
Robert F. answered 06/30/15
Tutor
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A Retired Professor to Tutor Math and Physics
Nomenclature:
P1=probability of faces 1 to 4.
P6=probability of faces 5 to 6.
Probabilities must sum to 1.
4(P1)+2(P6)=1
Problem specifies:
(P1)=3(P5)=3(P6)
These are two equation in two unknowns.
4[3(P6)]+2(P6)=1
14(P6)=1
(P6)=1/14
(P1)=13/14
Probability of rolling a 4 is 13/14.
Probability of rolling a 6 is 1/14.
Sav S.
probability of rolling a 4 is 3/14 because its P(1) = 3*P(6) = 3 * 1/1409/25/20