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21st May 2009, 09:40 PM | #1 |
Philosopher
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Need help with math homework- calculator needed
There are a group of people that know a secret. The probability that any one person leaks the secret is 0.03.
a) If there are 2 people in the group, what is probability that someone leaks the secret? What if there are 5 people in the group? b) How many people in the group do you need to guarantee that someone leaks the secret with 90% probability? (Rounded to decimal places.) Thank you for saving my neck. Also- A multiple choice exam has ten questions. Each question has five possible answers. If a student randomly guesses, what is the probability that she gets 70 percent? I got 7.86432 multiplied by 10 to the power of - 4. Just want to make sure that's right. |
21st May 2009, 10:03 PM | #2 |
Philanthropic Misanthrope
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1) Figure this out as 1 - the probability that nobody does.
In general, 0.97^n = probability of nobody spilling the beans. So for 2 people: 0.97 x 0.97 = 0.9409 1 - 0.9409 = 0.0591 Solve for n = 5 then make n the variable in the second question. Round up to the next whole person when you get an answer. 2) Since this is for a specific outcome, we're figuring out the probability of getting 7 correct and 3 incorrect (0.2 ^ 7 x 0.8 ^ 3) and then multiplying that by the number of ways we could get 7 correct and 3 incorrect (10!/(7!3!) So yes, you're right. |
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21st May 2009, 11:59 PM | #3 |
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Quote:
x is between 75 and 76. In other words if there were 76 people in the group there is > 90% chance (actually 90.1223916%) that someone would leak. With 75 people there is slightly less than 90% (89.8168986%). |
22nd May 2009, 12:08 AM | #4 |
Zygoticly Phased
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2 people can keep a secret.
3 people can keep a secret if two of them are dead. |
22nd May 2009, 12:20 AM | #5 |
Zygoticly Phased
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That's surprisingly high (I'm not disputing the value though). .03 is about 1 in 33 so, naively one might expect an answer in the 30's. That the number needed is more than twice that is a surprise -- 90% confidence is a high hurdle.
33 people give a 63% confidence that it's been leaked -- ooh, I recognize where that's come from -- another surprise |
22nd May 2009, 12:45 AM | #6 |
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Yeah, I actually got exactly 30 people when I sat down and looked at the problem again.
30 times 0.03 = exactly .90. Are you sure your calculations are correct, rjh01? I crossed out my answer and wrote down yours...and I have to give this in by the morning of the 22nd, so you had better be right! |
22nd May 2009, 02:12 AM | #7 |
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Just had another look and am sure that I am right. What everyone is overlooking is that given 75 people it is probable that 2.25 people will give away the secret. This is only an average and there is only a 10% chance that this will be out by over 2.25 people.
If you want to confirm my answer get your scientific calculator out and put the calculation below in it. You can cut and paste into the window's version. This will give you the probability that 75 people can keep a secret given that any one person has a .97 chance of keeping it. Do this by multiplying .97 by itself 75 times or shortcut - .97 x^y 75 = |
22nd May 2009, 03:49 AM | #8 |
Zygoticly Phased
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rjh01 is correct. What you are overlooking in doing 30 * 0.03 is that there are many ways in which the secret could be leaked, but only one way in which it could be kept.
To make things easier, let's just consider 2 people. For the secret to be kept, both of them must keep the secret. the probablility of that happening is .97 * .97. Let's call this Pkept. the secret can be given away in the following 3 ways: *) A keeps the secret and B gives it away: probability .97 * .03 *) B keeps the secret and A gives it away: probability .97 * .03 *) both A and B give the secret away: probability .03 * .03 Let's call the sum of these 3, Pgiven. As you might guess, this gets increasingly more complicated as more people are added -- with 3 people, 2 could give it away, or 1 could or they all could. But, there's a short cut. We know that the probability of the secret being kept + probability of it being given away is 1. So Pkept + Pgiven = 1. It's easy to calculate Pkept, so we can use that to figure out Pgiven rather than do the complicated summation. ETA: I'm changing the bit below in light of bjornart's comment below: Your simpler 30 *.03 is counting the third way of giving the secret away (both A & B giving it away) twice -- once in A's contribution of being a blabber-mouth, and once in B's contribution. |
22nd May 2009, 03:56 AM | #9 |
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What you're doing is adding together the probability of each person spilling the secret. If you double the number of people and observe that 60 people have a 180% chance of spilling the secret you see that this approach is obviously wrong.
The only time you add probabilities together is when each represents a different outcome, like Nathan does to get his Pgiven. |
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22nd May 2009, 04:46 AM | #10 |
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22nd May 2009, 09:07 AM | #11 |
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RJH01's calculation is correct.
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22nd May 2009, 09:14 AM | #12 |
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22nd May 2009, 09:35 AM | #13 |
Philanthropic Misanthrope
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Yeah, I got 5.9%, 14%, and 75.something people for the answers, 76 as the correct answer. I just didn't want to give the guy all of the answers to his homework.
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Sandra's seen a leprechaun, Eddie touched a troll, Laurie danced with witches once, Charlie found some goblins' gold. Donald heard a mermaid sing, Susie spied an elf, But all the magic I have known I've had to make myself. - Shel Silverstein |
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22nd May 2009, 04:25 PM | #14 |
Penultimate Amazing
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Of the two people, how many are girls?
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23rd May 2009, 01:45 AM | #15 |
Zygoticly Phased
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23rd May 2009, 09:32 AM | #16 |
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23rd May 2009, 10:23 AM | #17 |
Zygoticly Phased
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In the two person case, that can happen in exactly 2 ways -- A stays silent and B blabs (probability .97 * .03) or, vice versa (same probability). Thus the probability of either A or B, but not both, blabbing is 2 * .97 * .03.
For N people, the probability of exactly one person blabbing is that of N-1 people keeping quiet and one blabbing, which is: and there are exactly N ways this can happen, giving Notice that that doesn't increase without bound. There will be an N that maximizes the probability of exactly one person blabbing. |
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