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Five cards labeled with numbers: 0, 1,2, 3,and 4 respectively. Two cards are selected one after the other, without replacement Each card is being equally likely to ...

Question

Five cards labeled with numbers: 0, 1,2, 3,and 4 respectively. Two cards are selected one after the other, without replacement Each card is being equally likely to be selected. Define the random variable X(s) Sum of numbers appeared on the two selected cards The number of elements in the range of X (Rx) is2. Ihe value ol P(X = 3) 60400.1429 Qo;00.2

Five cards labeled with numbers: 0, 1,2, 3,and 4 respectively. Two cards are selected one after the other, without replacement Each card is being equally likely to be selected. Define the random variable X(s) Sum of numbers appeared on the two selected cards The number of elements in the range of X (Rx) is 2. Ihe value ol P(X = 3) 604 00.1429 Qo; 00.2



Answers

Two cards are drawn in succession from a deck without replacement. Find the probability distribution for the number of spades.

Okay, This question gives us a numbered set of cards from 1 to 5, and we're gonna take a set of two cards once they're shuffle. So part A just asks how Maney two cards sets are possible. So for this one, we have five total cards and we want an a Norden subset of to because we just care about the hand, not the order. So that's just five factorial divided by two factorial times. Three factorial because five minus two street. So if we do that, we're just going to get an answer of 10. So there's 10 different hands possible in Part B. Says how Maney to card hands contain a number less than three. So we want a number less than three. So what I'm going to say for this one is let's just find the number of sets with a number not less than three, and then subtract that from the total. So we know we have 10 total sets and let's see how many of those contained numbers that are not less than three. So we have three cards not less than three, and we need two of them to make our set, not follow the rules and three choose to is just equal to three. And this is the number of cards. Sorry. Number of sets without a card less than three. So the number with less than three is just the total minus that. So 10 minus three, which is seven. And I just chose to do it this way because I found it easier to just find the number of sets that we don't want and subtract that from the total which we already found and then again for this combination.

In problem. five. We select five cards. Five cards at random without replacement from an ordinary deck of playing cards. An ordinary deck contains 52 cards. These 52 cards is divided by four. We have four colors. Each color contains the same 13 cards, but in different colours for birth. Eight. We want to find the probability mass function for the random variable X. Where X represents the number of parts in the five courts first, let's let's get the total number of combinations. To select five Out of 52 cards. To get the probability we divide by the total space for the experiment. The total is based We have 52 cards. We choose five. And here let's define our problem. We want to find a number of combinations To have the number of hearts in the five cards. What are the possible number of hearts to have in the five cards? The first possible option Is to have zero cards. 0 hard cards. Heart is one of the colors we have four colors. We have bids, Hearts, diamonds and clubs. These four hearts speeds, diamonds and the clubs. Then We can have the five cards to be one of the other three cars, The other three colors from speeds the diamonds to clubs. Then the first possible option that X equals you or one Or two or 3 or four or 5. Let's take it as an ex. What are the number of combinations to have? And select X. Numbers out of the hearts. Hearts contains 13 court. Then we have a number of combinations of 13, choose X. X. Might be 012 or three or four. Or fight. This is for the number of course let's Assume that it's one. Then we should multiply by the number of combinations for the four other places. Because we select five cards here, we only selected one. What are the other combinations? To select the other four? It's the remaining number of courts Which is 13. multiplied by 13. 30. multiplied by three, sorry, Or 50 to -13 Which is 39. These are the remaining course. That are not hurts and we choose of them for if we have one we choose for and if we have to which was three. This means we choose X five minus X. If X is one here which is four and so on. This represents the probability mass function for the random variable X. Where X represents the number of fourth in the five course who brought me? We want to determine P of X to be smaller than records one. It equals the submission for B exotics. Where X starts at zero because it is a minimum value of X. Until we ritual then in just the probability we substitute here by X equals zero, plus X equals one. Then It's 13, Choose one, deployed by 39, choose four, Divided by 50 to choose five. Close 13, choose 2-0. Sorry, multiplied by 39 Shoes, five Divided by 50 to choose five. By creating this, we get the required probability equals 2109 Divided by 3332.

In problem. five. We select five cards. Five cards at random without replacement from an ordinary deck of playing cards. An ordinary deck contains 52 cards. These 52 cards is divided by four. We have four colors. Each color contains the same 13 cards, but in different colours for birth. Eight. We want to find the probability mass function for the random variable X. Where X represents the number of parts in the five courts first, let's let's get the total number of combinations. To select five Out of 52 cards. To get the probability we divide by the total space for the experiment. The total is based We have 52 cards. We choose five. And here let's define our problem. We want to find a number of combinations To have the number of hearts in the five cards. What are the possible number of hearts to have in the five cards? The first possible option Is to have zero cards. 0 hard cards. Heart is one of the colors we have four colors. We have bids, Hearts, diamonds and clubs. These four hearts speeds, diamonds and the clubs. Then We can have the five cards to be one of the other three cars, The other three colors from speeds the diamonds to clubs. Then the first possible option that X equals you or one Or two or 3 or four or 5. Let's take it as an ex. What are the number of combinations to have? And select X. Numbers out of the hearts. Hearts contains 13 court. Then we have a number of combinations of 13, choose X. X. Might be 012 or three or four. Or fight. This is for the number of course let's Assume that it's one. Then we should multiply by the number of combinations for the four other places. Because we select five cards here, we only selected one. What are the other combinations? To select the other four? It's the remaining number of courts Which is 13. multiplied by 13. 30. multiplied by three, sorry, Or 50 to -13 Which is 39. These are the remaining course. That are not hurts and we choose of them for if we have one we choose for and if we have to which was three. This means we choose X five minus X. If X is one here which is four and so on. This represents the probability mass function for the random variable X. Where X represents the number of fourth in the five course who brought me? We want to determine P of X to be smaller than records one. It equals the submission for B exotics. Where X starts at zero because it is a minimum value of X. Until we ritual then in just the probability we substitute here by X equals zero, plus X equals one. Then It's 13, Choose one, deployed by 39, choose four, Divided by 50 to choose five. Close 13, choose 2-0. Sorry, multiplied by 39 Shoes, five Divided by 50 to choose five. By creating this, we get the required probability equals 2109 Divided by 3332.


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