I have no idea on this one

Suppose you know that the probability she has to stop on exactly one of the days Tuesday and Wednesday of next week is 0.33.

What is the probability of her having to stop on any single day? You'll find that there are two correct answers here, i.e. there are two different possibilities for what the probability might be. Give both answers, with the smaller one first.

smaller solution = ? larger solution = ?

11/16/2006 8:26:20 PM

All I did for this one was use an equation.

Like it says, you kind of do it using a similar method as part (c) but opposite.

I used my TI-89 and solved the equation below.

2*(x)*(1-x) = 0.33

You will get two answers for the x value in that equation. Hope this helps.

11/16/2006 11:18:46 PM

Has anyone finished 5.7 yet?

Thanks.

11/20/2006 8:21:43 AM

yeah anyone?

[Edited on November 21, 2006 at 7:07 PM. Reason : d]

11/21/2006 7:06:51 PM

Roulette is one of the most common games played in gambling casinos in Las Vegas and elsewhere.

A roulette wheel has slots marked with the numbers from 1 to 36 as well as 0 and 00 (the latter is called "double zero"). With each spin of the wheel, the ball lands in one of these 38 slots.

Half of the slots marked 1 to 36 are colored red and the other half are black. So if Mike bets on "red", he wins if the outcome is one of the 18 red outcomes, and he loses if the outcome is one of the 18 black outcomes or is 0 or 00. So there are 18 outcomes in which Mike wins and 20 outcomes in which he loses. Therefore Mike's probability of winning is 18/38 or 9/19 if he bets on red. If he wins, then the payoff is equal to the amount of money he bets. If he loses, then he loses the money he bets.

Mike goes to the casino with $300 and decides to bet $50 on red. What is the expected value for the amount of money he will have after he has made this bet (and either won or lost)? (Give your answer in $ without including the "$" symbol. Your answer should be accurate to the nearest cent.)

Expected Value = $________

11/21/2006 9:20:55 PM

Do you still need to know how to do this problem?

11/22/2006 7:04:18 AM

can anyone tell me how they got the answer to the second part of 5.7, #10? I ended up getting the right answer, but it was by trial and error and I'm not fully understanding how to do it. I used the quadratic and all that, but my AX# came out to be the answer to the question, however I don't know if that's how it was supposed to come out.
Thanks so much!

11/23/2006 1:41:49 PM

I'm still wondering about this ^^^ if anyone has any ideas, I would greatly appreciate it!
Thanks.

11/25/2006 11:11:47 PM

For the roulette problem, take the amount of money that Mike bets on red ($50) and multiply it by the probability of winning if he bets on red (18/38). That's the expected value if he wins. But, the question asks for the expected value if he wins OR loses. As a result, you have to subtract the expected value for winning (50 times 18/38) by the expected value for losing. That would be the amount of money bet ($50) times the probability of losing if Mike bets on red (20/38).

So, (50)*(18/38) - (50)*(20/38) equals the expected value if he bets $50 on red. But again, the question adds that he enters the casino with $300, so you've got to take that into account in your expected value. You do this by adding or subtracting $300 by the expected value for winning or losing if he bets $50 on red, and that should be your answer.

Is #10 the free throw problem?

[Edited on November 26, 2006 at 7:56 PM. Reason : -]

11/26/2006 7:50:47 PM

Hey, thank you so much! I didn't mean to point to the roulette problem, I accidentally hit the arrow key two times too many! But, yes, number 10 is the free throw problem. Here it is:

A one-and-one free throw situation in basketball occurs sometimes when a player is fouled. The player gets to take a shot from the free-throw line, and if he hits it then he gets another shot. So if he misses the first shot, that's the end and no points are scored. If he hits the first attempt, he scores a point and gets another shot. So when it's all over, he will have scored 0, 1 or 2 points.
Suppose Sally is a 88% free throw shooter. And let's suppose her free throw attemps are independent trials, i.e. every time she shoots there will always be a 0.88 probability that she hits the shot.

What is the expected value for the number of points Sally will score if she is fouled in a one-and-one situation?

Expected value for number of points scored = 1.63680[1.6544]

--------------------------------------------------------------------------------
Now suppose that Charlie is a basketball player who averages scoring 1.25 points whenever he has a one-and-one free throw situation. Let's use that information to calculate what his free-throw shooting percentage is. Let's use the symbols p and E to represent what we're talking about:
p = success probability whenever Charlie shoots a free throw
E = expected value for number of points scored in a one-and-one situation
In Sally's case, we knew that p = 0.88, and we used that information to solve for E. In Charlie's case we know that E = 1.25, and we want to solve for p. To do this, use exactly the same procedure that you did with Sally, except instead of working with the number 0.88, simply use the symbol p. This should lead you to an equation involving p and E. Since you know the value of E in Charlie's case, plug in that value and solve for p. (It's a quadratic equation, so you'll need to use the quadratic formula.)

