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Suppose we have the following two events,

$A$, which denotes "a number less than $5$ is rolled" and,

$B$, which denotes "a number of at least $4$ is rolled".

Suppose then in our trial we roll a die once.

What is the probability of getting at most $4$ and a number of at least 4 when you are guaranteed that a number less than $5$ will be rolled.

I know this sounds like a relatively straight forward probability question but I'm not sure I'm understanding it.

We want to know $Pr(AB)$. An alternative rewording of the question would be "what is the probability of getting exactly $4$", and we know this should be $\dfrac{1}{6}$. In this instance we cannot take the product of their probabilities since one is dependent on the other, so we have that $Pr(AB)$ is given by $Pr(B|A)P(B)$.

We know $P(B) = 1$ since we are guaranteed it will happen. Except for $Pr(B|A)$ I'm a little confused, isn't it $\dfrac{1}{4}$ since we've rolled a number less than $5$, not $\dfrac{1}{6}$? Therefore $Pr(B|A)P(B) = \dfrac{1}{4}$. Where have I gone wrong in reasoning?

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The probability of AB is $P(B|A)P(A)$, not $P(B|A)P(B)$. When you write $P(B|A)P(A)$ you read it as probability of B given A times the probability of A. $P(A)=\frac 23$ because the die can be $1,2,3,4$. Then $P(B|A)=\frac 14$ because there are four cases where A is true and one of the four makes B true. The product is $\frac 16$ as it should be.

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  • $\begingroup$ Oh, wow. I didn't even notice that stuff up with conditionals. Thanks @Ross Millikan! $\endgroup$ – Retty Mar 21 '17 at 23:36

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