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.


Should read:


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.