find-each-probability-if-13-cards-are-drawn-from-a-standard-deck-of-cards-and-no-replacement-occur-4

Question

Find each probability if 13 cards are drawn from a standard deck of cards and no replacement occurs.
$P(	ext { no kings })$

Charles Carter · Accepted Answer

this problem wants us to calculate the probability of being no kings when drawing 13 cards with no replacement for a standard deck of cards. Eso these This is ah, Israel dependent events depending on the draw before. So we&#x27;re gonna have 13 fractions or 13 probabilities. Okay, we&#x27;re going to find the pattern here. So the first draw is out of 52 cards and there are 48 cards which were not kings and therefore kings in the debt. And then as we draw, uh, our denominators denominator will count down. 52 51 50 49 48 7654321 40. Okay, And then our memories will count down as well. It&#x27;s our last number will be 36 and we can Ah, simple like counting in multiply like this were used factorial so you can just plug this into the calculator on the way. I would type this in is 48 factorial and we wanted to stop at 36. So we just divided by 35 pictorial have everything less than 36 cancel in the denominator will have 52 factorial and then divided by 39 factorial. So you want to cancel everything? Less than 40. Okay. And then punched that into a calculator and we&#x27;ll get a probability of 0.30 20.304 which is around, uh, 38.4%. Okay. Thank you.

Charles Carter · Answer

problem. It wants us to find the prophet of drawing all one suit, Uh, in 13 on draws. Yes, we have 13 drawings and cards. So we have 123456789 10 11 12 13. I&#x27;m just doing this to kind of organize them thinking so the first draw. We don&#x27;t know what suit we&#x27;re drawing yet, so it doesn&#x27;t matter what we draw. So as long as we draw a card seeking the right, 52 out of 52 are just drawn the cards. It&#x27;s gonna be one either way. So now that we know, what suited is, there&#x27;s only gonna be 12 left and 51 cards left and then successfully, we have to draw the rest of that suit now down to the last card and 51 will count down to 40. Okay. And then how I would put this in a calculator if I had the factorial button, I would use that for sure. So I have 12 factorial on the top, about about 51 factorial on the bottom and then divided by 39 factorial. So the previous problem of drawing all hearts, this should just be that value Times four. Because now we have four suits to choose from. So it&#x27;s the same probability for each one and then said because there&#x27;s four of them because multiply that probability, but for, um, But our final answer is just gonna be 6.299 I&#x27;m staying to the negative 12. Okay, thank you very much.

Charles Carter · Answer

this different problem, drawing all red in 13 draws of the center for playing cards. So for 13 draws, we&#x27;re gonna have these are all dependent on each other because we&#x27;re not replacing one of 13 fractions. And so this is just to kind of side notice out. So the first draw is gonna be out of 52 cards and there aren&#x27;t 26 red cards, half the debt, and then each one will be counting down Torre. Last row will have out of 40 cards left and 26 25 24 23 22 2120 1918 17 16 15. Last Roby, only 14 red cards left. Okay, so when you use a calculator to find this safety, there&#x27;s multiply down and divine. One helpful thing is, if you have a factorial button on your calculator, you could do 26 factorial, which is just 26 times everything down to one. We&#x27;re gonna divide it by 13 factorial and then on the bottom, right? We&#x27;re starting at 52 factorial. They were divided by 39 factorial. Just the next number down after our we only stopped on skins are final probability here in Houston Calculator. It&#x27;s just 1.637 1.638 I&#x27;m sorry. Just get the rounding. Right? It&#x27;s times 10 to the negative death. All right. Thank you very much.

Charles Carter · Answer

and this problem&#x27;s is defined the probability of drawing all 13 hearts and around. But oh, just drawing 13 cards. I think so. For this event, we&#x27;re gonna have 13 probabilities there are gonna be dependent on each other. Okay? And then for our first, um, we&#x27;re just drawing a single heart. That just be the probability of 13 and 52 became our second probability because it was dependent on the first. Because no replacement is occurring are total cars goes on the 51 are total hearts, goes down to 12. This pattern is gonna continue all the way down s so we&#x27;re gonna have 13 times, 12 times 11 all the way down to one and then 52 counting down all the way down to 39. Okay, so we&#x27;re gonna need some calculator for this. The best way I can think of to put the senate calculators. If you have the factorial button, 13 factorial just means 13 times everything down the line, which is exactly what we need. And then I&#x27;m gonna do 52 factorial and we need 52 51 50 49 48 47 46 45 4 or 3 42 41 40 Yes. The last one should have been 40 chances to 40. I&#x27;m gonna make sure did this right. Okay, so which means we&#x27;re gonna divide out 39 factorial that way. It just gives us 52 times everything down to 40 because we&#x27;re gonna take out the 39 factorial with that 39 factorial piece. Okay, so the probability of this happening put it in a calculator, it&#x27;s just gonna be 1.574 That times 10 to the negative 12. So it&#x27;s very unlikely that this particular action would happen on drawing 13 cards and getting all 13 hearts. Okay, thank you very much.

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

Charles Carter · Answer

find the probability of drawing an ace and then a king, um, out of a state a deck of cards. If no replacement occurs, that means that these two events are depended on each other, so we have to do these in order. So drawing the ace first is ah, like a four. And there&#x27;s four aces and a standard Dex. That&#x27;s a four and 50 to chance, Um, which we can reduce the minutes seconds and then because we&#x27;re not replacing it are total amount of cards goes down to 51. And but we haven&#x27;t changed our number of King, so we&#x27;re still gonna have four kings. Okay, So reducing that first function on a four out of 52 just 1 13 I still don&#x27;t want about four out of 51. Since they did not share common factors. It&#x27;s our final answer. Well, before is ah well calculator. So 13 comes 51. It&#x27;s 6 63 since there were no common factors and the floor there aren&#x27;t common factors Now it&#x27;s the probability of getting an ace, and then a king without replacement is four out of 663. Thank you very much.

Problem

For Exercises $46-48$ , use the following informa…

Problem 45 Hard Difficulty

Find each probability if 13 cards are drawn from a standard deck of cards and no replacement occurs.
$P(\text { no kings })$

Answer

0.304

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