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The probability function
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 |
| $1$ | ? |
| $2$ | ? |
| $3$ | $1/6$ |
| $4$ | ? |
| $5$ | ? |
| $6$ | ? |
- 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 |
| $1$ | $1/6$ |
| $2$ | $2/6$ |
| $3$ | $1/6$ |
| $4$ | $1/6$ |
| $5$ | $0$ |
| $6$ | $1/6$ |
| 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 |
| $1$ | $1/6$ |
| $2$ | $2/6$ |
| $3$ | $1/6$ |
| $4$ | $1/6$ |
| $5$ | $0$ |
| $6$ | $1/6$ |