11/26/2006 8:13:57 PM

When you did the first part of question ten, you found the expected value by adding up the independent trials of hitting zero, one, or two free throws. The formula, (0)*(.12) + (1)*(.1056) + (2)*(.7744), can be simplified (with the success probability being p) as the formula (0)*(1-p) + (1)*([1-p]*p) + (2)*(p^2), which further simplified is p + p^2. This is set equal to E, the expected value.

You use this same formula in the second part. Set the formula equal to zero by subtracting E over to the left side, and you should have p^2 + p - E = 0. Since they gave you E, plug it in, and you've got p^2 + p - 1.25 = 0.

After this, you can graph it to find the roots or use the quadratic formula. Either works just as well, although graphing it on a graphing calculator is normally much easier and less time-consuming than working it out through the quadratic formula.

But, you asked how to do it using the quadratic formula, so in doing that, remember that the equation you set equal to zero is basically (1)*(p^2) + (1)*(p) - 1.25. In working with the formula, you're going to have a numerator of -1 plus or minus the square root of (1^2 - [4*1*-1.25]). Simplified, this should be -1 +/- sqrt(6). The denominator will be 2 times 1, which is 2. So, your formula will be -1 +/- sqrt(6) divided by 2. When you swap between addition and subtraction, you should get .72474 (adding) and -1.72474 (subtracting). Only one of these two answers makes sense in the context of the question, and since you can't have a negative probability of something happening, the only possible answer is .72474.

The question's basically busy work, because I don't remember a question like this being on our exam last semester (they might've changed that, but I doubt it). Still, if you don't want to use the quadratic formula, a graphing calculator makes this one much easier.

[Edited on November 26, 2006 at 10:06 PM. Reason : -]

11/26/2006 9:55:13 PM

Three FM radio stations (A, B, and C) are competing for customers in the same market area. Through an aggressive advertising campaign, station A is capturing 12% of station B's customers and 5% of station C's customers each month, while losing only 2% of its customers to B and 2% to C each month. And 1% of B's customers switch to C and 1% of C's customers switch to B each month. This gives you all the information you need to set up the Markov chain for this situation.
Initially A and B each have 30% of the listeners and C has 40%.


What fraction of the listeners will A have after 3 months? (Enter as fraction or decimal number.)
What fraction of the listeners will B have after 3 months? (Enter as fraction or decimal number.)

What fraction of the listeners will C have after 3 months? (Enter as fraction or decimal number.)


how do i go about doing this? i just need my intial matrices setup. i tried setting them up but it doesnt work. this is markov chains btw

12/1/2006 11:13:04 AM

Do you still need to know how to do this problem above?

If you do let me know and I can help ya out. Have you gotten any of the other problems for this hw assignment done. I forgot all about this assignment since I have been working on my senior design project so much.

Thanks.

12/1/2006 5:43:22 PM

I NEED HELP BIG TIME ON THIS ONE


Pete will be working for a company in the Midwest. His boss will send him each week to either Chicago, Detroit, or Milwaukee.

If he is sent to Chicago one week, he will always be sent to Detroit the next week. When he is in Detroit, the probability that he will be sent to Milwaukee the next week is 0.78 and otherwise he will remain in Detroit the following week. When he is working in Milwaukee, he knows that he will be sent to Chicago or Detroit the following week, but he doesn't know what the probabilities will be. So let's use "p" to be the probability he gets sent to Chicago the following week and "1-p" will then be the probability he gets sent to Detroit. (Since we don't know the value of p, you can just carry along p as a symbol in all your calculations. So all your answers will be expressions that involve p. Enter your answers in standard calculator notation using the four arithmetic operations +,-,*,/ and parentheses as needed. So an answer might look something like this: 2*p/(3+5*p).

Find the fraction of the time that Pete will spend in each of the three cities in the long run.

Chicago--
Detroit--
Milwaukee--

12/5/2006 4:28:42 PM

did you use her help page? There is another link on her help page that takes you to a more in-depth explanation of how to do this one. I'm still working on it or else I would help more.
Sorry!

12/5/2006 10:45:55 PM

Ya i found the help page, thanks

12/7/2006 3:21:56 PM

who's done the last one due?

12/8/2006 4:46:31 PM

Who did all of this assignment? I have been sick so I didn't get to do it. If someone can help me out please let me know.

Thanks.

