# probability proof requiring conditional probability

Consider the experiment where two dice are thrown. Let $A$ be the event that the sum of the two dice is 7. For each $i\in\{1,2,3,4,5,6\}$, let $B_i$ be the event that at least one $i$ is thrown.

(a) Compute $P(A)$ and $P(A\mid B_1)$.

(b) Prove that $P(A\mid B_i)=P(A\mid B_j)$ for all $i$ and $j$.

(c) Since you know that some $B_j$ always occurs, does it make sense that $P(A)\neq P(A\mid B_i)$? (After all, if $E$ is an event with $P(E)=1$, then for any event $F$, $P(F\mid E)=P(F)$. What is going on? Does this seem paradoxical?)

I am stuck on this question especially the proof. Can anybody help me out.

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Hint:

$\qquad\qquad\text{Event }A\\\begin{array}{c|c|c|c|c|c|c|c|} & 1 & 2 & 3 & 4 & 5 & 6\\\hline 1& & & & & & \bullet \\\hline 2& & & & & \bullet & \\\hline 3& & & & \bullet & & \\\hline 4& & & \bullet & & & \\\hline 5& & \bullet & & & & \\\hline 6& \bullet & & & & & \\\hline \end{array}$                $\qquad\qquad\text{Event }B_1\\\begin{array}{c|c|c|c|c|c|c|c|} & 1 & 2 & 3 & 4 & 5 & 6\\\hline 1& \bullet & \bullet & \bullet & \bullet & \bullet & \bullet \\\hline 2& \bullet & & & & & \\\hline 3& \bullet & & & & & \\\hline 4& \bullet& & & & & \\\hline 5& \bullet & & & & & \\\hline 6& \bullet & & & & & \\\hline \end{array}$

$\qquad\qquad\text{Event }B_2\\\begin{array}{c|c|c|c|c|c|c|c|} & 1 & 2 & 3 & 4 & 5 & 6\\\hline 1& & \bullet & & & & \\\hline 2&\bullet & \bullet & \bullet& \bullet & \bullet & \bullet \\\hline 3& & \bullet & & & & \\\hline 4& & \bullet & & & & \\\hline 5& & \bullet & & & & \\\hline 6& & \bullet & & & & \\\hline \end{array}$                $\qquad\qquad\text{Event }B_3\\\begin{array}{c|c|c|c|c|c|c|c|} & 1 & 2 & 3 & 4 & 5 & 6\\\hline 1& & & \bullet & & & \\\hline 2& & & \bullet & & & \\\hline 3& \bullet & \bullet & \bullet & \bullet & \bullet & \bullet \\\hline 4& & & \bullet & & & \\\hline 5& & & \bullet & & & \\\hline 6& & & \bullet & & & \\\hline \end{array}$

$\qquad\qquad\text{Event }B_4\\\begin{array}{c|c|c|c|c|c|c|c|} & 1 & 2 & 3 & 4 & 5 & 6\\\hline 1& & & & \bullet & & \\\hline 2& & & & \bullet & & \\\hline 3& & & & \bullet & & \\\hline 4& \bullet & \bullet & \bullet & \bullet & \bullet & \bullet \\\hline 5& & & & \bullet & & \\\hline 6& & & & \bullet & & \\\hline \end{array}$                $\qquad\qquad\text{Event }B_5\\\begin{array}{c|c|c|c|c|c|c|c|} & 1 & 2 & 3 & 4 & 5 & 6\\\hline 1& & & & & \bullet & \\\hline 2& & & & & \bullet & \\\hline 3& & & & & \bullet & \\\hline 4& & & & & \bullet & \\\hline 5& \bullet & \bullet& \bullet & \bullet & \bullet & \bullet \\\hline 6& & & & &\bullet & \\\hline \end{array}$

$\qquad\qquad\text{Event }B_6\\\begin{array}{c|c|c|c|c|c|c|c|} & 1 & 2 & 3 & 4 & 5 & 6\\\hline 1& & & & & & \bullet\\\hline 2& & & & & & \bullet \\\hline 3& & & & & & \bullet \\\hline 4& & & & & & \bullet \\\hline 5& & & & & & \bullet \\\hline 6& \bullet& \bullet & \bullet &\bullet & \bullet & \bullet \\\hline \end{array}$

