1.1.4 (a) The possible outcomes are 123 132 213 231 312 321
(b) i. The event consists of the outcomes 213 and 231, for a probability of 2/6.
ii. This event is 132 and 231, again with probability 2/6
iii. The event is 231, probability 1/6.
iv. The event is 213, 231 and 132, with probability 3/6.
v. This event is 231 and 312, with probability 2/6.
(c) X(123)=3
X(132)=X(321)=X(213)=1
X(231)=X(312)=0
(d) X= 3 with probability 1/6
1 with probability 3/6
0 with probability 2/6
using the results of part (c).
1.1.7 (a) i. {1,2}, {2,3}, {2,4} with probability 3/6.
ii. {1,3} and {2,3} with probability 2/6
iii. {1.2}, {1,3}, {2,3} with probability 3/6
(b) X({1,2})=3 with probability 1/6
X({1,3})=4 with probability 1/6
X({1,4})=X({2,3})=5 with probability 2/6
X({2,4})=6 with probability 1/6
X({3,4})=7 with probability 1/6
1.1.8 (a) i. 12, 21, 23, 32, 24, 42 with probability 6/12
ii. 13, 31, 23, 32 with probability 4/12
iii. 12, 21, 13, 31, 23, 32 with probability 6/12
(b) X(12) = X(21) = 3 with probability 2/12
X(13)=X(31)=4 with probability 2/12
X(14)=X(41)=X(23)=X(32)=5 with probability 4/12
X(24)=X(42)=6 with probability 2/12
X(34)=X(43)=7 with probability 2/12
Note: The probabilities and the possible values of X are the same as in 1.1.7 (a) and (b).
(c) i. 31, 32, 34 with probability 3/12
ii. 13, 23, 43 with probability 3/12
1.1.9 (a) i. 12, 22, 21, 23, 32, 24, 42 probability 7/16
ii. 13, 23, 32, 31, 33 probability 5/16
iii. 12, 13, 23, 33, 32, 31, 22, 21, 11 probability 9/16
(b) X(11) = 2 with probability 1/16
X(12)=X(21)=3 with probbility 2/16
X(13)=X(22)=X(31)=4 with probability 3/16
X(14)=X(23)=X(32)=X(41)=5 with probability 4/16
X(24)=X(33)=X(42)=6 with probability 3/16
X(34)=X(43)=7 with probability 2/16
X(44)=8 with probability 1/16
Note that X now can take on more possible values than in the previous
two problems.
2.1.6 .4 x .28 + .43 x .75 + .17 x .42 = .112+.3225+.0714 = .5059