Find the probability of each outcome when a biased die is rolled, if rolling a 2 or 4 is three times as likely as rolling each of the other$\dots$ Question:Find the probability of each outcome when a biased die is rolled, if rolling a $2$ or $4$ is three times as likely as rolling each of the other four numbers on the die and it is equally likely to roll a 2 or a 4.
My Attempt:
Let $O_1, O_2, \dots, O_6$ be the out come of dies.
Since rolling $O_2$ or $O_4$ is twice as likely, then
$$P(O_2) = P(O_4) = 3P(O_1) = 3P(O_3) = \dots = 3P(O_6)$$
Thus,
$$P(O_1) + P(O_2) + P(O_3) + P(O_4) + P(O_5) + P(O_6) = 1$$
$$P(O_1) + 3P(O_1) + P(O_1) + 3P(O_1) + P(O_1) + P(O_1) = 1$$
$$10P(O_1) = 1$$
$$P(O_1) = \dfrac{1}{10}$$
Therefore,
$$P(O_1) = P(O_3) = P(O_5) = P(O_6) = \dfrac{1}{10}$$ and,
$$P(O_2) = P(O_4) = \dfrac{3}{10}$$
Problem:
The book gave a different answer such that,
$$P(O_1) = P(O_3) = P(O_5) = P(O_6) = \dfrac{1}{16}$$ and,
$$P(O_2) = P(O_4) = \dfrac{6}{16}$$
Frankly, is $P(O_2) = P(O_4) = 6P(O_1)$ now? I think I'm missing something fundamental here.
 A: Your confusion stems from the obscure description of the question at hand.
The actual meaning of the question is apparently $\displaystyle P(2,4) = 3 \cdot P(1,3,5,6)$.
Therefore, $\displaystyle P(2,4) = \frac{12}{16} = 3 \cdot \frac{4}{16} = 3 \cdot P(1,3,5,6)$ implies the following:


*

*$\displaystyle P(2) = \frac{1}{2} \cdot \frac{12}{16} = \frac{6}{16}$

*$\displaystyle P(4) = \frac{1}{2} \cdot \frac{12}{16} = \frac{6}{16}$

*$\displaystyle P(1) = \frac{1}{4} \cdot \frac{ 4}{16} = \frac{1}{16}$

*$\displaystyle P(3) = \frac{1}{4} \cdot \frac{ 4}{16} = \frac{1}{16}$

*$\displaystyle P(5) = \frac{1}{4} \cdot \frac{ 4}{16} = \frac{1}{16}$

*$\displaystyle P(6) = \frac{1}{4} \cdot \frac{ 4}{16} = \frac{1}{16}$



In other words, you should change this attempt:


*

*$P(2) = P(4) = 3P(1) = 3P(3) = 3P(5) = 3P(6)$


To this attempt:


*

*$P(2) + P(4) = 3P(1) + 3P(3) + 3P(5) + 3P(6)$


Where:


*

*$P(2) = P(4)$

*$P(1) = P(3) = P(5) = P(6)$

A: It seems to me that your solution is correct. 
the book you are referring to solve the problem assuming that $P(O_2) = P(O_4) = 6P(O_1)$.
A: I agree the answer in the book is wrong. The key point in the solution is the interpretation of word "or" and thinking of events as sets. Recall that the logical "or" corresponds to "union" in the language of sets. Let $O_i$ be the event that the outcome is $i$. Then the event corresponding to rolling a $2$ or rolling a $4$ is the union $O_2\cup O_4$ so the probability of rolling a $2$ or a $4$ is $P(O_2\cup O_4)=P(O_2)+P(O_4)$. In the last equality we use the fact that the events $O_2$ and $O_4$ are disjoint. By the information given we thus have $P(O_2)+P(O_4)=3P(O_i)$ for each $i=1,3,5,6$. Consequently,  $P(O_i)=P(O_1)$ for every $i=1,3,5,6$ and $$1=P(O_1)+P(O_2)+P(O_3)+P(O_4)+P(O_5)+P(O_6)=4P(O_1)+3P(O_1)=7P(O_1).$$
This proves that $$P(O_1)=\frac{1}{7}=\frac{2}{14}.$$ Moreover since it is assumed that rolling $2$ has the same probability as rolling $4$ we aslo have
$$P(O_2)=P(O_4)=\frac{1}{2}\cdot 3\cdot \frac{1}{7}=\frac{3}{14}.$$
