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Solve the problem: 1) Suppose there are 6 roads connecting town A to town B and 4 roads connecting town B to town C In how many ways can a person travel from A to ...

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Solve the problem: 1) Suppose there are 6 roads connecting town A to town B and 4 roads connecting town B to town C In how many ways can a person travel from A to € via B? A) 16 ways B) 10 ways C) 36 ways D) 24 waysFind the indicated probability_ Round your answer t0 6 decimal places when necessary _ 2) A bag contains 6 red marbles, 3 blue marbles, and 5 green marbles If a marble is randomly selected from the bag what is the probability that itis blue? A) 3 B) 5 D)3) Two 6-sided dice are rolle

Solve the problem: 1) Suppose there are 6 roads connecting town A to town B and 4 roads connecting town B to town C In how many ways can a person travel from A to € via B? A) 16 ways B) 10 ways C) 36 ways D) 24 ways Find the indicated probability_ Round your answer t0 6 decimal places when necessary _ 2) A bag contains 6 red marbles, 3 blue marbles, and 5 green marbles If a marble is randomly selected from the bag what is the probability that itis blue? A) 3 B) 5 D) 3) Two 6-sided dice are rolled. What is the probability that the sum of the two numbers on the dice will be 52 A) 9 8 5 D) 4 Solve the problem. 4) In a certain town, 5% of people commute to work by bicycle. If a person is selected randomly from the town, what are the odds against selecting someone who commutes by bicycle? A) 19 : 1 B) 1 : 20 19 : 20 D) 1 :19 Determine whether the events A and B are independent: 5) 12 jurors are selected from pool , of 20 Event A: The first person selected is woman Event B: The second person selected is a woman A) Yes B) No Find the indicated probability. Round your answer to 6 decimal places when necessary- 6) An IRS auditor randomly selects 3 tax returns from 49 returns of which contain errors. What is the probability that she selects none of those containing errors? A) 0.6758 B) 0.0011 0.0018 D) 0.6698 7) A family has five children: The probability of having girl is 0.5_ What is the probability of having 3 girls followed by 2 boys? A) 16 32 D) 720 Evaluate the factorial expression: 10! 8! 21 A) 1 B) 10 D) 45



Answers

Determine whether or not the random variable $X$ is a binomial random variable. If so, give the values of $n$ and $p$. If not, explain why not. a. $X$ is the number of black marbles in a sample of 5 marbles drawn randomly and without replacement from a box that contains 25 white marbles and 15 black marbles. b. $X$ is the number of black marbles in a sample of 5 marbles drawn randomly and with replacement from a box that contains 25 white marbles and 15 black marbles. c. $X$ is the number of voters in favor of proposed law in a sample 1,200 randomly selected voters drawn from the entire electorate of a country in which $35 \%$ of the voters favor the law. d. $X$ is the number of fish of a particular species, among the next ten landed by a commercial fishing boat, that are more than 13 inches in length, when $17 \%$ of all such fish exceed 13 inches in length. e. $X$ is the number of coins that match at least one other coin when four coins are tossed at once.

Question number 16. It's from a given in equally then the financial binomial probability Be off X equal key equal in CK the people key the one minus b power in minus K equal factorial in over factorial key Multiply boy factorial in minus key The people k dot one minus p or in minus key. One off, one out off. Five answers is correct. The probability the number off favorable outcomes divided by the number off possible outcomes be equal 1/5 equally point to below it at K equal thin. So probability off X equal 10 equal Factorial 10 over. Factorial 10 Multiply Boyton minus 10 story In that 0.2. Parton the one minus point to Burton minus 10 equal. Almost zero question number we given an equal. Then we would use the above formula. So 11 out off five answers is correct. The probability is a number off favorable outcome divided by the number off possible outcomes The equal 1/5 equal point to value it AT T equals zero probability Off X equals zero equal factorial 10 over. Factorial zero boy didn't minus zero factorial that 0.2 point zero that one minus point to for them minus zero almost equal 0.17 four equal 10 point 10.74% Question Number. C Result. Port P Prosperity of X equal zero equal 00.1074 equals 10.4 at a 10.74 person A p equal 4.2 addition rule for multiple exclusive events. The probability off E or me equal probability off e plus probability Off be did remind the probability using this rule and table toe in the appendix. So probability off is bigger than or equal one equal probability of X plus X equals one plus the probability of X equal to and so on toe Probability of X equal 10 equals coin 268 plus 2680.302 plus 0.20 On plus 0.8 eight plus 80.26 plus 0.6 plus 0.1 plus zero plus zero plus zero equal 0.892 Equal 89.2 percent Compliment rule Peak off. Not a equal one minus B off A. Did you mind the probability you think the compliment rule? So probability Off X is bigger than or equal one equal one minus probability off X equal zero equals one minus 10.1 074 equal 0.89 to 6 equal 89.26%. Thes probability should be equal. They aren't equal, but the difference is most likely due to rounding errors in the use deep.

