Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP
Posting-Version: version B 2.10 5/3/83; site houxz.UUCP
Path: utzoo!linus!philabs!cmcl2!floyd!clyde!burl!spanky!hocda!houxm!houxz!halle1
From: halle1@houxz.UUCP (J.HALLE)
Newsgroups: net.math
Subject: New Probability problems: answers
Message-ID: <516@houxz.UUCP>
Date: Fri, 23-Sep-83 15:16:58 EDT
Article-I.D.: houxz.516
Posted: Fri Sep 23 15:16:58 1983
Date-Received: Mon, 26-Sep-83 05:33:30 EDT
Organization: Bell Labs, Holmdel NJ
Lines: 43

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.