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Intoduchon you bet on black; both bets on de roulette and bet on red at Supntv [ atd Wc both lose? probability thal What is the least one of us wins? ' probabi...

Question

Intoduchon you bet on black; both bets on de roulette and bet on red at Supntv [ atd Wc both lose? probability thal What is the least one of us wins? ' probability that at b) What is the that at least one of us loses? Wat is the probability ; shuffled and the top two cards are deah deck of 52 cards Suppose pairs of cards could possibly result as outconi How many ordered Assuming cach of these pairs has the same chance; calculate: b) the chance that the first card is an ace; anSWcf E b"

Intoduchon you bet on black; both bets on de roulette and bet on red at Supntv [ atd Wc both lose? probability thal What is the least one of us wins? ' probability that at b) What is the that at least one of us loses? Wat is the probability ; shuffled and the top two cards are deah deck of 52 cards Suppose pairs of cards could possibly result as outconi How many ordered Assuming cach of these pairs has the same chance; calculate: b) the chance that the first card is an ace; anSWcf E b" the chance that the second card is an ace (explain your argument as well as by counting); the chance that both cards are aces; the chance = of at least_ one ace amono the [WO cards.



Answers

Involve a standard deck of 52 playing cards. In such a deck of cards there are four suits of 13 cards each. The four suits are: hearts, diamonds, clubs, and spades. The 26 cards included in hearts and diamonds are red. The 26 cards included in clubs and spades are black. The 13 cards in each suit are: $2,3,4,5,6,7,8,9,10,$ Jack, Queen, King, and Ace. This means there are four Aces, four Kings, four Queens, four $10 \mathrm{s},$ etc., down to four $2 \mathrm{s}$ in each deck. You draw two cards from a standard deck of 52 cards without replacing the first one before drawing the second. (a) Are the outcomes on the two cards independent? Why? (b) Find $P(\text { Ace on } 1 \text { st card and } \text { King on } 2 n d)$ (c) Find $P(\text { King on 1st card and Ace on } 2 \mathrm{nd}$ ). (d) Find the probability of drawing an Ace and a King in either order.

In this problem, we are going to determine the probabilities of certain events. Now to fair dice are rolled a green one and a red one. In the first problem we need to find the probability of getting a sum of seven. Now let the event of getting a sum of seven be denoted by. Let us first find out what outcomes are in this event. Now the event is getting a sum of seven. So first of all we can get a one on the green dye and uh six on the red dye and some will be one plus six equals to seven. So this is one outcome. Another outcome is a two on the green dye and five on the red dye, so that's 25 That's another outcome. Similarly, we can have another outcome of 34. There's another one for free, 52 and 61 in each of these ordered pairs. The first number represents the number on the green day and the second number represents the number on the red. So this is our event. Now, the probability of this event, the probability of getting a sum of seven will be the number of elements in this set A And there are six elements in the set A. So we left six and we need to divide that by the total number of outcomes. Now there are two days and each day's has six outcomes. So the six outcomes for one days and for two guys will be six square. So the probability will be six divided by six square, which is six divided by 36 which is one by six. So the probability in this case is won by six. And the second problem, What is the probability of getting a sum of 11? Now let the event of getting a sum of 11 b. The event b. Now the outcomes in this event will be five comma six and six comma five because it is not possible to get an 11 in any other ways. Five comma six represents that five is on the Green Day and six is on the red one and six comma five represents that six is on the Green day and five is on the red one. Now the required probability will be P L P. Mhm. This will be equals to the number of outcomes in B. That's two divided by the total number of outcomes, which is six square. So that is two by 36 or one by 18. The last problem is what is the probability of getting a sum of seven or 11? Now, in order to do this? First of all, we need to find e intersection. We we need to find the common elements to both of these are events. So Bs 56 and 65 And he has 162534435 to 61 Nothing is common, so a intersection B is just the non set. Because of this, we can see that E&B are mutually exclusive. So the outcomes are mutually exclusive. So the probability of getting a sum of seven or 11 is the key of E or B and they're mutually exclusive. We can use the additional rule of probability to say that this is equals to be of a. Let's be A B. So that will be won by six Plus one x 18. So the least common denominator is 18. We'll have three over here plus one. So that is equals two, four by 18, which is equals to two by nine. If we reduce the numerator and denominator by nine, so the probability of getting a seven or 11 is two by nine, and these two outcomes are mutually exclusive because it is not possible to get a sum of seven and also a sum of 11 in the same rule of two days.

