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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