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Thacisajur containing tuce bluc balk nnd two ycllow balls. You will drw 4 bell fom this jer and rcordthc color of thc balLYou will draw anothcr ball from thc rcmain...

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Thacisajur containing tuce bluc balk nnd two ycllow balls. You will drw 4 bell fom this jer and rcordthc color of thc balLYou will draw anothcr ball from thc rcmaining balls in tc jar and rccord thc color: You do not Iplacc thc first ball in thc jar_ At thc cnd, you will huavc two balls drwn from thejer in ordcr:Answcr thc following - qucttionsDraw bcc dingram show all possiblc outcomncs of this two-stagc cprrimant Provid 4 complctc tTec dingmm wulh probabilitics cach branch and thc probabilitic

Thacisajur containing tuce bluc balk nnd two ycllow balls. You will drw 4 bell fom this jer and rcordthc color of thc balLYou will draw anothcr ball from thc rcmaining balls in tc jar and rccord thc color: You do not Iplacc thc first ball in thc jar_ At thc cnd, you will huavc two balls drwn from thejer in ordcr: Answcr thc following - qucttions Draw bcc dingram show all possiblc outcomncs of this two-stagc cprrimant Provid 4 complctc tTec dingmm wulh probabilitics cach branch and thc probabilitics ncxt t0 cach outconk What probability that the two balls arc different colons. What thc probability that the balls arc the same color' Show your work.



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An urn contains 3 green balls, 2 blue balls, and 4 red balls. In a random sample of 5 balls, find the probability that both blue balls and at least 1 red ball are selected.

First question, there are six balls in a in A on on their identical except for color. So I read three are blue and one is yellow. So that's six in total. Now you are to draw a ball from the own notes is color on setting aside? Then you drawn on that ball from its own unnoticed color. So you draw twice. Now, the first question wants you to make a tree diagram that shows all possible outcomes of the experiments and level the probability association with each stage off the experiments on the appropriate branch. So I'm going to be joining three drug 83 diagrams on We're going to start with the starts, so I'm gonna have a start right here, and there should be 2/6. Then we'll have a circle right here, which is our then should go right here. They have enough circle year, which is our then the other woman. Come right here. And this is why. And this is one of us six and this is gonna lead to that one right here. Which is why and then from here we can draw the branches. This is gonna be 3/5. And it's going to be big, which is Blue Manju. Aris red base blue wise yellow. So this is one about five. And then it's going to have another branch over here as yellow and one about five. At least 1/15. So now we're going Thio, draw the side. So to create space, I'm going to take this for that down. So just to create space, how am I going to have Why? Here? This is one bull. This is 11 ball I'm gonna have. Is everybody here? We're gonna have zero right here, too. Them are going to draw. Good lie. I will have Be right here on is one of attain on the next branch. This is 3/5. Next brush is to over five and this is Red. I would have won over 15 Now, that's for that. Now, the media portion. I might be taking it out a little bit because of space, but this is 3/6, and you could have put it in there. But because of the space I brought out the group. This is blue. And then the branches would look like this. It's so over five over here and red over here. One of our five. The meter branch should look like this blue over a year on the next branch. You looked like this. What about five over a year And this is yellow. 1/10. So this is what the three the tree diagrams would look like. We all the probabilities and possible outcomes showing. And that's the answer. Now the other part of the question probability, extension, Um, compute the probability for each outcome off the experiment. So from this apple tree diagram, we can show that it shows all the possible outcomes and displace the probability for the upper pitch branch. So the next question if am be after independence events, then probability off A and B is equal to the probability of a once supplied by a probability off. Be right. So probability off read. First on red Second is equal to probability. Off read multiplied by. Probably see off Ridge. So this is equal to two of us six multiplied by one of our five, which is equal to 1/15 now probably are red first on blue. Second is it called the probability Off Red. What's better probability of blue giving. Read on this because to two of US six multiplied by 3/5 on this is 1/5. That probability of red fist on yellow Second is it cost probability off bridge multiplied by the probability off yellow giving red. And this is equal to two of us six multiplied by one of our five. This recalls 1/15. So moving on the problems off blue first multiple animal and read second is because the probability off blue one supplied by probability off red given blue and this is three of us six months by by two or five on this is one of our five non probability off blue first on blow Second is equal. Superb asi or blue multiplied by privacy off blue giving blue. Now this is equal to 3/6 multiplied by two or five which is one of our sites, not probability of blue. First on why second, is it cause the probability of blue multiplied by probability off? Why giving blue on this? Because to 3/6 month supply by 1/5, which is close to 1/10, not probably see off, why first on our second red in the course. Probability of yellow multiplied by the probability off our giving yellow on this. Because the one of us six months supplied by to about five on because toe 1/15. Now, the last one, the probability off yellow first on blue Second is it called a priority off yellow multiplied by the probability off. No giving yellow. Now, this is the call to one of US. Six multiplied by 3/5. And this is equal to want over 10. So this is the This is a question that completes the publicity for each outcome off the experiment.

