Conditional probability - Random number of dice

[Alexsandro]

Can someone help me with this question ?

A random number N of dice is thrown. Let Ai be the event that N = i, and assume that P(Ai) = 1/(2)^i, i >= 1. The sum of the scores is S. Find the probability that:

(a) S = 4, given N is even;
(b) the largest number shown by any die is r, where S is unknown.

 

[EnumaElish]

(a) P(S=4 \text{ and N is even})/P(\text{N is even})=
P\left(\left\{N=2 \text{ and }(S=1+3 \text{ or }S=2+2 \text{ or }S=3+1)\right\}\text{ OR }\{N=4\text{ and }S=1+1+1+1\}\right)\left/ \sum_{k=1}^\infty 2^{-2k}\right.

If the world is just, then the denominator should add up to 1/2. Because P(odd) = P(even) and P(odd) + P(even) = 1, so P(odd) = P(even) = 1/2.

So calculating P(S=4 and N is even)/P(N is even) is a matter of calculating the numerator and then multiplying it with 2.

 

[NateTG]

For part b:

The chance that r is the largest value shown is going to be
\sum_{i=1}^{\infty} P(A_i) \times p(r,i)
where
p(r,i)[/itex]<br /> is the probability that r is the largest value shown by i dice.<br /> <br /> Now, the chance of all of the dice being less than or equal to r, assuming that r\in{1,2,3,4,5,6} is going to be<br /> \left(\frac{r}{6}\right)^{i}[/itex]&lt;br /&gt; and of those&lt;br /&gt; \left(\frac{r-1}{6}\right)^{i}[/itex]&amp;lt;br /&amp;gt; don&amp;amp;#039;t contain any r&amp;lt;br /&amp;gt; So we have&amp;lt;br /&amp;gt; \sum_{i=1}^{\infty} \frac{r^i-(r-1)^i}{12^i}=\sum_{i=1}^{\infty}\left(\frac{r}{12}\right)^i - \sum_{i=1}^{\infty}\left(\frac{r-1}{12}\right)^i=\frac{r}{12-r}-\frac{r-1}{13-r}=\frac{13r-r^2-12r+r^2+12-r}{156-25r+r^2}=\frac{12}{r^2-25r+156}&amp;lt;br /&amp;gt; So&amp;lt;br /&amp;gt; r=6 \rightarrow \frac{12}{156+36-150}= \frac{12}{42} = \frac{2}{7}&amp;lt;br /&amp;gt; r=5 \rightarrow \frac{12}{156+25-125}= \frac{12}{56} = \frac{3}{14}&amp;lt;br /&amp;gt; r=4 \rightarrow \frac{12}{156+16-100}= \frac{12}{72} = \frac{1}{6}&amp;lt;br /&amp;gt; r=3 \rightarrow \frac{12}{156+9-75}=\frac{12}{90}= \frac{2}{15}&amp;lt;br /&amp;gt; r=2 \rightarrow \frac{12}{156+4-50}=\frac{12}{110}=\frac{6}{55}&amp;lt;br /&amp;gt; r=1 \rightarrow \frac{12}{156+1-25}=\frac{12}{132}=\frac{1}{11}

 

[Alexsandro]

The question of the part (a) is P(S = 4 | N is even) and not P(S = 4 and N is even | N is even).

 

[Galileo]

The question of the part (a) is P(S = 4 | N is even) and not P(S = 4 and N is even | N is even).

That's what EnumaElish did. By the definition of conditional probability:

P(A|C)=\frac{P(A \cap C)}{P(C)}

Which is, by the way, trivially equal to P(A\cap C|C).