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Need help with math homework- calculator needed

EeneyMinnieMoe

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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. :blush:

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.
 
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.
 
b) How many people in the group do you need to guarantee that someone leaks the secret with 90% probability? (Rounded to decimal places.)
Click to expand...

That is .97^x = 0.1 (this is the probability that the secret is safe).
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%).
 
2 people can keep a secret.

3 people can keep a secret if two of them are dead.
 
rjh01 said:
That is .97^x = 0.1 (this is the probability that the secret is safe).
x is between 75 and 76.
Click to expand...

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 ;)
 
nathan said:
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 ;)
Click to expand...

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! :p
 
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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 =
 
EeneyMinnieMoe said:
30 times 0.03 = exactly .90.

Are you sure your calculations are correct, rjh01?
Click to expand...

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.
 
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EeneyMinnieMoe said:
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! :p
Click to expand...

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.
 
bjornart said:
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.
Click to expand...

But you have an expected value of the secret being reveled 1.8 times by that group.

So with that group the most common outcome would be that the secret was revealed by two people.
 
nathan said:
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.
Click to expand...


Thanks for that. There's a bit of ambiguity in the OP where it says "someone". If this is read to mean exactly one person gives away the secret, how would this affect your analysis?
 
sphenisc said:
Thanks for that. There's a bit of ambiguity in the OP where it says "someone". If this is read to mean exactly one person gives away the secret, how would this affect your analysis?
Click to expand...

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:
[latex] .97^{N-1} * .03[/latex]
and there are exactly N ways this can happen, giving
[latex]N * .97^{N-1} * .03[/latex]

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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