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Suppose we have two fair dice and rolled them.

Let

  • $A$ be the event "the sum of the two dice is equal to $3$";
  • $B$ be the event "the sum of the two dice is equal to $7$";
  • $C$ be the event "at least one dice shows $1$".

How to calculate $P(A \mid C)$?

In this case can we say that $A$ and $C$ are independent? Can we say that $B$ and $C$ are independent?

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  • $\begingroup$ I don't see how $A \cap C$ and $B \cap C$ can be independent, since they're mutually exclusive and each has non-zero probability. Did you mean something else? $\endgroup$ – Brian Tung Mar 13 '19 at 7:18
  • $\begingroup$ Dice is already plural. The singular is die. So it's one die, 2 dice, 3 dice... $\endgroup$ – Davor Mar 13 '19 at 11:01
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$P(C)$ is actually $\frac{11}{36}$$11$ of the $36$ possible rolls show at least one 1 (don't forget to consider the double-1 case!). $P(A\cap C)=\frac2{36}$ since only 1-2 and 2-1 have at least one 1 and sum to $3$. Thus $$P(A|C)=\frac{P(A\cap C)}{P(C)}=\frac{2/36}{11/36}=\frac2{11}$$

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  • $\begingroup$ Thank you. Can I say that $P(B|C)=\frac{2}{11}$? $\endgroup$ – Ali J. Mar 13 '19 at 6:44
  • $\begingroup$ @AliJ Yes, since $P(B\cap C)=P(A\cap C)$. $\endgroup$ – Parcly Taxel Mar 13 '19 at 6:48
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If one of the dices shows $1$ then there are only two ways to get $3.$ If both the dices show $1$ then there is no chance of getting $3.$ So the required probability is $\frac {2} {11}.$ Because when $C$ has already occurred then the reduced sample space has $11$ elements where at least one of the events is $1$ which are $$\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(3,1),(4,1),(5,1),(6,1) \}$$ and only two of them suit your purpose which are $$\{(1,2),(2,1) \}.$$

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  • $\begingroup$ You mean that $P(A|C)=\frac{1}{18}$? $\endgroup$ – Ali J. Mar 13 '19 at 6:16
  • $\begingroup$ May you please help about $P(B|C)$? $\endgroup$ – Ali J. Mar 13 '19 at 6:16
  • $\begingroup$ Same thing happens for $P(B \mid C).$ $\endgroup$ – Dbchatto67 Mar 13 '19 at 6:18
  • $\begingroup$ Can't we use the rule $P(A|C)=\frac{P(A.C)}{P(A)}$ $\endgroup$ – Ali J. Mar 13 '19 at 6:19
  • $\begingroup$ Try to think it on your own. If the event $C$ has already occured then in how many ways can the event $B$ occur? $\endgroup$ – Dbchatto67 Mar 13 '19 at 6:19

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