Inverse Probability and conditional probability.

Question

An unbalanced die (with 6 faces, numbered from 1 to 6) is thrown. The probability that the face value is odd is 90% of the probability that the face value is even. The probability of getting any even numbered face is the same. If the probability that the face is even given that it is greater than 3 is 0.75. What is the probability that the face value exceeds 3? How to solve this question?

Answer 1

Information:

• $P_1+P_2+P_3+P_4+P_5+P_6=1$
• $P_1+P_3+P_5=\frac{90}{100}(P_2+P_4+P_6)$
• $P_2=P_4=P_6$
• $P_4+P_6=\frac{75}{100}(P_4+P_5+P_6)$

Calculations:

• $P_2+P_4+P_6=3P_2$
• $P_1+P_3+P_5=1-3P_2$
• $P_1+P_3+P_5=\frac{90}{100}\cdot3P_2$
• $1-3P_2=\frac{90}{100}\cdot3P_2$
• $P_2=\frac{10}{57}$
• $P_4=\frac{10}{57}$
• $P_6=\frac{10}{57}$
• $P_5=\frac{20}{171}$

Hence the probability that the face value exceeds $3$ is $\frac{10}{57}+\frac{20}{171}+\frac{10}{57}=\frac{80}{171}$

Answer 2

Hint: $P(\texttt{"number 1-6"})=1=P(\texttt{"even number"})+P(\texttt{"odd number"})=P(\texttt{"even number"})+0.9\cdot P(\texttt{ "even number"})$