Question

Suppose that a six-sided die is "loaded" so that any particular even-numbered face is three times as likely to be observed as any particular odd-numbered face. (a) what are the probabilities of the six simple events? (hint: denote these events by o1,..., o6. then p(o1) = p, p(o2) = 3p, p(o3) = p,..., p(o6)= 3p. now use a condition on the sum of these probabilities to determine p. answer as an exact fraction or round your answers to three decimal places.)

Answer 1

p = 1/12 = p(o1) = p(o3) = p(o5)

3p = 1/4 = p(o2) = p(o4) = p(o6)

Explanation:

p(o1) +p(o2) +... +p(o6) = 1 . . . . the condition on the sum of probabilities

... p +3p +p +3p +p +3p = 1 . . . . substitute values

... 12p = 1 . . . . simplify

... p = 1/12 . . . . divide by 12

Then ...

... 3p = 3/12 = 1/4