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(For the questions asking for probability, put your numerical answer in percentage and rounded to 2 decimal places For example_ if the answer is 0.22575, you have t...

Question

(For the questions asking for probability, put your numerical answer in percentage and rounded to 2 decimal places For example_ if the answer is 0.22575, you have to put 22.58% in the box. DO NOT FORGET TO PUT THE PERCENT %) SIGN AT LAST. Putting only 22.58 without the trailing percent sign will be assessed as a wrong answerl You must NOT type any other symbol, space, operator; fraction etc in the box)Suppose you are repeatedly picking one random card from a deck of cards with replacement (that

(For the questions asking for probability, put your numerical answer in percentage and rounded to 2 decimal places For example_ if the answer is 0.22575, you have to put 22.58% in the box. DO NOT FORGET TO PUT THE PERCENT %) SIGN AT LAST. Putting only 22.58 without the trailing percent sign will be assessed as a wrong answerl You must NOT type any other symbol, space, operator; fraction etc in the box) Suppose you are repeatedly picking one random card from a deck of cards with replacement (that means you pick a card randomly, see it, put the card back to the deck again and shuffle. You keep doing it repeatedly) What is the probability that you get your first Diamonds card on the 3rd pick? Unanswered ii_ What is the probability that you']l get AT MOST 2 clubs after picking 5 times? Unanswered



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You are playing with an ordinary deck of 52 cards by drawing cards at random and looking at them. a) Find the probability that the card you draw is (i) the ace of hearts (ii) the ace of hearts or any spade (iii) an ace or any heart (iv) not a face card. b) Now you draw the ten of diamonds, put it on the table and draw a second card. What is the probability that the second card is (i) the ace of hearts? (ii) not a face card? c) Now you draw the ten of diamonds, return it to the deck and draw a second card. What is the probability that the second card is (i) the ace of hearts? (ii) not a face card?

In this question, it says, What is the probability of selecting a random position Inserting the card with the result that the card is inserted correctly? I can accept the card in four different ways, and there is only one correct positions. So the answer for the first one is one by four. Now for option for Sorry for Part B. I want the car to be inserted in correctly in the first attempt. So the probabilities three by four and correctly in the second attempt, which is one by four. Or this turns out to be three by 16 three by 16. And the last question says, how many random selections are required to be absolutely sure that the card works because it is inserted correctly. And how many different ways can I insert the card for possible ways? So this is my answer to see these are my answers

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.

Should resolve this question. Critical numbers typically have 16 days. It's and not all of them are random. But not all of them are. And what is the probability of randomly generating 16 digits and getting your MasterCard number? Well, there are 10 numbers. Alright, there are 16 places. There are 16 places, and on every place there are 10 numbers. So which means there are 10 raised to 16 possible combinations. It means the probability off me getting my master card number just by guessing is one by 10. Raise to six. Yes, receives often show the last four digits of the credit card number. If only those last four digits are known. What is the probability of randomly generating the other digits off your MasterCard number? Well, now I know the last four digits, which means I still have to guess that 12 remaining digits. So my probability in this case is going to be won by 10 Raised to 12. The discovered cards begin with digits 611 If you know the first four digits and you also the excuse me and you also know the last four digits. What a discover card. What is the probability off randomly generating other digits and getting all of them correct. This means I know eight visits in all I know the first four digits. And I also know the last four digits. Now there are eight. They just in the middle that I wanna guess. So what is the probability of getting them right? One divided by 10 raised to it. This is my answer for the last part. Well, is this something to worry about? Well, I wouldn't think so, because, you know, still, this is a very small probability of getting it right in the first time. But yes, I mean, it is only obvious the more digits you reveal, the higher the probability gets.

This question is asking us to calculate the probability of getting a certain five card hand. Specifically, we're trying to get three of a kind when you're just being dealt that five card hand from a standard 52 card deck. And the the four parts of this question are really just leading us through the big answer because this is a pretty complicated probability. So looking at part, we want to know how many ways can five cards be selected from a 52 card deck. So for a part a I've got 52 total cards and I'm choosing a group of five. I'm using a combination here because the order of these cards doesn't matter, you're just getting a group of five cards when you're dealt that hand. And so the combination remember uses the first the big totals, the first number, so 52 cards in total. And we are essentially counting groups of size five, so that's where the five is coming from. So this number comes out to be 2,598,960. So that's how many total five card hands there are. That's going to be the denominator of our eventual probability calculation. Now, in part B, we have to be able to figure out how many ways you can get three of the same card selected from the deck. So you want to imagine that you're kind of rifling through the deck of cards and you are picking your group of three of one kind. And so the way, that's how I like to think about it. And so imagine, you know, first you have to decide um which which kind do you want? Which three of a kind? You want? Three aces? Do you want three kings? Do you want 35? And the number of choices you have for that is 13 Because there are 13 different kinds. You can also phrase this as 13, choose one Because of the 13 different kinds. You're just picking initially one of them to be the three of the kinds that you're going to want. So once I've, let's say decided on queens, I want three queens. Now I have to figure out OK, how many groups of three queens can be chosen because there are four total. So I'm going to multiply this by of the four queens. I now need to choose groups of three. Okay, so this is going to count the number of three of the kinds that you can get Just with the three cards. So this comes out to be 52 possible groups of three of a kind. Now for part C. I have to now get two other cards to complete my five card hand to imagine right, I've got my three queens in my hand. I had a lot of ways to choose those 52 ways in fact, but now I have to have two other cards in my hand to complete it. And to be three of a kind, you should have two different cards. So if I already have three queens, I can't have another queen, and the other two cards also can't match each other. They have to be two completely different cards, like a king and a three or something like that. So first thing I want to do here is decide which two other kinds of cards do I want. So I've got the queens and now I need to other kinds of cards, so of the remaining kinds, right? There's only 12 left since I've already used up the queen's, I need to pick two other kinds and once I've decided on those kinds, maybe I've picked kings and uh fives, I now need to pick which king in which five. So of the four kings, I have to pick one And of the four or 5, I have to also pick one. So this calculation now tells me how many ways I can have two other different cards. And it comes out to be 1,056. So now I've completed my five card hand and I'm ready to calculate the final probability in part D. So in my numerator, I want to have the number of ways for three of a kind, which remember also means those two other different cards. And then I'll divide that by the total number of hands. And so my numerator will be the product of 52 and 1056 because that's how I'm getting the three of the kind also with the two different cards. And then I'll divide it by that 2.5 million number from Part A. Because that's how many total five card hands there are. And so then this calculation comes out to be about 2.1%.


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