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#1
08-26-2002, 04:16 AM
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More Challenging Problems

http://bsuvc.bsu.edu/~d004ucslabs/probability.html

Since the answer to one of the problems is not given, and the answer to another is completely wrong, my answers will appear in a post below.
#2
08-26-2002, 06:02 AM
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1. The probability of not getting heads in 2 tosses is 3/4. To get two heads in a row on exactly the 10th toss, we have to NOT get two in a row in 4*2 flips, and then get two heads on throws 9 and 10. The probability of this is

(3/4)^4(1/2)(1/2) = 7.9%.

2. This result is counterintuitive. The posted solution is fine except the line:

Thus, the probability of both coins coming up with the given results is 2/27.

Thus, the probability of both coins coming up with the given results is (1/2)(4/9) = 2/9.

3. (My solution) The posted solution would be correct only if the names were replaced in the hat after each draw. That isn't the way this kind of thing works, or some people might not get any presents, and some might get several. The correct solution is to note that if f(n) is the number of ways n people can draw names without drawing their own, then f(n) satisifes:

f(2) = 1

f(n) = (n-1)f(n-1)

Thus, f(n) = (n-1)!

So the probability is 9!/10! = 1/10.

4. The posted solution is clever. Mine is to note that you can only see two sides of something if one side slopes away from you less than 90 degrees. Since the angles of a pentagon are 108 degrees, the sides "stick out" 18%. We can see 3 sides as long as we are not more than 18 degrees off center to any one side in either direction for a total of 36 degrees per side. Since there are 5 sides, the probability is 36*5/360 = .5.

5. (My solution) The tallest person will always be visible. The second tallest person will be visible if and only if he is in front of the tallest person, which has probability 1/2. The third tallest person will be visible if and only if he is ahead of the tallest and second tallest people, which has probability 1/4, etc. Now notice that since one person OR MORE will be visible all the time, two people or more will be visible half the time, and three people or more will be visible 1/8 of the time, that means that exactly 1 person will be visible 1/2 the time, exactly 2 people visible 1/4 of the time, etc. So the expected value of the number of visible people is:

1/2 + 2/4 + 3/8 + ... (10 terms)

We know this sums to approximately 2 people visible on average.

5. Perhaps this is more intuitive. 50 errors were found by both proofreaders, and this was 50/70 of the errors found by the second reader alone. So if P(A) and P(B) are the fractions of the total errors found by readers A and B, then

P(A)P(B) = (50/70)P(B) so P(A) = 50/70. Since A found 50/70 of the errors and he found 60 errors, there must be (70/50)(60) = 84 errors total, so 4 errors were undetected.
#3
08-26-2002, 06:08 AM
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correction to 1

Ignore my explanation for 1, and look the one on the other website. You can't just consider the individual pairs since you could get a head at the end of one pair and at the beginning of the next.
#4
08-26-2002, 07:26 AM
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Nice problems, thankyou *NM*

#5
08-26-2002, 09:41 AM
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#3 is wrong ?

Nice Q's !

But you are wrong on #3 ! I guess ...

#6
08-26-2002, 11:20 AM
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Re: correction to 1

Great Problems BruceZ. I have some comments below.

Question 1 :

I agree that the posted solution to #1 is correct, but it is not very enlightening. It would be very hard to extend the given solution to n=10000. I will post my solution to problem #1 separately.

Question 3 :

I agree that the posted solution would be correct if the names were replaced in the hat after each draw.

I disagree with your solution to #3. You claim that

f(2) = 1

f(n) = (n-1)f(n-1)

where f(n) is the number of ways n people can draw names without drawing their own. Obviously f(2) = 1. It is easy to write out the 6 permutations to check that f(3) = 2 = 2*1 which agrees with your formula. When I wrote out the 24 permuatations for f(4), I got f(4) = 9, which does not obey your formula.

Question 5 : I like your reasoning on this one. I got slightly different numbers. You state that the 3rd tallest person would be visible 1/4 of the time. I think that he would be visible 1/3 of the time. There seem to be 6 equally likely possibilities to me: 123, 132, 213, 231, 312, 321. In 1/3 of these possibilities the 1 can be seen. Similarly, I think the nth person can be seen with probability 1/n, so then, using your reasoning, the average number of people seen in an N person line would be

1+ 1/2 + 1/3 + 1/4 + ... + 1/N

#7
08-26-2002, 11:25 AM
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Re: #3 is really the math geek problem

I think problem #3 is essentially the same as the problem posted by math geek.

math geek -- Saturday, 17 August 2002, at 5:26 a.m

If they are the same problem, then the answer would be the same as the posted solution : 0.36787946428572.

#8
08-26-2002, 12:42 PM
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correction to 5 theatre line probem

Seeing the 3rd tallest person doesn't mean you can see 3 people or more. You might only see the 3rd tallest and the tallest, with the rest obscured behind the tallest. Also, the probability of the 3rd tallest being visible is 1/3 not 1/4. The probability of the nth tallest person being visible is (n-1)!/n! = 1/n.

With these corrections the problem is actually eaiser. Notice that if we just add the probabilities for seeing the tallest, 2nd tallest, etc. people, we will get the desired average since we will be counting all the times we see 2 people exactly twice, all the times we see 3 people 3 exactly 3 times, etc. So the series is:

1 + 1/2 + 1/3 + ....+ 1/10 = 2.93

So almost 3 people would be visible on average.

#9
08-26-2002, 12:49 PM
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yes you are correct

I don't know what the hell I was thinking. My recursion doesn't work because after the first person chooses a name, although there are n-1 names left, it is not the same as n-1 people choosing n-1 names because one of their names may already be taken.

This is a little better than the math geek problem because since there are only 10 names, you have to use inclusion-exclusion or your answer will be off. If you assume indpendence or you will be off by 2%.
#10
08-26-2002, 12:51 PM
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Re: correction to 1

I fixed #5 above before I saw you answer. This is correct.

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