So we know that he has 10 Multiple choice questions and five alternatives. So the probability that he gets one right by guessing is 12 out of one out of five or point to. And we know I'm going to let our stand for the number that he gets correct As my random variable. So on part a we want to know what's the probability that he gets all 10 correct? And we know that in the binomial setting that I'm gonna have 10 shoes 10 and I'm going to have point to to the 10th and I'm going to have .8 to the zero. And when we do that, we end up getting 1.24 times 10 to the negative seventh power. So it's extremely unlikely for that to happen now. What's the chance that he doesn't get any of them right? So he gets all of them wrong? And likewise, we're going to have 10-0 and we're going to have point to to the zero. It and we're going to have .8 to the 10th power And that .8 Yeah. Let me just change my question here when we do this. We he has a About a 10.7% chance of that happening now. We want to find the probability that he gets at least one right and we know that's going to be one minus the probability that he gets them all wrong. So one minus that answer is going to end us. Because this is that answer is going to give us .89-6. And then in part D we want to find the probability that he gets at least half of them right? And we can find 01234 and take one minus that. Or find 56789 10. I am going to use my Binomial CDF button And I have 10 and I have point to as my success and The events that this does not include. R zero up through four. So I want to find this. And so when I take one minus Binomial Cd app With 10 trials point to is the probability of success and four. That gives me A zero, So he doesn't have a good chance of guessing and getting half of them right.

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.

A job condense mind mambo Somebody one told nine, two models are members and looked into one time about what is the probability that ah, why me? Submitted fist in Mayan subjective embassy. So they're number one to my so is want to bang but are simple space my marbles So to mumbles under my selected without this man So what is the probability that while is selected fist and mine is selected So from ability Vault one and my it's an extinct consecutively it is probability of getting on There is only 11 So it's one of a man Ah, trying So providing that, um, one has been chosen. It means while it's no longer there, what do we have now? Eight balls and we have one off, huh? One or more? A. So the probability of that happening off the one off us of it, too. Uh, then it's some off the numbers selected is for So the sum is for and the combination that gives the sum of four will be one and please Oh, two and two. And these are the only possibilities that we have. So what we want do is it's aimed that we pick one and two. Oh, we do twin one. All we do. Ah. Okay, then, true into just the same person. So about oneness where we pick three in one pleases so probability or some fool is equal to I would pick one and And what is the probability of that? I pity it's 1/72. Oh means plus to end too again. It's all right. If you'll pick two fists, then that is no longer days on this one. Both This doesn't wake because can only big to once. And we're not suppressing it if it's my place or they only two cases, which is this and that. So the other one, Nobu on over 72 again because it's the same person is just a test. So this one will be to off us event, too. We choose one or clown 36. Ah, then the last one is the some off the numbers. Invective is life. So get five. You need to have one and four or you can therefore and one Oh, it could be three and two Oh ju And so therefore well, such pieces. So probability off some five is about to 1/72. Oh! Oh, Swan. That's for almost 70 King, Um, and four blossom into trees, one over for you to save me. It's long and then scooting main. Go three. Fall into 32. It's a judge.


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