- #1

mathmari

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MHB

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We consider the following game:

We can roll a dice until a number appears twice. We can then write down as many points as the times that we rolled the dice.

Let $X$ be the random variable that describes the points that we get at each game. I want to calculate the probabilities $P(X=x_i)$, where $x_i\in \{2,3,4,5,6,7\}$.

We have the following:

- $P(X=2)$ is the probability that we get $2$ points, i.e. that we roll twice a dice and get the same result.

When we roll a dice twice, we get the set $\Omega = \{ (a\mid b) \mid a,b \in \{1,2,3,4,5,6\} \}$ and so the number of possible results is equal to $6\cdot 6=36$.

$6$ of these $36$ possible results have twice the same number.

So we get the probability $P(X=2)=\frac{6}{36}$.

- $P(X=3)$ is the probability that we get $3$ points, i.e., we roll three times a dice and a number appears twice and the second time is at the third place.

When we roll a dice three times, we get the set $\Omega = \{ (a\mid b\mid c) \mid a,b,c \in \{1,2,3,4,5,6\} \}$ and so the number of possible results is equal to $6\cdot 6 \cdot 6= 216$.

The favorable results are contained in the set $M=\{ (a\mid b\mid c) \mid a,b,c \in \{1,2,3,4,5,6\} , a\neq b, c=a \text{ oder } c=b\}$, right?

So $M$ contains $6\cdot 5= 30$ favorable results, or not?

So, we have $30$ favorable results.

Therefore, the probability is $P(X=3)=\frac{30}{216}$.

- $P(X=4)$ is the probability that we get $4$ points, i.e., we roll four times a dice and a number appears twice and the second time is at the $4$th place.

When we roll a dice four times, we get the set $\Omega = \{ (a\mid b\mid c\mid d) \mid a,b,c,d \in \{1,2,3,4,5,6\} \}$ and so the number of possible results is equal to $6\cdot 6 \cdot 6\cdot 6= 1296$.

The favorable results are contained in the set $M=\{ (a\mid b\mid c\mid d) \mid a,b,c,d \in \{1,2,3,4,5,6\} , a\neq b\neq c, d=a \text{ or } d=b \text{ or } d=c\}$.

So $M$ contains $6\cdot 5 \cdot 4\cdot 3= 360$ favorable results.

Therefore, the probability is $P(X=4)=\frac{360}{1296}$.