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If two successive cards are drawn from standard 52-card deck (without replacement) what is the probability of the following?(a) the first card is two and the second...

Question

If two successive cards are drawn from standard 52-card deck (without replacement) what is the probability of the following?(a) the first card is two and the second card is ten or better (including aces) (b) the first card two and the second card is spade (Watch out for the two of spadesl)(a) The probability that the first card is two and the second card is ten or better (including aces}663(Type an integer or simplified fraction:)(b) The probability that the first card is two and the second card

If two successive cards are drawn from standard 52-card deck (without replacement) what is the probability of the following? (a) the first card is two and the second card is ten or better (including aces) (b) the first card two and the second card is spade (Watch out for the two of spadesl) (a) The probability that the first card is two and the second card is ten or better (including aces} 663 (Type an integer or simplified fraction:) (b) The probability that the first card is two and the second card is spade (Watch out for the two of spadesl) is 663 (Type an integer or simplified fraction )



Answers

A card is drawn from a deck and replaced, and then a second card is drawn.
(a) What is the probability that both cards are aces?
(b) What is the probability that the first is an ace and the second is a spade?

In this problem. We've got some card problems. Um, so in part, a question want What's the probability of drawing the A supports? Well, there's only one in a deck of 52 so this will be one out of 52 and part two. We want either the ace of hearts. So there's one option there or the spades. There are 13 options there, so this will be 14 out of 52. And when we reduce it Ah, this will be 7/26. In Part three. We want an ace or a heart. There are four aces and there are 13 hearts, but we've counted one of them already in the aces. So this will be, Ah, four plus 12 which will be 16 out of 52. And if we divide by four, this will be 4/13 non face cards. So the cards that are not Jack, Queen or King so that's taking out 12 cards 50 to minus 12 will leave us with 40 out of 52. And ah, when we reduce that, it will be 10 out of 13 in Part B. We've drawn out of 10 of diamonds already now, what's the probability of drawing the ace of hearts? Will we have one less card? So that will be one out of 51? Because the 10 has been drawn with the non face cards again. Remember, we have 12 face cards, so that was 40 non face cards, but we've already taken out a 10 so that will be 39 out of 51 and I think we can divide by three. So that'll be 13. Divide by three would be 17 on the bottom in part C, and we drawn the 10 of diamonds and then we put it back in the deck. What's the probability of drawing? The is a farts? Well, nothing's changed then, because we've put it back in. That will be won over 52 and, ah, non face cards would be the same as part A. Because we've put the card back in. That will be 40 out of 52 which reduces to 10/13

In this question has been on a date. Total number of cards, couple number of cards is equal to Simple Space is equal to 52. So as we know date number of yes cars in this for two is equal to four. So the probability of getting this card will be go to mm four they went by 52 and it's not the number of number card number dogs Is equal to as little 36 because This color content four cards, 39 cards number 30 From 2 to 10. So probability of getting member card will go to 36 D one x 52. Now we can find the probability of getting is in number power, probability of getting discard and number card will be good to probability of getting this card will take my property of meeting number card which is good too. 36-1 x 52 multiplied by 41 by 52 for four For 1352 For 9, 36 for 1352. So when it's obviously get 90 right by 169 is the property of getting yes and number card, which is the answer this question. Yeah. Thank you for watching video.

