Question 14 ptsAbox contains black; 3 red and 6 green marbles. Three marbles are drawn from the box at random: What is the probability (rounded to 4 decimal places)...

The College Board reports that 2$\%$ of the two million high school students who take the SAT each year receive special accommodations because of documented disabilities Angeles . Times, July $16,2002 ) .$ Consider a random sample of 25 students who have recently taken the test.
(a) What is the probability that exactly 1 received a special accommodation?
(b) What is the probability that at least 1 received a special accommodation?
(c) What is the probability that at least 2 received a special accommodation?
(d) What is the probability that the number among the 25 who received a special accommodation is within 2 standard deviations of the number you would expect to be accommodated?
(e) Suppose that a student who does not receive a special accommodation is allowed 3 hours for the exam, whereas an accommodated student is allowed 4.5 hours. What would you expect the average time allowed the 25 selected students to be?

Get in this case was simply need to find the probability. Were marbles are off. Different collect. Okay, so for that case, we need to understand this question carefully. There are a total of one green, two yellow and three red marbles. Right. So the cases which will be formed here, is the marbles of different Cal average mints. You have one green, then you have. When you look, you're hungry. Then you have 100. Okay, then you have one yellow. Then you have one grade. All right? Oh, are Yeah. Only this cases will arise, which is mathematically treason. Treason, Those three. So one of those cases, one green and one yellow, one green and one yellow means that is simply one C one names off, Bruce. Even one green once even. Okay, on one rate, that is three C one yellow. It is 27 on one red, which is three C two. So we just need to count this 17 is one to see to you see to this two factor Really worried weapon factorial times One factory, which is to itself again. This is three itself. Does even is two and three, sir, to history. So two times 36 of their total of 236 Right. If I just add the simply consult a bit. 11 pieces. Right? So there are a total of 11 cases here. Hence the required provided it is equals to 11 over 62. Because you are drink two marbles, That is It was toe 11 of us six times, five on, then in tow. Okay, so 11 times two is 22. When did you? Ok, this cancels right on three times to the 11 by 15. The final answer is 11 Bip of being all right.

So in this question we are basically given the sample space of rolling a two sided um to dies basically. And for us to first identify certain events. So the event end is basically the event that the sum is at least nine. So we're going to look At all of the events in the entire sample space of Rolling two Dice. So for the sum to be at least nine, nothing in our first role, in our second role nothing as well. Third role, We have one event 3, six, 4th row, we have 4,5, 5th rule, we have 54 and six through. We have 63 So that constitutes all of the events where the sum is at least nine. Now T at least one of the dies is a truth at least one is it too? So we have 1 to at least one of the devices are too and then we have 21 two, two or three 4 to 5 26 We have 32 again and then we have 42 52 Nothing else over there. And then six two. And then we have the third event at For at least one of the dies is up five at least one is five. So we have 15 to fly 35 four or 5. And then we have my warning 525354. Fine fine. 56 and six. Bye. So we have our three events, N. T. And F. Now first we are asked to find the probability of and so the probability of end is basically 1234 over. There's 36 events Which is 1/9 and that's equal to point 11 The probability that the sum is at least part B. The probability of N. Even X. Is basically the probability according to conditional events. The quality of the intersect with death divided by the probability of. So we are given the event F. If it's basically this event, and if we have 123456789 10, 11. So we're just going to Use events. So we're going to take our denominators 11. And the probability of any given f. So we're going to see where the intersect. So the intersect at 45 and five, so that's true out of 11 and that's equal to point one heat. Now, the third we asked the probability of and given T. So we're given the event T. And we have 1234567 8, 9, 10, 11 Events in T. And the probability of any given T. So that let's see how many intersect here. So we have Let's see if 36 intersex. So it doesn't for 55 for 60, none of them, because at least one of the ties has to be true and that's never going to happen with this event. So We basically have zero. Our third Cancer here is zero. And were asked to determine from the previous answers whether or not the events N and F are independent and whether or not, and and t are independent. So we're going to compute here. So to see an N. F are independent. We're basically going to take the probability of 10 times the probability of f. So a probability of N. We have 1/9 probability of event F. So we have 123456789 10 11 11/36. So Times 11/36. So 1/9, times 11/36 is approximately point for three or now the probability of and intersect with that and intersect with F. We basically had two events out of 36. So that's two out of 36 Which is find 0.56. And so we say that they are not independent because the probabilities are not this is not equal to this. So then an F. Or not independent. Now R. N. And T. Independent. Let's take a look. So we have event. And here we have event. See here now the probability of event and intersecting with event T. was basically zero. So the probability of an intersect with teen was zero. Now the probability of end we know is 1/9. The probability of T. We know is 234567 You know it's 11 over 36. And we know that these are not equal. So again they are not independent. So and until they are not independent but they appear because their intersection is zero, they appear to be mutually exclusive. So we have our answer to part mm B. C. Andy

Already this particular question. Two marbles are drawn without replacement on the back. Contains one green, uh, do yellow and three red marbles. Right. Then we just need to find the probability that both the marbles are red. So probability that but what marble are right. So here we will be using the concept up combination security number element of the sample spaces three plus two plus one, which is six OK, on board the marble separate. So bored the marble are rape. If I just write all those cases rate the number of cases where both the marbles. All right, which means that you are drink to mumbles out of the spirit, Mumbles. In that case, you will be getting to read mumbles. Right. So there's three c two, so therefore the required probabilities equals 23 C two divided by 60 to get three settlements three times to by two aan den. This is too here on six times five to into cancels out three cancers on two times You want to cancels out. So this values equivalent to one by five. All right,

Okay and in this case was implanted to find the probability that both marbles area boot marvel are yellow. So again, there are a total number of some elements of the sample spaces. One green marble, two yellow marble and spirit marbles with simply comes out with their acquittal of six mumbles. So the final answer, Boluda quite probability is equals to your bring to mumbles. That should be a which means that that will be from here itself. So you see, two divided by 62 using the concept off combination toe to see toe is one and 62 6 times five and a little bit too. So this is one by three times five. That is equivalent when by 15 all right.