And problem. Nine. Where you're looking at three different earns. A has to whites four reds. The probability of white there being one third B has eight whites and four reds. Probability of white there being two thirds and then, in turn, see one white and three reds. Probability of white. They're being quarter now. We want to find the probability that the first earn a produces a white, given the fact that there's exactly two whites chosen, others one ball chosen from every single one. So just one from a one from B one from See, What's the probability that A is white, given the fact that on Lee to of the entire three selections are white. So exactly two or white? Well, in order to figure this out because we're looking at a given, we want to know what is every single possibility of exactly two whites. So if we look at a BNC, we want to whites. We could have white, white, red, we could have white, red, white, or we could have red, white, white. These are all the possibilities of having exactly to whites. We could look at this in terms of probability. One third time's two thirds times 3/4 the 3/4 come from that's the probability of it not being white. Which is the other end, um, 1/4 white. So 3/4 not white. That gives us 6/36 the second situation. We have one third time's one third times 1/4 which is 1 36 and then the final situation two thirds times, two thirds times 1/4 which give us 4. 36. These air, all the possibilities and the probabilities of them occurring. If we can add those together, we get to some of 11/36 so 11/36 times we get exactly to whites. Now, since we know are given, we know the denominator. Let's figure out out of these occurrences. When is the first draw a white? That happens in the first and the second time so we can trace those over to their probability. 6/36 1/36. And we could just simply combined those two. Make that 7/36. We could divide 7/36 to that about 11/36 which, in actuality, just cancels out the denominators. So we get 7/11 7. About about 11 is 110.6364

So in this problem we have two jars. Jari contains very red balls and four white balls. Darby contains five red balls into white balls. So which one of the following ways of randomly selecting balls gives the greatest probability of drawing two red balls? So the 1st 1 is drawing to balls from Darby, which if we draw two from ready, we'll have five out of seven squared. So this is for I, which would be the same as saying 25 out of 49 for two draw one ball from each jar. So if we do that, we'll have five out of seven. So hold on a second. This isn't right because we don't have replacement. So this will be five out of seven times four out of six. So could be 23 actually end up having Linda having 10 of 21. And so in this one will have the five out seven from Darby and from the IRA will have three out of seven, so I'll be 15 0 49 So currently are most is from from I. And now put all the balls in one jar and then draw two balls so if we did that, what will end up doing is we'll have 14 in a single jar and then we will have five red balls plus three red Bulls, which will give us eight. We'll have seven out of 13 for the next one, which thank God, no. Cancel that out kind of nicely. Then we'll get eight out of 26. So quick check with the calculator. This is about 0.3077 in the 10 out of 21 is about. I should probably but wavy about 0.4762 So the 10 hour 21 will by far be our best chance. The 15 49. It's not a contender.

Hair for the solution. The total number off balls in the you are in is 12 out off 12 balls, eight balls are white and four balls are black. Let a be the event off the first and the third balls are white. Okay, let be the event at the simple drone. Contain exactly three white Paul. Now, the conditional probability is that it be off a by B call toe p off a be divided by P off B Yeah. Now a intersection be the 1st and 3rd balls are white and the sample contains exactly three white balls. The probabilities are W that is white, black, white, white and white, white, white, black W for white and before black be The sample contains exactly three white balls. The probabilities are black, white, white, white, white, black, white, white, white, white, black, black, white, white, black, white, white, white, white, black. For the case, one with replacement calculate the conditional probability that the first and the third both drone will be white. Given that simple drone contains exactly three white balls that p off a by B equal toe p off a be divided by pure B S. O help by substituting. Developing. Get and by 12 for my 12 8 by 12 ft by 12 plus eight by 12 IT by 12 8 by 12 4 by 12 Divided by four whole multiple off eight by 12 4 by 12 8 by 12 8 12 and we simplify this we get to 048 divided by 20736 plus 2048 divided by 20736 divided by four multiplied by 2048 Divided by 20736 Again by calculating this, we get 4096 divided by 207 36 divided by four multiple off 2048 divided by 20736 And from here, by solving this, we get one bite, too. Therefore, the conditional probability that the 1st and 3rd balls drone will be white, giving their the sample drone contains exactly three white balls. In the case off white rip placement is one by do Uh huh Now case toe without replacement. Calculate the conditional probability that the first on the third Balls drone will be white, given that the sample drawn contained exactly three white balls, so P off a by B equal topi off a be divided by PFP here by putting the values we get. A by 12 Bracket eight minus one divided by 12 minus one in Bracket eight minus two. Divide by 12 minus two IN Bracket four by 12 minus three plus eight by 12 Bracket four DIVIDED by 12 minus one in bracket eight minus one, divided by 12 minus two in bracket eight, minus studio about two well minus three whole, divided by four multiple off eight by 12 Sundby 11 6 by 10 and four by nine AGAIN By simplifying this we get eight by 12 7 by 12 6 by 10 4 by nine plus eight by 12 4 by 11 7 by 10 6 by nine HOLE DIVIDED by four THE multiple off whole eight by 12 7 by 11 6 by 10 and four by nine By simplifying this, we get 11344 DIVIDED by 1180 plus 134 41 by 1188 zero WHOLE, divided by four multiple off 134 41 by 11 80 by simplifying this we get to multiple off 1344 divided by 1180 Divided by four multiple off 1344 divided by 11800 And by simplifying this, we get two by four and then one by two. Therefore, the conditional probability that the first and the third balls drone will be white. Giving dead The simple tone contains exactly three white balls in the case. Off without replacement is one by two. So this is a complete explanation of the solution. Please go through this step by step in detail.


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