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Mathematics 6 Online
OpenStudy (anonymous):

So Malik is considering investing $900 in a certain company. Financial advisors forecast that there is a 30% chance that the stock will increase in value by 150%, and a 50% chance that he will lose his initial investment. Determine if Malik should make the investment, and find the expected value of the investment. I don't even know where to start

OpenStudy (whpalmer4):

Is that all of the information given? Those two possibilities only add up to 80%...what happens the other 20% of the time?

OpenStudy (anonymous):

Yes thats whats getting me I tried 9450(.3)+-9000(0.5) but that doesn't seem to make sense

OpenStudy (whpalmer4):

To find the expected value, you multiply each percentage by the outcome associated with that chance, then add up all the products. For example, if you play a coin flip game where you have a 25% chance of winning $100, and a 75% chance of losing $50, the expected value would be 0.25 * $100 + 0.75 * (-$50) = $25 - $37.50 = -$12.50 — clearly a losing game for you on average.

OpenStudy (whpalmer4):

I wonder if we are supposed to assume the investment stays constant the other 20% of the time?

OpenStudy (anonymous):

yes so he makes no money how does that change it?

OpenStudy (whpalmer4):

Well, each outcome is weighted by its probability, so it does matter.

OpenStudy (whpalmer4):

$900 investment, if we hit the jackpot it is worth $900*(1+1.5) = $2250. 30% chance. $900 investment, if we crap out, it is worth $0. 50% chance $900 investment, worth $900. 20% chance. $2250*0.3 + 0.5*$0 + 0.2*$900 = $855. $855 < $900, so it appears that this is not a good investment, if our assumption about what happens the other 20% of the time is correct.

OpenStudy (anonymous):

ok now that makes sense

OpenStudy (whpalmer4):

Earlier you wrote: "I tried 9450(.3)+-9000(0.5)" — how did you come up with 9450 and 9000?

OpenStudy (anonymous):

150% was 9450 -9000 was losing all his money

OpenStudy (anonymous):

Can you help me set this one up? > You are taking a ten-question multiple choice test. Each question has five choices. If you answer a question correctly, he earns one point; if he answers incorrectly, he loses half a point. You answer the first six questions correctly, then guesses on the last four. What is the expected value of your test score? Round the answer to the nearest whole percent.

OpenStudy (whpalmer4):

Except that the investment was 900, not 9000, and 150% gain on $900 is an additional $1350 (100% gain is $900, 50% gain is $450, 150% = $900 + $450 = $1350) on top of initial investment

OpenStudy (anonymous):

Thats typical me adding an extra 0 thanks

OpenStudy (whpalmer4):

Okay, let's see here...you've got the first 6 correct, so that's 6 correct answers * 1 pt/correct answer = 6 points. For the other 4, you guess, with a 1 in 5 chance of being correct. If you guess correctly, you get 1 point, and if you guess incorrectly, you lose 1/2 point. What is the respective probability of a correct guess and an incorrect guess?

OpenStudy (anonymous):

1/4 of gaining a point and 1/4 of losing 1/2 point

OpenStudy (whpalmer4):

Hmm. Want to try again? :-)

OpenStudy (whpalmer4):

The probabilities have to add to 100% (1)...

OpenStudy (anonymous):

yes am I multiplying 4 by .25?

OpenStudy (whpalmer4):

Remember that each question has 5 possible answers.

OpenStudy (anonymous):

so 4x.20 then by the 1 point added to the 4x.20 times -0.5

OpenStudy (whpalmer4):

Look at just one question.

OpenStudy (whpalmer4):

4 out of 5 choices are incorrect, and 1 out of 5 is correct.

OpenStudy (anonymous):

this is of the 4 he guessed on right?

OpenStudy (whpalmer4):

Yes, but each question is a separate event. What you do on one question has no impact on the others.

OpenStudy (anonymous):

So first Id like to find the expected probability of a correct guess

OpenStudy (whpalmer4):

Sure.

OpenStudy (anonymous):

4 questions 5 possible answers each question worth 1 point

OpenStudy (whpalmer4):

No, only one question at a time.

