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Iftwo persons are randomly selected without replacement; What is the probability of the first selected person is a left-handed men and the second selected person i...

Question

Iftwo persons are randomly selected without replacement; What is the probability of the first selected person is a left-handed men and the second selected person is right-handed men?What is the probability of the first selected is left-handed women and the second selected person also left-handed women?

Iftwo persons are randomly selected without replacement; What is the probability of the first selected person is a left-handed men and the second selected person is right-handed men? What is the probability of the first selected is left-handed women and the second selected person also left-handed women?



Answers

About $13 \%$ of the population is left-handed. If two people are randomly selected, what is the probability that both are left-handed? What is the probability that at least one is right-handed?

This question asked us to answer the various probabilities we know for part. A six of the 30 students are left handed and five of the remaining 29 students are left handed. Therefore, what we know is that if we multiply these fractures together, we end up with 1/29 for part B. What we know by looking at this is that 24 of the 30 students are right handed. 23 of the remaining 29 are right handed Multiplying these Together we get 92/1 45 for part C. We know that 24 of the 30 students are right handed. Six of the remaining 29 students are left tended multiplying these together we end up with 24/1 45.

In this problem, we are dealing with um data that was introduced in example 4.21 On page 1 81. Um about students from Professor Weiss is introductory statistics class. So the first thing you're going to want to do is just go refresh yourself on that example and remember what we're discussing here. Um and what we're trying to do here is calculate various probabilities using the general multiplication rule for three events. And just as a refresher, this is the general multiplication rule for three events. So the probability of A. And be and see is equal to the probability of a. Times the probability of being given A. Times the probability of C. Given A. And B. Now also just to refresh the example. So we're dealing with a sample that contains 17 males and 23 females. Yeah. And this equals 40 total. So let's see what we're being asked. So suppose now that three students are selected without replacement, What is the probability that the first two students chosen? Or female? And the third is male? Oh my God, okay, so um we want to find two female. Sorry, let me to female. And the third is male. Two female. One male. And so let's just look at this equation. Um so we're going to find the probability of female for the first one and female for the second one and mail for the third one. And so this is equal to the probability of choosing the first student being female, which is simply the number of females over the total number of students multiplied by the chance of choosing another female, given that the first one was female. So I want to think we're selecting without replacement. So what happened between the first and the second selection? So the answer there would be that we now have one less student and we have one less female. So it will be 22/39. And then we're going to multiply by the probability that the third is male, given that the first two were female. So what has changed? Well, once again, um there is one less student. So now there are 38 students, but there are still 17 males because none of them have been selected. So we'll multiply that by 17/38 and this gets you 8602 over um 64,280 which is approximately 0.145 point or 14.51%.

So obviously our class data will be different, but in this class there are six females and four males left. Two females are left handed. For right, no males are left handed, and for our right, so the probability of a female Would be six out of the total of 10. Left handed in general will be two out of 10. A female and left handed would be Two out of 10 and neither feminine or female nor left is four out of 10.

So there are 15% in a particular room in that we are given. Our six are left handers, and obviously nine will be right handers. How many ways we can choose seven people from 15 people? It's a number of ways of choosing seven from 15. Cable is but, uh, formula of NZ R. It is 15 c seven, 15 seventies 15 factorial by eight factorial into seven factors. So this is the value of 15 c seven. No, let's find the probability of individual questions in these seven people. Three should be left handers and four should be right handed. So how many were saying it? I can select three left handers from six left handers, 63 us and how many years I get select four right handers from nine right handers 94 we need to multiply those because selecting left handers and selecting right handers are independent. So the probability that there are three left handers and four right handers will be 63 into 94 divided by 15 c seven. So this you can calculate 63 is 2094 so 94 will be nine factorial by five factorial into four Factorial divided by 15 factorial by eight factorial into seven Factory So this gets simplified 20 into nine factorial by five factorial into four factorial into eight. Factorial seven factorial divided by 15 Factory So you can, uh, use, uh, calculator if you want. So 28 to 9 C four divided by 15 c seven so that will be around 0.39 So this is the probability. Okay, next to part B, but be what they're asking is all should be right handers in all the seven. Whatever the seven people you select all should be right handers. So how many right handers we have? Nine right handers. So all the, uh, seven. What you select should be from the nine people. So how many ways you can select seven from 9 97 since we need probability divided by total possibilities 15 c seven. So 97 divided by 15 c 79 c seven will be 36. Do I advise 15 c seven So 15 c seven will be 6435 So the answer will be 36 file 6435 So that will be an old, uh 5.59 Sorry. 5.59 into 10. Power minus three. Okay, Next party. All left handers. Do you think this is possible? Because there are only six left handers. But you want to select seven or can all be left handers, so there should be at least one right handed. So the probability of this is zero. Because this is an impossible event. Next last one is very good question. When you select seven, they are saying that, Harry and bigger the left handers. So Harry and Peck are the left handers. They should be fixed in the seven people. That means two people have already taken, so remaining combinations should be There are four left handers remaining and nine right handers remaining. And we need to select five people. So five people. What are the combinations we have? I can select zero left handers, Fire right handers or one left hander for right handers to year, two year, three year, two years or four years. One year. Okay, so what is the possible combinations here? 40 Because any zero left hand is from four. Only not 640 into 95. Is this answer and for this one it is 41 into 94. And for this one we need two left handers, 93 in three right handers and this is four. C three into 92 this is four C four into 91. So let's calculate this. So foresees 40 into 95. So 95 will be 1 26. 40 is one, and this value will be four in 29 c four. So that will be 504 and this will be 42 16 to 93. That is 504 Jimmy 1 44 is this one, and this one will be nine. So let's add all the numbers. 1 44 plus nine plus 504 plus 5 four plus 1 26 total possibilities. +1287 So now probably will be 1287 divided by 15 c seven so divided by 15 c seven. So the answer is zero point. That's it


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