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 |
$$1$$ |
? |
$$2$$ |
? |
$$3$$ |
$$1/6$$ |
$$4$$ |
? |
$$5$$ |
? |
$$6$$ |
? |
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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 |
$$1$$ |
$$1/6$$ |
$$2$$ |
$$2/6$$ |
$$3$$ |
$$1/6$$ |
$$4$$ |
$$1/6$$ |
$$5$$ |
$$0$$ |
$$6$$ |
$$1/6$$ |
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 |
$$1$$ |
$$1/6$$ |
$$2$$ |
$$2/6$$ |
$$3$$ |
$$1/6$$ |
$$4$$ |
$$1/6$$ |
$$5$$ |
$$0$$ |
$$6$$ |
$$1/6$$ |
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