12/9/2006 4:30:55 PM

I can help you with all but the roulette problem (I'm still working on that one), so let me know what you need help with.

12/9/2006 7:39:45 PM

Can you please explain any of these problems to me. I don't understand this section at all.

Thanks.

12/10/2006 1:04:18 PM

Post some of the questions.

12/10/2006 4:05:51 PM

The roulette problem.

12/10/2006 4:09:21 PM

Alright... For that one, you've got to set up an absorbing Markov Chain with six states. The six states are the amounts of money that Mike has, from $0 to $500. This is because Mike will stop playing after he's reached either of these levels of money. You'll also need to use the Matrix Algebra tool for it, unless you want to be doing some heavy matrix multiplication.

The problem says that Mike's probability of winning is 9/19 if he bets on red, and that Mike is going to the casino with $400. His plan is to bet $100 on red nine consecutive times or until he either has increased his total to $500 or lost all his money.

Your absorbing Markov Chain will the have rows and columns in this order, where the 0 and 500 are squared off and are the 2x2 identity matrix: $0 $500 $100 $200 $300 $400

The rest of your matrix will have the probabilities 10/19 and 9/19 in them. So, think each bet through for the respective dollar amounts. If Mike has $400, then he's got a 10/19 chance of losing and going back to $300, and a 9/19 chance of winning and going to $500. So, put 10/19 in the $300 column and 9/19 in the $500 column, both on the $400 row. If he's got $300, then he's got a 10/19 chance of losing and going back to $200, and a 9/19 chance of winning and going to $400. Same thing, here. 10/19 goes in the $200 column and 9/19 in the $400 column of the $300 row. Continue doing this until you've labeled all of the parts of the matrix (filling in zeros for the rest).

Put this matrix into the Matrix Algebra Tool. Since he's betting nine times, raise the matrix to the ninth power. When you see the resulting matrix, look at the number in the $0 column and the $400 row (it'll be in the bottom left corner). After nine bets, this is the probability that Mike will go bankrupt, given that he starts out with $400.

To find his long-term probability of going bankrupt, raise the matrix to some high number (like 5000). Once again, the result you're looking for will be in the $0 column and $400 row, in the lower left corner. This is the probability of him going bankrupt in the long-term, given that he starts out with $400.

[Edited on December 10, 2006 at 4:45 PM. Reason : -]

12/10/2006 4:36:24 PM

Ok got it.

Thanks.

12/10/2006 4:50:04 PM

anybody taken the exam previous semesters and remember it?

12/14/2006 12:45:04 PM

Anyone taking this right now? i am having trouble with 1.5 #2

1/26/2007 4:39:11 PM

5.4 help?

4/9/2007 1:45:46 PM

What part of 5.4 do you need help with? Post the question(s).

[Edited on April 9, 2007 at 5:24 PM. Reason : -]

4/9/2007 5:23:59 PM

Martha and Antoine are part of a group of 12 women and 6 men who will participate in a chess tournament. They will be paired up randomly into teams of two. (Martha is a female and Antoine is a male.)
What is the probability that Martha and Antoine will be paired together as partners?

What is the probability that Martha will be paired with a woman?

What is the probability that each of them will be paired with a person of the same sex, i.e. Martha with a woman and Antoine with a man?

What is the probability that each of them will be paired with a woman?

4/9/2007 7:56:05 PM

Never seen that problem before, so none of this might work at all...

Alright, well I think the first part is done with combinations and logical reasoning. You're going to take the combination that'll describe either Martha or Antoine being picked and then someone else being picked. So, C(17,1) = 17 different ways to pick someone after we assume Martha or Antoine are picked. Out of those 17 outcomes, only one of them picks the other person (again, Martha or Antoine), so theoretically (hopefully), the probability is 1/17 that they'll be paired together. It's either that or 1/153, which is Martha and Antoine being paired together over the total number of outcomes, C(18,2). I'd be more inclined to go with the second one, since the first looks a lot like a conditional probability.

For the second one, use combinations to find the number of ways that Martha can be picked and her partner be a woman (just change C(17,1) in the first part to reflect you picking out of the group of women, only) and then place that over the total number of outcomes, C(18,2).

For the third, use the same process as the second part, except multiply the combinations of Martha and a female parter with Antoine and a male partner. Put that product over the total number of outcomes, C(18,2).

Do the same thing for the fourth, except do it with Antoine and a female partner this time. Make sure to keep track of how many people you're counting in the first part of the combination, and taking care to get rid of Martha and her female partner and Antoine in the combination of him and his female partner.

Still, I'm flying a little blind on that problem. Hopefully, I made some sense and that problem will work out alright. Good luck, man.

4/9/2007 8:56:54 PM