- Inicio
- Distributions
- Independent event
- Ejercicios
Independent event
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 $B$?
- 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.
$$P(A=6 \ \& \ B=2) = 0 \cdot \Big(\dfrac{2}{6}\Big) =0$$
| 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 |
| $1$ | $1/6$ |
| $2$ | $2/6$ |
| $3$ | $1/6$ |
| $4$ | $1/6$ |
| $5$ | $0$ |
| $6$ | $1/6$ |
$$P(A=1 \ \& \ B=3) = \Big(\dfrac{2}{6}\Big) \cdot \Big(\dfrac{1}{6}\Big) =\dfrac{1}{18}$$
$$P(A=6 \ \& \ B=2) = 0 \cdot \Big(\dfrac{2}{6}\Big) =0$$