Assignment 12

Problem 2: 
a) 1/12, since 1 of 12 sides sports a 5
b) 1/2, since half of the 12 numbers are even
c) 2/12=1/6, since 2 of 12 sides show 11 or 12.

Problem 4: P(not victim) = 1-P(victim) = 1 - 19650/265284000 = 0.999926

Problem 6: (4 red)/(6+4+7) = 4/17

Problem 8:
a) 13 spades, 52 cards: 13/52 = 1/4
b) 13 hearts, 13 clubs; hence 26/52=1/2

Problem 9: P(not ace) = 1- P(ace) = 4/52 = 1/13

Problem 11: five of eight regions show a 2. Assuming that all of the
regions are equally likely, the probability is 5/8.

Problem 12: one of four regions show a 1. Assuming that all of the
regions are equally likely, the probability is 1/4.

Problem 15: you have 6 chances out of 870, or 6/870 = 1/145

Problem 16: P(lose) = 7/9; hence P(not lose) = 1 - P(lose) = 1-7/9 = 2/9.

Problem 17: The births could happen in eight different ways:
BBB
BBG
BGB
BBG
GBB
GBG
GGB
GBG
of which 3 result in a single girl. Hence, assuming that they're equally
likely, P(one girl) = 3/8.

Problem 18: We could rephrase the question this way: if the custodian
grabs four keys to try, what's the probability that the correct key is among
them?

This formulation has the advantage of not suggesting that the order in
which we try the keys matters.

There is a 4/15 chance that the correct key is in the first four tested
randomly. (It's like buying four lottery tickets for a lottery with 15 tickets
total: you have a 4/15 chance of winning the lottery - that is, getting the
right key.)

Problem 19: there are 36 outcomes possible using two ordinary dice, given
 in the table on page 189. Of these, 4 sum to 9. Hence

             P(sum of two dice = 9) = 4/36 = 1/9

Problem 20: there are 36 outcomes possible using two ordinary dice, given
 in the table on page 189. Of these, 5 sum to 8. Hence

             P(sum of two dice = 8) = 5/36

Problem 21: there are 36 outcomes possible using two ordinary dice, given
 in the table on page 189. Of these, 5 sum to 6. Hence

             P(not sum of two dice = 6) = 1 - P(sum of two dice = 6) = 1 - 5/36
 = 31/36

Problem 22: 
Here's the universe:
1 1
1 2
1 3
2 1
2 2
2 3
3 1
3 2
3 3
Nine total outcomes. Of these, five have at least one 2. Hence

           P(lands on two at least once) = 5/9

Problem 23: You have a 1 in 5 chance (.2) of getting the right answer.