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5 straws of equal length and 1 shorter straw. The probability of the second person drawing the short straw,$P(B)$ would seem to be $1/5$ since if the first person drew it there would be no second draw. If the first person didn't draw it, then there are 5 straws remaining, 1 of which is short. Thus, $P(B)=1/5$.

If I define the probabilities as follows: P(A) = the probability of the 1st person drawing the short straw, then $P(A)=1/6$. P(not A) = $5/6$.

$P(B)$ = probability of the second person drawing the short straw. Now, I use my formula: $P(not A \cap B)=P(not A)P(B|not A)=(5/6)(1/5)=1/6$

It would seem to me that $P(notA\cap B)$ would be the same as $P(B)$ if $P(B)$ is defined as the probability of the second person drawing the short straw. Are these two things, $P(B)$ and $P(not A \cap B)$ different?

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    Your first calculation is off. Consider the straws in order. The short straw is in one of the slots with equal probability for each....hence $P(B)=\frac 16$. this is also the probability that the third, fourth, fifth, and sixth person gets it. – 2017-01-14

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Your first computation of $P(B)$ is flawed. What you described in the first paragraph is indeed $P(B| \text{not } A)=\frac15$.

\begin{align} P(B)&=P(B|A)P(A)+P(B|\text{not } A)P(\text{not } A)\\ &= 0+ \frac15 \left(\frac56 \right) \end{align}

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    Ah. I see this now. The probability of any of the people drawing the short straw is 1/6....even the last person. However, if you change the question to be what is the probability of drawing the short straw given the other 5 have not drawn it then the prob would be 1. It is more of a terminology thing. – 2017-01-14
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    I understand the equations for $P(A \cup B)$ and for $P(A \cap B)$. Can you direct me to where I can understand the equation you wrote? – 2017-01-14
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    It is known as the law of total probability. https://en.wikipedia.org/wiki/Law_of_total_probability – 2017-01-14