Trick dice Probability

Question

I was given this exercise:

We have 2 dice, the first is normal $\Omega = \left\{1,2,...,6 \right\}$ and the other is a trick dice, with sample space $\Omega = \left\{2,3,4,5,6,6 \right\}$. If there is probability $ \frac{1}{3}$ of choosing the first dice and $\frac{2}{3}$ of choosing the second dice what is the probability of:

1)Getting a number $\leq 3$.

2)Getting a number >3.

3) Choosing the trick dice if we got a number < 4.

My answers are $\frac{7}{18}$, $\frac{11}{18}$ and $\frac{7}{18}$. I would like to know if I'm correct. Thanks everyone.

Answer 1

The first two are correct, the third is not. Denoting with $D$ the result of the dice, with $D_1$ the event to pick the fair dice and with $D_2$ the (complementary) event to pick the tricky dice you have that:

• For the 1). By the law of total probability $$P(D\le 3)=P(D\le3|D_1)P(D_1)+P(D\le3|D_2)P(D_2)=\frac{3}{6}\cdot\frac{1}{3}+\frac{2}{6}\cdot\frac{2}{3}=\frac{7}{18}$$
• For the 2). Since $D>3$ and $D\le 3$ are complementary you have that $$P(D>3)=1-P(D\le 3)=1-\frac{7}{18}=\frac{11}{18}$$
• For the 3). By Bayes rule $$P(D_2|D<4)=P(D_2|D\le 3)=\frac{P(D\le 3|D_2)P(D_2)}{P(D\le 3)}=\frac{\frac{4}{18}}{\frac{7}{18}}=\frac{4}{7}$$

Answer 2

As a complement (and compliment) to Stefanos's answer, I'll describe a simpleminded approach I take to problems of this type.

Because of the two-to-one odds of choosing the trick die, I imagine having rolled each side of the normal die once and each side of the trick die twice, for a total of $18$ rolls. If you do so, you get the numbers $1$, $2$, and $3$ once each from the normal die, and the numbers $2$ and $3$ twice each from the trick die, for a total of $3+4=7$ rolls. That gives the answer $7/18$ for part 1).

Part 2) is just the complementary probability $1-(7/18)=11/18$.

For part 3), we're focusing on the $7$ rolls described above, which resulted in numbers less than $4$, and asking, in essence, how many of them came from the trick die. We thus get the conditional probability $4/7$.