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Point) A deck of cards has; 6 cards of suit A. cards of suit B 9 cards of suit C A hand is a subset of 4 cards_ How many hands are there?(b) What is the probability...

Question

Point) A deck of cards has; 6 cards of suit A. cards of suit B 9 cards of suit C A hand is a subset of 4 cards_ How many hands are there?(b) What is the probability of getting a hand with: 1 card of suit A 1 card of suit B, and 2 cards of suit C? Please round to at least 6 decimal places_

point) A deck of cards has; 6 cards of suit A. cards of suit B 9 cards of suit C A hand is a subset of 4 cards_ How many hands are there? (b) What is the probability of getting a hand with: 1 card of suit A 1 card of suit B, and 2 cards of suit C? Please round to at least 6 decimal places_



Answers

A standard deck of cards has 52 cards divided into four suits: clubs, diamonds, hearts, and spades. Each suit has 13 cards consisting of an ace, a king, a queen, a jack, and cards numbered from 2 to $10 .$ Assume that one card is selected from a standard deck. For Exercises $37-48$, find the probability of selecting the indicated card. A heart or a 6

Let feel you or b denotes the ability off selecting a card with number five or with number 10 which is equal to P off a plus p off be were p off a you know, double belt e off selecting the card with number five and p off. We do you know the operability off? Selecting a card with number 10 which is equal to the number off cards with number five upon total number off cards plus B O. B is equal to the number off cards with number 10 a bone number off total number of cards which is equal to four upon 52 plus four upon 52 which is again equal to it upon 52 which is equal to two upon today in this is our answer.

Okay, so this farm's got a multi part solution to it. So part of me just wants us to to describe the sample space of drawing five cards from a standard deck of cards and on Lee recording the suits said to describe it, the first thing we can do is we can get hearts and get all five to be hards can, uh, the 2nd 1 we could get four hearts in a club. 3rd 3 spades, two hearts. Um, so basically, we're drawing cards without replacement, so each heart that we grabbed would be less likely to get the next one. Um, and so our sample space could be fairly large. Um, and so the sample space size eyes, uh, gonna eat? I'm gonna call it s. That's gonna be We have a choice of 52 for the first slot. 51 50 49 48. Ah, and that number, it's relatively large, is around 312 million. Uh, so it's It's a fairly large number of the size of this sample space. The 3 11 8 75 200 that is the size of our samples base. Okay, Part B. What's the probability that the set of five cards you draw consist of two spades, one heart, one diamond and one club on draught in any order. Okay. So to calculate that particular case, we have out of the 311 hit 75 200 ways. Um, actually, okay, we we we do want to draw this in any order. Okay? So the number of hearts to choose from on the first draw, let's just say the first draw has to be, um the most limiting is that we have to go to spades. So we're gonna have 13 spades to choose from and then 12 on. And then we have one heart, which is 13 one diamond and one club. So each one of those only has 13 All right choices, because we didn't have to do any, Didn't have to draw a second time for anyone. It does. Okay, so the probability of this happening his 13 times, 12 times, 13 times 42,732. Okay. And then we're dividing by our 311 million. The likelihood of this happening is around 0.1 zero, like 0.1% of the time. This will happen. Okay. I think a lot of this is just there are only 342,732 ways to do this out of the 300 almost 12 million. Okay. Part. See, uh, what's the problem? That exactly two of the five card draw are from the same suit? Um, exactly two. So capture the probability of getting too out of the same suit. We're still gonna have that large sample space of the 3 11 number and then on top, um, our first choice. Weaken draw pretty much anything. So our first choice can be out of 52. All right. Let's, uh I'm still kind of half thinking about listen. Okay, so, um, let's say we get a heart. It's a draw. The first thing we're gonna draw is yeah, I just one card of any suit. Um, So in order to do that way, have 52 choices. Uh, this second card we're gonna draw. Oh, my gosh. This is hard to visualize. Give me one second. It was tried this way. There are 13 ways to get a spade. Okay, that our next card We have 13 ways to get a heart. You may have 13 ways to get a diamond in 13 ways to get a club. So then after that, we want exactly one of them two pair. It doesn't really matter what we draw because the next car that we draw will pair with one of those. So of the cards that are left, there are 48 cards left. So that'll get our total number of ways to draw exactly one pair that repaired. Just buy the suit. So that probability is going to equal around 0.4 1.4% 0.44%. And that's all. Okay, thank you very much.

Let and denote the number off face. Carl's is equal to hello and and no festive note. Don't total number of cars which are equal to 52. Now we need to find the probability for selecting a face card which is equal toe be, or five, which is equal to a number of face cards divided by the total number off cars physical to hello upon 52 which is again equal to three. Upon putting, this is our answer.

Lead B o A. Or B denotes the ability of selecting ah card off diamond old card with number two, which is equal to the P off a Robel tee off selecting a diamond card plus probe lt. Or selecting a guard with number two minus probe. Lt off selecting uh oh God! With number two off diamond, which is equal to number off diamond cards divided by the total number of cars plus number off cars with number two do I do by the total number of cars minus number off diamond card with number two and divided by the total number of cards. So we're putting the value of and no fe. We have 13 and by putting the value of an office, it is equal to 52. We have 13 upon 52 now, by putting the value Pandolfi and manifest, it becomes four upon 52 my putting the value open off a intersection B a bone and no fest. We have one upon 15 to now by simply find it, we have 16 upon 50 do, which is equal to four. Upon starting that is about cancer


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