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Thu 23 Dec, 2004 09:16 am
Hi, i just signed up. i'm quite interested in riddles. I found one but i do not know the answer. I need your help.
Mrs. mathemagician has three children. If atleast one of her children is a boy, what is the probability that she has three sons?
Another one is
10% of the people in a certain population use an illegal drug. A drug test yeilds the correct result 90% of the time, whether the person uses drugs or not. A random person is forced to take the drug test and the result is positive. What is the probablity he uses drugs?
If you can help me out i appreciate it.
Hi DragonRiddler.
Q1: answer = 0.25 = 25% = 1-in-4. The prob. that child #1 is a boy is 100%, because we are told so. The prob. any child (of unknown gender) is a boy is 50%. The prob. that they are all boys is therefore: 100% x 50% x 50% = 25%
And I'm working on the 2nd riddle.
I'm not certain about #2 but I think the accuracy of the drug test is a "red herring".
Whether or not he passes or fails the test does not have an effect on whether or not he actually takes the drug - I think the correct term for this is "mutually exclusive events".
If this is the case, then the chance that ANY random person takes the drug is still 10%, whether they pass or fail the test.
And welcome to A2K. Good riddles!
#2 has something to do with Bayesian statistics, I think. Going to look that up now.
http://www.kevinboone.com/bayes.html
Quote:Another example
Here is another example of Bayes' theorem at work. Consider a system that tests whether computer components are working or faulty, based on the temperature they run at when they are working. Suppose, for example, that 99.5% of components that are faulty run at over 50 degrees. Now, if we take a particular component and test it, and the temperature if 53 degrees, how likely is it to be a faulty component? If you can answer that question with a number, particular if that number is `99.5%', you need to read the first part of this article again, because your answer will be wrong. The correct answer is `I don't know'. I simply haven't given you all the information you need to answer the question.
What you need to know is the probability of the component's being faulty, given that runs too hot. We can write this as `p(faulty|hot)'. But what I've told you is the probability of a faulty component running hot -- p(hot|faulty). These two figures are related by Bayes theorem:
p(faulty|hot) = p(hot|faulty) * p(faulty) / p(hot)
p(faulty) is the overall probability of a component's turning out to be faulty, whether or not it runs too hot. Suppose, for example, that 1 in 100 components is faulty overall. So p(faulty) is 0.01. p(hot) is the probability of a component's running hot, whether or not it is faulty. Suppose, for example, that 1 in 20 components runs too hot, whether or not it is faulty. So p(hot) is 0.05. This gives us:
p(faulty|hot) = p(hot|faulty) * p(faulty) / p(hot)
= 0.995 * 0.01 / 0.05
~= 0.2 or 20%
So in fact, in these circumstamces, if a component tests as faulty the probability that it is really faulty is only 20%, not 99.5% at all. This could have profound consequences if you're using this test to determine which components to sell and which to put out with the rubbish.
I still can't figure it out, though, since it seems we need to know the probability of testing positive for anyone. But it looks like we are looking for p(uses|positive) or the probability that the person uses drugs given that they've tested positive.
Well, here's what I get.
If the test is right 90% of the time then 90% of the 10% of drug users will test positive while 10% of the 90% who don't will test positive. Thus 18% of the population will test positive for drugs.
So:
p(uses|positive) = p(positive|uses)*p(uses)/p(positive)
= .9 * .1 / .18 = .5
So the answer is 50%?
Grand Duke wrote:Hi DragonRiddler.
Q1: answer = 0.25 = 25% = 1-in-4. The prob. that child #1 is a boy is 100%, because we are told so. The prob. any child (of unknown gender) is a boy is 50%. The prob. that they are all boys is therefore: 100% x 50% x 50% = 25%
Had she said, "My first child is a boy," this would be correct. However, doesn't her statement merely eliminate one out of eight possibilities (G,G,G)? That would make the probability 1/7.
I'm a little lost Markr!
It doesn't only eliminate the (G,G,G) option, it also takes out the BGG, BBG - basically anything that isn't (B,B,B).
Because you know that you know one is definately a boy, this becomes conditional prob'y!
So this would leave 1/4!!
At least one boy only eliminates all girls. You know she has a boy, but you don't know which one is the boy.
I don't think it matters whether the boy we are told about is the first, second or third child. What we are really trying to calculate is the prob. that 2 children of unknown sex are both boys. And that is 25%.
I agree, order doesn't matter in this case.
It's all a matter of interpretation. Here's a good discussion on the topic.
http://www.wiskit.com/marilyn/boys.html
I just re-read my last post - it didn't come across too clear!
What it meant to mean was that the only thing we're interested in is (B,B,B), so we can eliminate all others.
One of the B's is certain (as we are told), the other two are not. So given that the probabilities are independant of each other, P(B & B) = P(B) * P(B) = 1/2 * 1/2 = 1/4 or 0.25 or 25%!!!
That's all it is!
Oh and while I'm in a maths train of thought:
FreeDuck is correct that 50% is the correct answer for the drugs question. You can work it out using a simple tree diagram if you really wanted to!
+ve = result was drug user,
-ve = result was clean
The no. drug users who are +ve = 9%
the no. drug users who are -ve = 1%
the no. clean who are -ve = 81%
the no. clean who are +ve = 9%
So if we know the answer was +ve, this limits our results to 18% of the population. What % of that 18% were drug users? Simply half of them hense 50%!
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