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Suppose that you have 11 cards. 6 are greenand 5 are yellow. The 6 green cards arenumbered 1, 2, 3, 4, 5, and 6.The 5 yellow cards are numbered 1, 2, 3,4, and 5. Th...

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Suppose that you have 11 cards. 6 are greenand 5 are yellow. The 6 green cards arenumbered 1, 2, 3, 4, 5, and 6.The 5 yellow cards are numbered 1, 2, 3,4, and 5. The cards are well shuffled. You randomly drawone card.• G = card drawn is green• Y = card drawn is yellow• E = card drawn is even-numberedPart (a)List the sample space. (Type your answer using letter/numbercombinations separated by commas. Example: G1, Y1, ...) Part (b)Enter the probability as a fraction.P(G) = Part (c)Enter the

Suppose that you have 11 cards. 6 are green and 5 are yellow. The 6 green cards are numbered 1, 2, 3, 4, 5, and 6. The 5 yellow cards are numbered 1, 2, 3, 4, and 5. The cards are well shuffled. You randomly draw one card. • G = card drawn is green • Y = card drawn is yellow • E = card drawn is even-numbered Part (a) List the sample space. (Type your answer using letter/number combinations separated by commas. Example: G1, Y1, ...) Part (b) Enter the probability as a fraction. P(G) = Part (c) Enter the probability as a fraction. P(G | E) = Part (d) Enter the probability as a fraction. P(G AND E) = Part (e) Enter the probability as a fraction. P(G OR E) = Part (f) View Previous QuestionQuestion 9 of 26



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Suppose that you have eight cards. Five are green and three are yellow. The five green cards are numbered 1, 2, 3, 4, and 5. The three yellow cards are numbered 1, 2, and 3. The cards are well shuffled. You randomly draw one card.
• G = card drawn is green
• E = card drawn is even-numbered
a. List the sample space.
b. P(G) = _____
c. P(G|E) = _____
d. P(G AND E) = _____
e. P(G OR E) = _____
f. Are G and E mutually exclusive? Justify your answer numerically

And this question, we are given a special deck of 16 cars. So we have four blue, four yellow for green and four red cards. And basically here The four cars are numbered from one through 4. So in part they were asked for the probability that the car drawn is a two or 4. So Two or four basically there's two in each category, so there's 2468 And there are 16 total cars. So that's .5, part B. Were asked for the probability that the card is a two or four given that it's not a one. So we're told that the card is not a one, which basically leaves us with 234 so 369 12 And were asked for a policy that it's either two or four. So that's basically two instances and use of 2468 Divided by four. We have to over three which is Approximately .67 now part C. Whereas the probability that the card is a two or four given that it is either a two or three. So we're given that it's either two or three. So we're told there's two cases here's the two or 32468. Probably that it's a two or a four so we don't have any four. So it's just a probably that is too so it's 1234 so that's 0.5 and then the last one the probability that the card is two or four given that it is red or green. So we're told it's either red or green so that it gives us it possible cards probably there is either two or four so that's 24 to 4. So that's four instances. And the probability here it's .5 and we have nothing Francis.

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.

Total number off cards in her deck is equal to 52. The first part of the question say's probably off or nine. We know that there are two black nines and two red nine. So four C one upon 52 weeks, even four by 52 which is equal to one upon Gordon. Now, the second part say's probably be off blood. There are 26 black guards and 26 red guards in her deck. So probably deal of black would be 26 c one upon 52 7 26 by 52 which is he will do one by toe. The third part say's probably off black nine in our deck that the wounded nine and two black names. So the probability of a black nine would be, Do you see? One upon 50 does even so to upon 52 which is equal to one upon 26. The fourth part says, probably off hard inner deck. There are four times off cards, their hearts, diamonds, spades and trees. So there are 13 hearts, so probably died of a heart would be 13 C one upon 50 does even 13 by 52 which is equal do one by four. Now, the 50 parts is probably the awful face guard in the question. It is given that a face card is okay. Que gee off any sort. Now we know that there are four sorts on. We've been given three guards so foreign to Therese 12. Well, see one upon 52 was even which is equal to 12 by 52. So six by 23 which is equal to three upon 13. No, if part is given are probably tee off or red on up to the which is equal. Do you were there? Don't three. We have home in either guards. We have 2600 guards. So 26 c one Bless how many or trees we have four threes foresee one minus minus the card trajectory as well as their So that would be loose even upon 52 was Even so this would give us 26 plus four, which is 27 2130 30 minus two. So 28 upon for state to so 14 upon 26. Seven A born 14. Now the next parts is probably be off less than a Ford less than four on its. Given that we have to consider races as ones so less than a four include a one, two and three and therefore swords, So the favorable will be four into 4 16 So 16 c one upon 50 to see one 16 upon 52 or 8.26 which is equal to four by 13.

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


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