OpenStudy (whpalmer4):

You are a student in my class. Tests in my class have just 1 question, with 5 different choices. You don't study, so you're forced to guess. What is the probability you will get a correct answer on the one question?

OpenStudy (anonymous):

So you said they have to add up to 100%

OpenStudy (whpalmer4):

Yes, we'll get to that, I promise it will make sense :-)

OpenStudy (whpalmer4):

Or a full refund of everything you've paid me :-)

OpenStudy (anonymous):

So it is the 5-1/5 of negatives =4/5 or 80%

OpenStudy (anonymous):

haha you could work with Billy Mayes

OpenStudy (whpalmer4):

You have 1 chance in 5 of getting a correct answer. Right? That is 1/5 = 0.2 (or 20%). What is the chance of getting an incorrect answer?

OpenStudy (anonymous):

4/5= .8+.2=1

OpenStudy (whpalmer4):

Tough to work with him now, he's been dead for a few years :-)

OpenStudy (whpalmer4):

Right, so the two possible outcomes are correct answer (prob 0.2) and incorrect answer (prob 0.8) and the probabilities add up to 1. We've covered all the possible cases for guessing on 1 5-choice question. Now, if you get 1 point for a correct answer, and lose 1/2 point for an incorrect answer, what is your expected value for that guess?

OpenStudy (anonymous):

2.5? 5-2.5

OpenStudy (whpalmer4):

No, multiply each probability by the outcome for that probability, and add them up.

OpenStudy (anonymous):

.2(5)=1+

OpenStudy (whpalmer4):

What we're trying to do is get a weighted average. 20% of the time we'll get 1 point. 80% of the time we'll lose 1/2 a point. In no case are we going to get more than 1 point or lose more than 1/2 a point, right?

OpenStudy (whpalmer4):

0.2*(+1) + 0.8(-0.5) = ?

OpenStudy (anonymous):

60%

OpenStudy (anonymous):

BINGO

OpenStudy (whpalmer4):

0.2*1 = 0.2 0.8*0.5 = ?

OpenStudy (anonymous):

I think im getting it. Thank you

OpenStudy (whpalmer4):

No, not so fast, I'll tell you when you're getting it :-)

OpenStudy (whpalmer4):

0.8*0.5 = ?

OpenStudy (anonymous):

0.4 +.2

OpenStudy (whpalmer4):

Okay, but that's actually 0.8*(-0.5) = -0.4 (I dropped the - sign in case it was confusing you). 0.2 + (-0.4) = ?

OpenStudy (anonymous):

-0.008

OpenStudy (whpalmer4):

0.2 + (-0.4) = -0.2, meaning that we actually lose points on average by guessing! Every 5 problems we guess on, our score will go down by 1 point on average (5 * (-0.2) = -1).

OpenStudy (whpalmer4):

Now, we had 6 points due to our 6 correct answers on the first 6 problems. If we guess on the remaining 4 problems, with an expected value of -0.2 points per problem guessed, what is the expected value for the entire test?

OpenStudy (whpalmer4):

This is an example of why it is good to know if you are penalized for wrong answers, or merely don't get credit! In this case, the student would be better off leaving the last 4 blank...

OpenStudy (anonymous):

so It makes sense to me like this 6 +-0.2(-2)=1

OpenStudy (whpalmer4):

Hmm, that number sentence makes no sense to me :-) 6 + (-0.2)(-2) = 6 + .4 = 6.4, not 1.

OpenStudy (whpalmer4):

Where is that -2 coming from?

OpenStudy (anonymous):

theres an absolute value of 0.4 to make 100% of possible points

OpenStudy (anonymous):

added to the 6 right you can either gain a point or lose -2

OpenStudy (anonymous):

*gain 4 points

OpenStudy (whpalmer4):

For each question on which you guess, 80% of the time you guess wrong, 20% of the time you guess right. That's where the 100% comes into play. The value of the outcome doesn't have anything to do with 100%. Look at it this way. If I say that you have a 1 in 100 chance of buying a lottery ticket from me, 1 time out of 100 (on average) you'll buy a winning ticket, and 99 times out of 100 it will be a losing ticket. It doesn't matter whether a winning ticket pays $1 or $1,000,000, the odds are still the same. The expected value is what changes, but there's nothing that says the expected value has to add up to any particular value...

