Problems from 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 $B$ ?

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:

$P (A = 1 & B = 3) = (\frac{2}{6}) \cdot (\frac{1}{6}) = \frac{1}{18}$

Since it is impossible to observe $A = 6$ , the probability will be zero.

$P (A = 6 & B = 2) = 0 \cdot (\frac{2}{6}) = 0$

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

$P (A = 1 & B = 3) = (\frac{2}{6}) \cdot (\frac{1}{6}) = \frac{1}{18}$

$P (A = 6 & B = 2) = 0 \cdot (\frac{2}{6}) = 0$

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