2.A card is selected at random from a pack consisting of a totalof 52 cards, including 4 queens and 4 kings. The card is notreplaced. A second card is selected. Fin...

Suppose you are dealt 5 cards from a standard 52 -card deck. Determine the probability of being dealt three of a kind (such as three aces or three kings) by answering the following questions: (a) How many ways can 5 cards be selected from a 52 card deck? (b) Each deck contains 4 twos, 4 threes, and so on. How many ways can three of the same card be selected from the deck? (c) The remaining 2 cards must be different from the 3 chosen and different from each other. For example, if we drew three kings, the 4 th card cannot be a king. After selecting the three of a kind, there are 12 different ranks of card remaining in the deck that can be chosen. If we have three kings, then we can choose twos, threes, and so on. Of the 12 ranks remaining, we choose 2 of them and then select one of the 4 cards in each of the two chosen ranks. How many ways can we select the remaining 2 cards? (d) Use the General Multiplication Rule to compute the probability of obtaining three of a kind. That is, what is the probability of selecting three of a kind and two cards that are not like?

Ready for number 22. We're talking about the basics off a deck of cards, and so we're not replacing. So because we're not replacing, they're asking are the outcomes of drawing two cards independent. And the answer is gonna be No, because that when you draw the first card and you don't replace it, that's going to affect the second drawing. So the first draw and that probability will affect the second draw and that probability. Okay, so for part B, it says, find the probability of getting a three, then a 10. So if there's four threes in a deck that's four out of 52 we're gonna multiply. Um, and then a 10 is gonna be four tens out of now, 51. So that's gonna end up being 16 out of 2652 And that could be reduced. And it's a 0.6% or 0.6 probability. So then, for part C, they want us to switch probability of drawing a 10 than ah three. Well, it's gonna be the same exact thing here. It's gonna be four out of 52 of getting a 10 and then four out of 51 for getting a three. So that's gonna be the same exact answer, Um, and then for part D, it says, What's the probability of getting a 10 and a three in either order? So probability of getting a 10 and three in either order. So what we'll do is we'll add 16 out of 2652 plus 16 out of 2652. That will be 32 out of 2652 and that's approximately 26520.12 chance of happening.

Let's go ahead and choose two cards from a standard deck. Yeah. And the question is, what's the probability of getting one king and one queen? Well, let's look at the denominator first because Each pair of 52 cards is equally likely to be chosen. Our denominators 50 to choose to. So the 52 is for the number of cards in the deck, the tools for the number chosen, And then here you have four kings and you have four queens, and you want to choose one of each, so you have four kings. Four choose one the and becomes a product. So for each of those four choose one ways to choose the king. You also have poor, choose one way to choose a queen and then just go ahead and simplify that fraction there. So at this point you can use the definition of the factorial. So the denominator would be so you can simplify that out a little bit. So that would be a 26, Giving us a four and a four Go a little further here becomes 13

All right. So for number 21 we're talking about probability of cards. And so there's 52 cards in a standard deck, and we're not replacing. So this particular problem is not replacement. So without replacing the card once you draw the first time, Okay, so if that's the case, number 21 A is asking us, Are they independent? And the answer is gonna be? No, because once you draw the first card and then you don't replace it, that's going to change your probability for your second card. So the first event will affect the second events probability and the probabilities change. So that's important. Alright, So Part B is asking us to find the probability of getting an ace, then a king. So since thes air, not independent events, um, you're gonna be multiplying here. We're gonna have four out of 52 times four possible kings out of now 51. So when we do that, we're going to end up getting 16 out of 2652 which you could reduce that fraction or go get the decimal 16 out of 2652 is approximately 26520.6 So about a 0.6% chance of this occurring then in part C. If we switch it probability of getting a king on the first card, then an ace. It's not going to change your still going toe have 4 52 times four out of 51 which we know the answer is 510.6 already. So then what we dio in part D is it says, find the probability in either or they're either order. So the probability of a than K or the probability of a K, then an A. So that means we're just going to be doing 16 out of 26 52 plus 16 out of 26 52 which is 32 out of 26 52 which is approximately point 012 as a probability.

One card is selected at random from an ordinary deck of 52. We're going to let a stand for the event. A face card is selected. We're going toe. Let be stand for in the event that a king is selected and we're going toe. Let's see, be the event that a heart is selected. Okay? And with this information, we're going to determine various conditional probabilities. So let's start with part a part. A. Is the probability of event Be so keep in mind there are 52 cards, and probability is always going to be your favorable over your possible. So in this case, there are 52 possible cards, and the favorable is that we select a king and there are four kings in the deck. So the probability of event be would be four out of 52 which simplifies toe 1 13 and as a decimal, that is approximately equal 2.77 part B. The probability of event be given that we know a okay, so we know that we're dealing with a face card. So the face cards are the jacks, the queens and the kings of each suit. So we know there are 12 possible cards because there are 12 face cards. Of those 12 face cards, four of them are kings. We have the king of diamonds, the king of hearts, king of spades and the king of clubs, which simplifies down into one third or approximately equal toe a probability of 0.3 three three. Well, let's go to Part C. What's the probability of be given that we know? See? So this time we know that card is a heart. Well, there happened to be 13 hearts in the deck, and we want a king as the favorable Well, there's only one king of hearts, so therefore, the probability of be given see would be 1/13 which is approximately 0.77 Letter D. What's the probability of be given that not a so not a means? It's not a face card, so not a means that there are 12 less cards in the deck, so there are 40 cards that air not face cards. If they're not face cards, then we'll never get a king. So therefore, the probability of be not a is going to be zero letter E. What is the probability of a well again probability is favorable over possible. There are 52 cards in the deck, and there are 12 face cards as are favorable, and that will reduce down to three out of 13, which, as a decimal, is approximately 0.231 Let her act. What's the probability of a given B? So be Waas that it was a king. So there are four possible kings in the deck. So when we're drawing from those, what's the chances of them having or being a face card? Well, all four would be face cards, so the probability of a given B would be one part G. What's the probability of a given that we know? See, we'll see was a heart, and we know that there were 13 possible hearts of those 13 possible hearts. How many are face cards? Well, there's the jack of hearts, the queen of Hearts and the king of Hearts. So are favorable. Is three making our probability three out of 13, which is approximately point to 31 and the final part to this problem, Part H. What is the probability of a not be well, not be means it's not a face card. Sorry, not being means it's not a king, so that removes four cards from the deck. So we now have 48 possible cards. And if we've removed the four Kings from the deck, we have removed four of the face cards. So there's only eight possible fakes cards leftover that we could access, as are favorable cards, so that simplifies down into 16 which is approximately 0.1 six seven.