Okay, So 23 is talking about a standard deck of cards. There's 52 decks in 52 cards in a deck. Excuse me. And it says here that we're going Thio draw the first card. But before we draw the second, we're going to put the first one back and reshuffle. So that way, there's an equal chance of getting all 52 cards in both drawings. So, in part A, it's asked, Are those outcomes of the two cards independent? And this time it is, Yes. Because if we are replacing it, Okay, one drawing does not affect the other. Okay. Okay, the second drop. Okay, so now we're gonna go calculate some probabilities here. So here they're asking for the probability of drawing an ace on the first card, then a king on the second. So we're still gonna be multiplying. We're gonna have four chances out of 52 of getting an ace. But remember, now for that second time, we're going to be replacing that card, so we're still gonna have 52 cards and there's four kings to choose from, so there's a probability going to get multiplied, so it's going to be 16 out of 2704. Now, this fraction can be reduced, so check and see if you need to reduce it. Otherwise, if you want. The decimal 16, divided by 2704 is approximately 27040.59 Then for part C, we're switching it up. The probability of getting a king first, then a Nace is actually going to be the same exact set up here because it's still going to be a four out of 52 chance times a four out of 52 chance, which is still 0.59 approximately then for Part D. We want to know the probability of getting a king and an ace and either order. So that would be us adding 16 out of 2007 and four plus 16 at of 2007 and four. That's going to get US 32 out of 2704 which can be reduced a za fraction as a decimal. That's approximately 27040.118

Ready for number 22. We're talking about the basics off a deck of cards, and so we're not replacing. So because we're not replacing, they're asking are the outcomes of drawing two cards independent. And the answer is gonna be No, because that when you draw the first card and you don't replace it, that's going to affect the second drawing. So the first draw and that probability will affect the second draw and that probability. Okay, so for part B, it says, find the probability of getting a three, then a 10. So if there's four threes in a deck that's four out of 52 we're gonna multiply. Um, and then a 10 is gonna be four tens out of now, 51. So that's gonna end up being 16 out of 2652 And that could be reduced. And it's a 0.6% or 0.6 probability. So then, for part C, they want us to switch probability of drawing a 10 than ah three. Well, it's gonna be the same exact thing here. It's gonna be four out of 52 of getting a 10 and then four out of 51 for getting a three. So that's gonna be the same exact answer, Um, and then for part D, it says, What's the probability of getting a 10 and a three in either order? So probability of getting a 10 and three in either order. So what we'll do is we'll add 16 out of 2652 plus 16 out of 2652. That will be 32 out of 2652 and that's approximately 26520.12 chance of happening.

All right. So for 24 we're talking about general deck of cards. Um, we're drawing two cards, but we are replacing this time. So when we replace cards are first drawing in our second drawing, we're gonna have equal outcomes. So by definition, that would be yes, an independent situation. Because each outcome is equal, each outcome is equal, and the first one doesn't affect the other. All right, So for part B were asked to find the probability of drawing a three on the first card, then 10. Well, there's five. Uh, no. There are four threes in a deck out of the 52 there are four tens out of the 52. And since we are replacing this bottom number stays 52. So we're gonna end up getting 16 out of 2704 And then when part C asks us to switch the outcomes a 10, then ah, three. It's gonna end up being the same exact Set up four out of 52 times four out of 52 which equals 16 out of 27 04 If you do 16 out of 27 04 on your calculator. That's gonna be approximately 40.0 59 Part D is asking us to combine thes probabilities of getting a three and a 10 or a 10 than a three. And so that's what we're gonna take that 16 out of 27. 04 plus 16 at a 27 +04 And we're gonna end up getting 32 out of 27 04 Which, if we do 32 divided by 27 04 that's gonna be approximately 40.118


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