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Q: Probability *** ( Answered 3 out of 5 stars,   0 Comments )
Subject: Probability ***
Category: Science > Math
Asked by: alpa101473-ga
List Price: $4.00
Posted: 05 Sep 2003 19:49 PDT
Expires: 05 Oct 2003 19:49 PDT
Question ID: 252775
A Dir is tossed and the number of dots facing up is noted.

a.  Find the probability of the elementary events under the assumption
that all faces of the die are equally likely to be facing up after a
b.  Find the probability of the elementary events under the assumption
that the face with a single dot is twice as likely to be facing up as
the rest of the faces.
c.  Find the probability that the outcome of a toss is even under the
assumptions in parts a and b of this problem

Clarification of Question by alpa101473-ga on 05 Sep 2003 19:52 PDT
Give little explanation of each answer
Subject: Re: Probability ***
Answered By: mathtalk-ga on 05 Sep 2003 21:25 PDT
Rated:3 out of 5 stars
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

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

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

[Law of Total Probability]

regards, mathtalk-ga
alpa101473-ga rated this answer:3 out of 5 stars

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