halle1@houxz.UUCP (J.HALLE) (09/23/83)
1. The fact that 5 are clubs is immaterial, since you do not have any new
information that matters, so the probability that the lost one is a club is
1/4. However, if 13 are clubs, the probability is now 0.
2. P(right)=p P(both right)=p*p P(only one right)=2p(1-p)
(above only for non-flipper)
P(at least 2 right)=P(both)+.5P(one right)=pp+p(1-p)=p
3. 1/3 of rolls divisible by 3. 1/3*1/2+2/3*2/3=11/18
P(cake donut)=11/18
4. She will say white: .01(w)*.9(truth)+.99(b)*.1(lie)
P(white)=.009/(.009+.099)=1/12
5. V=value, N=#coins V/N=15 (V+10)/(N+1)=14
Solving:N=4, V=60 Thus only can be 2 nickles and two quarters
6. P(not Sunday)=6/7 (Assume all days equally probable, which is not true.)
The probability that n people were not born on Sunday is (6/7)**n. This
number first becomes <.5 when n=5. Consequently, on average, you will
meet 4 people who are not Sunday born, before meeting the 5th who is.
7. P(head)=p p**2=1-p Solving for p: p=(5**.5 -1)/2=.62...
8. P=1 I think (unsolved)
9. There are two answers, depending whether n is odd or even. Counting the
number of permutations where sum<=n:1 and n-1 numbers, 2 and n-2 numbers,...,
n-1 and 1. There are n*(n-1)/2 such permutations. However, there is some
double counting. For n odd, (n-1)/2 are NOT duplicated, for n even, n/2.
To get the total favorable cases, add this number to first one and divide by
2. The probability is this number divided by the total number of ways of
picking the numbers. The total is (n**2+n)/2, as above. Solving the algebra:
P(n even)=n/(2(n+1))
P(n odd)=(n-1)/(2n)
10. There are 44 such sums. 33 of those have product > 1000
(12,87;13,86;...;44,45) Therefore P=33/44=3/4
11. This is the sum of an infinite series. .5+.5**3+.5**5...
a=.5, r=.25 S=a/(1-r) S=.5/.75=2/3=Probability that first one wins.
Odds are 2 to 1 in favor of the first shooter.koch@stolaf.UUCP (Paul E. Koch) (09/26/83)
I'm sorry but the answer to your card problem is wrong. By Bayes theorem P(Lost club | 5 Clubs drawn)= P(5 clubs drawn | L(ost) C(lub))*P(LC) --------------------------------------- P(5 C|LC)*P(LC)+P(5 C | not LC)*P(not LC) = 12*11*10*9*8 1 ------------- * - 51*50*49*48*47 4 ------------------------------------------------ 12*11*10* 9* 8 1 13*12*11*10* 9 3 -------------- * - + -------------- * - 51*50*49*48*47 4 51*50*49*48*47 4 = .1702 Paul Koch
dms@dciem.UUCP (Dave Sweeney) (09/27/83)
Problem no. 6 (how many people will you next meet before you meet one born on Sunday?) should be answered in the following way: Let p = Prob(born on Sunday) = 1/7 on the equidistribution hypothesis, and q = 1 - p = 6/7. Let n = number of people met including the last who is born on Sunday. Then n = 1 with probability p; n = 2 with probability pq (first person is not born on Sunday, second person is); n = 3 with probability (q**2)p, and so on. The expected value of n is the mean of (n * probability of n), summed over all values of n from 1 to infinity: E[n] = sum from 1 to infinity of (n * (q**(n-1)) * p) = (p/q) * sum from 1 to infinity of (n * q**n) which reduces after a little algebra to E[n] = (p/q) * (q/p**2) = 1/p = 7. The interested reader can verify with little trouble that the probabilities themselves sum to 1. Thus one can expect on the average to meet 6 people not born on Sunday before meeting the 7th who is. Dave Sweeney ..!utzoo!dciem!dms
ntt@dciem.UUCP (Mark Brader) (09/29/83)
The answer previously posted to the "lost club?" problem is wrong; stolaf!koch (Paul Koch) has the right result, but there is a much easier way to obtain it. 5 cards were drawn and are known to be clubs. The remaining 47 cards including the lost card contain 8 clubs. We have no information about which of the 47 cards was lost. Therefore the probability that the lost card is a club is 8/47 (or .1702 for those who prefer approximations). The fact that the card was lost before the 5 cards were drawn is irrelevant, because we did not see it and therefore have no information about it. Mark Brader, NTT Systems Inc., Toronto