OpenStudy (whpalmer4):

Let's play a little game. You've already got 6 correct answers. I will choose 4 letters in the set A,B,C,D,E to represent the correct answers for the 4 remaining questions. You send me 4 letters as your guesses, and I'll tell you how many you got right, and we'll add up the points. We'll do this a handful of times to get an average. Okay?

OpenStudy (whpalmer4):

What is your first set of 4 guesses?

OpenStudy (anonymous):

abcd

OpenStudy (whpalmer4):

okay, correct answers were daaa so you gained no points and lost 4*1/2 = 2 points, giving you a total score of 4. Let's try again.

OpenStudy (anonymous):

so daaa would give a perfect score

OpenStudy (whpalmer4):

Yes, they would have, if you had only guessed daaa :-) I've got a new set of answers, try again.

OpenStudy (anonymous):

dbca

OpenStudy (whpalmer4):

okay, this time correct answers were ebbc, so you got 3 wrong and 1 right, for 3*(-1/2) + 1(1) = -1/2 point, giving you 5 1/2 points total. Let's do it again!

OpenStudy (anonymous):

ok aaaa

OpenStudy (whpalmer4):

This time my computer picked cede as the correct answers. 4 wrong, lose 2 points, total is 6-2 = 4 points. As you can see, if you are penalized enough for wrong answers, guessing becomes a losing game!

OpenStudy (whpalmer4):

Here's a variation: if we get 1 point for a correct answer, have 1/5 chance of getting a correct answer, what is the most we can be penalized for a wrong answer and still have it be worth guessing?

OpenStudy (anonymous):

yes I know So we've always got 6 points to start

OpenStudy (anonymous):

never

OpenStudy (whpalmer4):

Right, but actually that's irrelevant to the question of whether or not we should guess. We could have 6 points or 6 million points — if we lose 1/2 for a 4/5 probability and gain 1 for a 1/5 probability, it isn't worth guessing, because on average we'll lose 1/5 of a point per guess.

OpenStudy (whpalmer4):

Well, if we lose 0 points for an incorrect answer, it's clearly worth guessing, right? 1/5 * 1 + 4/5 * 0 = 1/5. If we set the probability of a correct answer * points for a correct answer equal to probability of a wrong answer * points lost for a wrong answer, we can find out the penalty value that makes it break even (no penalty or bonus for guessing). \[\frac{1}{5}*1 = \frac{4}{5}*x\] where \(x\) is the points for a wrong answer \[\frac{1}{5} = \frac{4x}{5}\]multiply both sides by 5 and we get \[1=4x\]or\[x=0.25\]which means if we lose 1/4 point for a wrong answer and gain 1 point for a right answer, it is neutral on average — we won't gain or lose points for guessing if we have a lot of questions.

OpenStudy (anonymous):

so the more trials the closer it is to the expected value?

OpenStudy (whpalmer4):

Yes, it's like flipping a coin and expecting it to come up heads. With only one or two tries, you might get them all right, or all wrong, but if you do it 1000 times, you'll get about 50% of them right. There's a different angle to all of this: we've figured out what we can expect on average. What are the extremes that we might encounter? Well, we could guess correctly on all 4 questions, which would give us 4 extra points, for a total of 10. That's the best possible outcome. We could also get all 4 wrong, and that would lose us 2 points, taking our score down to 6-2 = 4. With that scoring, if it was more important that we get at least 6 points (say that was the cutoff for a passing grade), and we somehow knew that we had the first 6 correct, we would want to not guess, because if we were a bit unlucky, we might drop below the magic 6 point threshold.

OpenStudy (whpalmer4):

On the other hand, if we knew our overall grade was safe, and we just wanted to get the highest possible score, to impress our friends (if we guessed right), then it might be worth guessing even with the average 0.2 points lost per guess, because we might get lucky and end up getting more points than we "deserve", just like we might flip the coin heads 4 times in a row.

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