Probability - Peculiar die

Question

A peculiar die has the following properties: on any roll the probability of rolling either a $4$, a $6$, or a $1$ is $1/2$, just as it is with an ordinary die. Moreover, the probability of rolling either a $1$, a $3$, or a $2$ is again $1/2$. However, the probability of rolling a $1$ is $5/16$, not $1/6$ as one would expect of an ordinary fair die.

From what you know about this peculiar die, compute the following:

a. The probability of rolling either a $5$, a $3$, or a $2$;

b. The probability of rolling a $5$.

My answer

Given $P(1) = \frac{5}{16}$, then

$$P(2) + P(3) + P(1) = P(2) + P(3) + \frac{5}{16} = \frac{1}{2} \iff P(2) + P(3) = \frac{3}{16}$$

Similarly, using $P(4) + P(6) + P(1) = \frac{1}{2}$, I find

$$P(6) + P(4) = \frac{3}{16}$$

b. $P(5) = 1 - \left(\frac{5}{16} + \frac{3}{16} + \frac{3}{16}\right) = \frac{5}{16}$

a. $P(5) + P(3) + P(2) = \frac{5}{16} + \frac{3}{16} = \frac{1}{2}$

Are my calculations correct?

Answer 1

I agree with your solution. I wrote a very similar solution below.

Let $E_j$ be the event that we roll a $j$. Then $$ P(E_2 \cup E_3) = P(E_1 \cup E_2 \cup E_3) - P(E_1) = \frac12 - \frac{5}{16} = \frac{3}{16}. $$ It follows that \begin{align} P(E_5) &= 1 - P(E_1 \cup E_2 \cup E_3 \cup E_4 \cup E_6) \\ &= 1 - P(E_1 \cup E_4 \cup E_6) - P(E_2 \cup E_3) \\ &= 1 - \frac12 - \frac{3}{16} = \frac{5}{16}. \end{align} Finally, we can put these pieces together to see that $$ P(E_2 \cup E_3 \cup E_5) = P(E_2 \cup E_3) + P(E_5) = \frac{3}{16} + \frac{5}{16} = \frac12. $$

Answer 2

Your calculations are correct but slightly more complicated than necessary. In a way, the order in which parts (a) and (b) are presented is a hint at a simpler solution.

Letting $R_{i,j,\ldots,k}$ be the event that we roll one of the numbers in the set $\{i,j,\ldots,k\},$ then we are given that $R_{1,4,6} = R_{1,2,3} = \frac12$ and that $R_1 = \frac{5}{16}.$

For part (a), simply observe that $\{2,3,5\}$ and $\{1,4,6\}$ are disjoint and that $\{2,3,5\} \cup \{1,4,6\}$ is the entire sample space. Therefore $$P(R_{2,3,5}) = 1 - P(R_{1,4,6}) = 1 - \tfrac12 = \tfrac12.$$

Now for part (b) you know that $$P(R_1) + P(R_{2,3}) = P(R_{1,2,3}) = \tfrac12$$ and $$P(R_{2,3}) + P(R_5) = P(R_{2,3,5}) = \tfrac12,$$ so $$P(R_{2,3}) + P(R_5) = P(R_1) + P(R_{2,3}),$$ and canceling $P(R_{2,3})$ on each side you have $$P(R_5) = P(R_1) = \tfrac{5}{16}.$$