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suppose you roll a fair die and $B$ is the event that the roll is even, while $A$ is the event the roll is a '$6$'. Clearly $A\subset B$ Thus $P(B\mid A)=1$

How is $P(B\mid A)=1$ calculated here?

I have that:

$$P(A)=\frac16 \text{ and } P(B)=\frac36$$

So I get

$$P(B\mid A)= \frac{P(B \cap A)}{P(A)} = \frac{P(B)P(A)}{P(A)} = \frac{\frac12\cdot \frac16}{\frac16}\neq 1$$

Where am I going wrong?

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Your mistake is assuming that $P(B \cap A) = P(A)P(B)$ which is only true when $A$ and $B$ are independent.

Since $A \subseteq B$, can you simplify $A \cap B$?

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Intuitively speaking, since we know $6$ is an even number, the "probability" of the roll being an even knowing that the roll is a $6$ (Which is obviously even and part of the set of even outcomes,) would be $1$.

Hope this helps.

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  • $\begingroup$ I understand the intuition behind it, I was mistaken in the math arriving at it. $\endgroup$ – EvaD Oct 29 '17 at 23:20

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