The owner of a casino fakes two dices so that in dice $$A$$ we can never get a $$6$$ (and get twice as many ones), and in dice $$B$$ we never get a $$5$$ (and twice as many twos).
- Fill in the following table of probabilities for every dice:
| result dice A | probability |
| $$1$$ | ? |
| $$2$$ | ? |
| $$3$$ | $$1/6$$ |
| $$4$$ | ? |
| $$5$$ | ? |
| $$6$$ | 0 |
| result dice B | probability |
| $$1$$ | ? |
| $$2$$ | ? |
| $$3$$ | $$1/6$$ |
| $$4$$ | ? |
| $$5$$ | ? |
| $$6$$ | ? |
-
What is the probability of obtaining $$1$$ with dice $$A$$ and $$3$$ with dice $$B$$?
- What is the probability of obtaining $$6$$ with dice $$A$$ and $$2$$ with dice $$B$$?
See development and solution
Development:
- The impossible events have zero probability $$(A=6, B=5)$$. As we are been told, there is twice the probability of observing events $$A=1$$ and $$B=2$$ (probability $$2/6$$):
| result dice A | probability |
| $$1$$ | $$2/6$$ |
| $$2$$ | $$1/6$$ |
| $$3$$ | $$1/6$$ |
| $$4$$ | $$1/6$$ |
| $$5$$ | $$1/6$$ |
| $$6$$ | $$0$$ |
| result dice B | probability |
| $$1$$ | $$1/6$$ |
| $$2$$ | $$2/6$$ |
| $$3$$ | $$1/6$$ |
| $$4$$ | $$1/6$$ |
| $$5$$ | $$0$$ |
| $$6$$ | $$1/6$$ |
- Since the dice are independent, probabilities of the events are multiplied:
$$P(A=1 \ \& \ B=3) = \Big(\dfrac{2}{6}\Big) \cdot \Big(\dfrac{1}{6}\Big) =\dfrac{1}{18}$$
- Since it is impossible to observe $$A=6$$, the probability will be zero.
$$P(A=6 \ \& \ B=2) = 0 \cdot \Big(\dfrac{2}{6}\Big) =0$$
Solution:
| result dice A | probability |
| $$1$$ | $$2/6$$ |
| $$2$$ | $$1/6$$ |
| $$3$$ | $$1/6$$ |
| $$4$$ | $$1/6$$ |
| $$5$$ | $$1/6$$ |
| $$6$$ | $$0$$ |
| result dice B | probability |
| $$1$$ | $$1/6$$ |
| $$2$$ | $$2/6$$ |
| $$3$$ | $$1/6$$ |
| $$4$$ | $$1/6$$ |
| $$5$$ | $$0$$ |
| $$6$$ | $$1/6$$ |
$$P(A=1 \ \& \ B=3) = \Big(\dfrac{2}{6}\Big) \cdot \Big(\dfrac{1}{6}\Big) =\dfrac{1}{18}$$
$$P(A=6 \ \& \ B=2) = 0 \cdot \Big(\dfrac{2}{6}\Big) =0$$