In this problem, we are going to determine the probabilities of certain events. Now in this problem, cars are built one at a time from a standard 52 card deck. In the first problem, it is said that the first two cards are both speeds and what we need to find is the probability that the next three cards are also speeds. Now, since the first two cards and spades, that means that there are out of a total of 13 spades, there are only 18 speeds left because in a standard 52 card deck, there are 13 spades. So since the first two cards in spades, now there are only 11 speeds left. So in order to find the probability, first of all, we need to find the total number of favorable outcomes and out of the 11 speeds that we have, We need to choose any three spades. So that's the total number of favorable outcomes. 11 C3 and The total number of outcomes. Since two cars have already been dealt over the 52 card check, that means there are only 50 cars left. So the total number of outcomes will be 50 C3. Because we are choosing any three cards out of 50 cars. So if we evaluate this 11 C3 is equals to 11 Times 10 times nine, Divided by one times 2 times three. And we divide this by 50 C three, so 50 C three's 50 times 49 times 48. And we divide that by one times two times three. So we can cancel out these three times and we are left with 11 times 10 times nine development 15 times 49 times 48. So the value of that can be calculated to be 0.0084. This is an approximate value. The value is approximately 0.0084. Now, in the second problem, it's said that the first three cards are all speed. What is the probability that the next two cards are also speeds? So, over here, the first three guards are speed. So if you want to find the probability out of 13 spades, three cards are already taken, three of them are speed. So we are left with only 10 speeds and we need to choose two cards, so that the next two cards and spades. So that means that the remaining 10 speeds, we need to choose any two of them. So 10 C two is the number of favorable outcomes And the total number of outcomes. Since it's a 52 card deck and three cards have already been drawn, that means there are 49 cars left and out of these 49 cards, you need to select two. So that's 49 C two. So if you calculate this will get 10 times nine divided by one times two and 49 C two as equals to 49 times 48 divided by one times two. So we can cancel out one times 21 times two. We are left with 10 times nine, divided by 49 times 48. And the value of that is approximately equals two. Zero point 38 It's approximately goes to zero Um Sorry 0.0383. This is the approximate value. 0.0383. That is are required probability. And in the last problem in the first four cards in spades and what is the probability that the next card is also spade? So out of 13 spades, four of them have been selected already. So we are left with only nine spades. And out of these nine spades, we have to choose just one. That's 91 and out of 52 cards for cards have already been selected. So there's a remaining number of cars, 48 out of that, we have to choose one car. So this is the probability 91 is equal to 9 48. C one is equals to 48. So the probability in this case will just be 1948 which is equals to zero point 1875.

In this problem, we are going to determine the probabilities of certain events. Now in this problem, cars are built one at a time from a standard 52 card deck. And the first problem, we need to determine the probability that if the first two cards are both space, then what, what is the probability that the next three cards are also spades? So in order to find the probability first of all, we need to find the number of favorable outcomes. Now in a standard 52 card deck, there are 13 speeds. And since the first two cards are both space, that means out of these 13 spades, the number of spades left is 13 -2, which is left. And since we need to find the probability that the next three cards of space, that means we need to choose three speeds out of the remaining 11 speeds. So the number of ways that can be done is equals to 11 C3. Now the total number of outcomes also needs to be determined. There are 52 cards and two cars have already been drawn, so that means that there are 50 cards remaining and out of that three cars need to be selected. So that means the total number of outcomes will be 50 C3. So if this is calculated then 11 C3 is equals to 165 and 50 C3 is equals to 196 00 So 1 65 divided by 19600 is approximately equals to 0.84 So that is the required probability in this case. Now, in the second problem, If the fullest three cards of space and what is the probability that the next two cards are also speeds? So we first of all need to find the number of favorable outcomes. Out of 13 spades, three spades have already been selected, so the remaining number of speeds is 13 -3, close to 10. And out of that we need to find the next two cards are speed. So out of those 10 remaining speeds, we need to select to space, so a number of ways that can be done is equal to 10 C three and the total number of outcomes since there are 52 cards in the deck and three of them have already been selected. So the remaining number of cards is 50 to minus three, which is equal to 49. And out of that we have to choose a total or two cards, so that's 49 C. To actually the numerator also needs to be that tendency to we need to choose two cards out of the remaining 10 spades and we need to choose two cars are the remaining 49 cards, So the value of 10 C. Two is equals to 45 and the value of 40 92 is equal to 1176. And the value of 451176 is approximately, it goes to 0.0383. So that is the probability in this case. And in the last problem, four cars are selected, They are all spades. What is the probability that the next card is also a spade? So in this case four cards and spades are 13 cars. Four of them have already been selected out of 13 spades, four have been selected, So the remaining number of spades will be 30 minus four, which is nine, so that's nine C. One because we only need to choose one more car. And Out of 52 cards, four cars have been drawn, so there are 48 cards in the deck, and out of that we have to choose one. So 48 C1 is the total number of outcomes, So 91 is nine and 48 C1 is 48 and nine, divided by 48 is equals to 0.1875. So that is the probability in this case.


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