MATH 55(31) WD SMITH HOMEWORK #8  DUE TUES 8 APR AT CLASS START
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Do these exercises:
7B #  1,3,7,11,13,17     (6 exercises total)
(all of these are odd prime numbers!)
ANSWERS

#1.
Independent events. So can use And-multiplication rule:
Prob all tickets are winners = 1/10 * 1/10 * 1/10 = 1/1000 = 0.001.

#3.
Dependent events since If you pick the first juror to be a woman, that affects the 
probability the 2nd juror will be a woman!
  Prob(given 12 men & 12 women, a 6-person jury is picked randomly that is 6 women)
FIRST SOLUTION:
       #ways(all 6 are women)                12C6       924     3
       -------------------------------    =  ---- =   ------ = ---  
       #ways(to pick 6 people from 24)       24C6     134596   437
SECOND (EQUALLY GOOD) SOLUTION:
     12 11 10  9  8  7    3
     -- -- -- -- -- -- = --- = 0.00686.
     24 23 22 21 20 19   437
because the prob the first one picked is a woman is 12/24.
Once that woman used up, there are 23 people left and 11 are women, 
so prob 2nd one picked is woman is 11/23.
Once they are used up, there are 22 people left and 10 are women, 
so prob 2nd one picked is woman is 10/22, and so on for 6 terms (since picking 6)
in all.
This is using the Prob(A AND B) = Prob(A) * Prob(B given that A happened)
dependent-events ANDing rule, repeatedly.

#7.
These events (8 coin tosses) all independent.
The prob we get 4 tails then 4 heads  in 8 consecutive coin tosses
is thus  (using AND-multiplication rule)
   (1/2) * (1/2) * (1/2) * (1/2) *   (1/2) * (1/2) * (1/2) * (1/2) 
= 1/256 = 0.00391.

#11.
These events non-overlapping, i.e. disjoint.
So prob(get 7 OR 8) = Prob(get 7) + Prob(get 8)
                    = 6/36 + 5/26 = 11/36 = 0.306.
(See example 8 unit 7A in text for table giving the 6/36 and 5/36
in case you did not know how to get them in the first place.)

#13.
These events overlap. 
Prob(woman OR democrat) = Prob(woman) + Prob(democrat) - Prob(WomanDemocrat)
                        = (25+25)/100 + (25+25)/100    - 25/100 
                        = 75/100 = 0.75.

#17.                                                6
Prob(get at least 1 six in 5 dice rolls) = 1 - (5/6)  = 0.598.
Using the "at least once" rule.