;s 32 times 51 is going to be 2652 So that&#x27;s going to be the dumb denominator for each of these fractions. So let&#x27;s just multiple use together. 12 times 11 is 1 32 26 times 25 is 6 56 times five is 30. So let&#x27;s combine these terms. 1 32 plus 6 50 is going to be 780 to minus 30 is going to give us 750 to over 2652. Now. If you put this into your calculator and you reduce it all the way, you should end up with a fraction of 188 over 663. That is your probability of finding both face cards or both black cards

Garrett Bess · Answer

Okay, So for this problem, we&#x27;re going to find the probability of both jacks being pulled from a deck of cards or both face cards. So let&#x27;s dissect this problem. So first, in order to do this problem, you have to know when to you multiply your fractions and went toe. Add your fractions when you see an or you&#x27;re gonna add. But when you see both jacks, it means Jack and Jack. When you see an and you&#x27;re gonna multiply so both face cards that means a face card and a face card. You&#x27;re gonna multiply there. So now that we know that, lets get started. So the probability of pulling a jack from the top of the deck is going to be four out of 52 now from the same deck. The probability from pulling a jack again is going to be 3/51 because there&#x27;s only 51 cards left in the deck and you&#x27;ve already pulled the jack. So then we&#x27;re going to add that because there&#x27;s an or here or both face cards. Well, there&#x27;s 12 face cards in a deck, so to be 12/52 multiplied by well one face cards already out, so it&#x27;s going to be 11/51. But since these events are not mutually exclusive there inclusive, we have to subtract the probability of getting both of these well, Jax or face cards. So every time you pull a jack, you&#x27;re going to get face cards. So you need to just subtract the probability of finding a Jack because Jax or just face cards. So for over 52 times 3/51 Now we can actually simplify this. We have a negative of this in a positive this so we can actually just cancel those out and just end up with this. So if you put that into your calculator, you should get 12 times 11 is 1 32 over 52 times 51 that is 2000 to 2652. And if you simplify that, you end up with the fraction of 11 over 221. So that is the probability of pulling either both jacks or both face cards

Trinity Steen · Answer

so for this problem was very similar to 44 except now we&#x27;re not having any replacement. So recall cards have 52 52 in them. And so, um, when we&#x27;re doing the probability of an ace, that is four out of 52. Now, when we are drawing a king, the fact that of a Senate can&#x27;t can&#x27;t exist on the same card means that there is still four kings available. But because we didn&#x27;t replace when we drew the first card are number now goes out of 51. So our top number is going to be 16 and then it&#x27;s going to be 2652 and want to reduce that by dividing both the top and bottom by four turns out ends up being four out, um, 663

Trinity Steen · Answer

So for this problem, we are trying to look at a center deck of cards, so recall that cards have 52 in them. And we&#x27;re trying to look at what happens if I draw an ace first than a king with replacement. So that means we&#x27;re drawing, um, putting the card back and then redrawing. So what that means is that when I am trying to pick an ace, the probability of me drawing an ace is four out of 52. Now, this is the con actuacion where we&#x27;re going to be multiplying because of the multiplication principle. And so because we&#x27;re replacing, that means we&#x27;re still gonna have 52 cards. And because Ace and King can overlap, they can&#x27;t co exist at the same being in the same car on the same card. There&#x27;s four kings, so that&#x27;s the probability of drawing a king. So we&#x27;re gonna go ahead and multi. We get 16 over 2704 and to go ahead and reduce that, um, you can divide both the top and bottom by 16 which gets us a probability, um, won over 1 69

Julie Silva · Answer

So we&#x27;re told we&#x27;re going to pick one card from Standard deck of 52 playing guard. So if we want to figure out, what&#x27;s the probability that the card that&#x27;s picked is an ace or a black seven? Well, because there&#x27;s 52 cards in the deck Are denominators 52 Because that&#x27;s our sample size. Well, let&#x27;s think about how many choices we actually have. Well, how many aces air in the deck? Well, there&#x27;s four. And how many black sevens air in the deck? Well, there&#x27;s two, so there&#x27;s a total of four plus two or six cards we could potentially pick from. Well, we can reduce this by dividing both the numerator and denominator by two on. When we do, we&#x27;ll be left with 3/26 so the probability would be

Problem

GAMES For Exercises $30-35$ , use the following i…

Problem 29 Medium Difficulty

Two cards are drawn from a standard deck of cards. Find each probability.
$P(\text { both either black or an ace) }$

Answer

$\frac{63}{221}$

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