Hi, alpa101473-ga:
Let's assume the "die" considered here is the typical kind with six
equal sides (a cube), so that each number of dots from 1 to 6 has a
chance of turning up. Although "fancier" dice are sometimes used in
various games, I would not suspect their being meant for this problem
without explicit directions to that effect.
a. The "elementary events" are the various numbers of dots on the
"up" face that can be observed. If the six possible values (1 to 6
dots) are equally likely, each elementary event has probability 1/6
(since their probabilities must sum up to 1).
b. Under the assumption that the "single dot" outcome is twice as
likely as any one of the other "faces" to come up, we would have these
probabilities:
Pr( single dot ) = 2p Pr( four dots ) = p
Pr( two dots ) = p Pr( five dots ) = p
Pr( three dots ) = p Pr( six dots ) = p
with the condition 2p + p + p + p + p + p = 7p = 1. Thus p = 1/7, and
the implications for the above elementary events' probabilities are
clear.
c. The likelihood of an even toss is the probability of the union of
three mutually exclusive (elementary) events, namely two dots up, four
dots up, or six dots up.
Under the assumption of part a., we have:
Pr( two dots OR four dots OR six dots ) = 1/6 + 1/6 + 1/6 = 1/2
Under the assumption of part b., we have:
Pr( two dots OR four dots OR six dots ) = 1/7 + 1/7 + 1/7 = 3/7
Note that it is a basic law of probability that the chance of the
union of mutually exclusive events is the sum of their individual
probabilities.
[Law of Total Probability]
http://www.biostat.wisc.edu/faculty/eickhoff/lecturenotes/lecture4.pdf
regards, mathtalk-ga |