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(a) There are $6\cdot6=36$ possible outcomes when two dice are thrown:

$$\langle 1,1\rangle,\langle 1,2\rangle,\dots,\langle 1,6\rangle,\langle 2,1\rangle,\langle 2,2\rangle,\dots,\langle 2,6\rangle,\dots,\langle 6,1\rangle,\langle 6,2\rangle,\dots,\langle 6,6\rangle\;.$$

Which are in the event $A$? $\langle 1,6\rangle,\langle 2,5\rangle,\langle 3,4\rangle,\langle 4,3\rangle,\langle5,2\rangle$, and $\langle 6,1\rangle$. Thus, $6$ of the $36$ equally likely outcomes are in $A$, and by definition $$P(A)=\frac6{36}=\frac16\;.$$

You can also reach this conclusion by noticing that no matter how the first die comes up, there is exactly one value on the second die that puts the roll in event $A$, and the probability of getting that one value is $\frac16$.

Now suppose that $B_1$ has occurred: you’ve rolled at least one $1$. The possible outcomes now are $\langle 1,1\rangle,\langle 1,2\rangle,\langle 2,1\rangle,\dots,\langle 1,6\rangle,\langle 6,1\rangle$: there are $11$ of them, and they’re equally likely. How many of them are in event $A$? Once you’ve ascertained that, you can finish the calculation just as I finished the first one.

(b) HINT: Show that no matter what the value of $i$, there are exactly $11$ rolls in event $B_i$; what are they? Then show that the number of them in event $A$ is the same for each $i$, so that the calculation in the second part of (a) ends up being the same for each $i$.

(c) HINT: If the events $B_1,\dots,B_6$ divided up the possible outcomes into equally likely disjoint events, then we could add probabilities to say that

$$P(A)=\sum_{i=1}^6P(A\mid B_i)P(B_i)=P(A\mid B_1)\sum_{i=1}^6P(B_i)=P(A\mid B_1)\cdot1=P(A\mid B_1)\;,$$

since we know from (b) that the probabilities $P(A\mid B_i)$ are all equal. However, the events $B_1,\dots,B_6$ are not disjoint. For instance, the roll $\langle 2,3\rangle$ is in both $B_2$ and $B_3$.

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We write a more or less formal proof for (b), though the result is almost obvious. Let us calculate $\Pr(A|B_1)$. By the usual formula, we have $$\Pr(A|B_1)=\frac{\Pr(A\cap B_1}{\Pr(B_1)}.$$

Now calculate the probabilities on the right. The probability that at least one $1$ is thrown is $1$ minus the probability that both throws are different from $1$. This is $1-(5/6)^2$.

Now we find $\Pr(A\cap B_1)$. This event happens if the red die has a $1$ and the green has a $2$, or the other way around. Thus $\Pr(A\cap B_1)=2/36$. Now we can compute $\Pr(A|B_1)$ by dividing.

Next we calculate $\Pr(A|B_2)$. We have $$\Pr(A|B_2)=\frac{\Pr(A\cap B_2}{\Pr(B_2)}.$$ It is clear that $\Pr(B_2)=1-(5/6)^2$, by the same argument as the earlier one for $B_1$. And the event $A\cap B_2)$ happens if we have a $2$ on the green and a $5$ on the red, or the other way around. So $\Pr(A\cap B_2)=2/36$. These are the same numbers as the ones we used to compute $\Pr(A|B_1)$. It follows that $\Pr(A|B_2)=\Pr(A|B_1)$.

Essentially identical arguments show that $\Pr(A|B_i)=\Pr(A|B_1)$ for all $i$.

If we were doing a similar calculation for numbers other than $7$, we would not get that all conditional probabilities are equal. For example, the probability that the sum is $4$, given that at least one throw is $6$, is clearly $0$. What's special about $7$ is that all the possible dice numbers $1$ to $6$ can be used to make $7$.

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