1
$\begingroup$

I have a probability question dealing with conditionality. This one is a bus example: Li = # people on the bus leaving the ith stop Bi = # people that board the bus at the ith stop Bus arrives at stop 1 empty. P(Bi=0) = 0.5, P(Bi=1) = 0.4, P(Bi=2) = 0.1 Prob of a person leaving bus = 0.2

The question is "Someone tells you that as she got off the bus at the second stop, she saw that the bus was empty. Find the probability that she was the only passenger when the bus left first stop."

I have the solution but just don't understand it.

We are given that L2 = 0 and we are also given that L1 > 0. Find: P(L1 = 1 | L2 = 0 and L1 > 0)
This equates to: P(L1 = 1 and L2 = 0) / P(L2 = 0 and L1 > 0)

This is the part I don't understand - how does one arrive from the first expression to the second?

$\endgroup$
0
$\begingroup$

This is just the law of conditional probability \begin{equation} p(A,B,C) = p(A|B,C)p(B,C) \end{equation} In your example, the events are \begin{equation} A \equiv \{L1 =1\} , B \equiv \{L2 =0\}, \text{and } C \equiv \{L1 >0\} \end{equation} Observe that \begin{equation} p(A,B,C) = p(A,B) \end{equation} because \begin{equation} A \cap C = A \end{equation}

$\endgroup$
  • $\begingroup$ - Thanks for the reply. As you might tell, I'm new to the probability game. I will study up on the law of total probability, but on the face of it, I'm still confused how the numerator and denominator of "P(L1 = 1 and L2 = 0) / P(L2 = 0 and L1 > 0) " is derived. $\endgroup$ – RawlinsCross Sep 3 '17 at 23:51
  • $\begingroup$ $P(L1 = 1 \text{ and } L2 = 0) = P(A,B) = P(A,B,C)$, where the last equality due to the specific nature of the events in your example and $P(L1 = 1 \text{ and }L2 = 0) = P(B,C)$. By definition of conditional probabilities $p(A|B,C) = p(A,B)/p(B,C)$. Does this clarify anything? Forget about total probability; see this link on conditional probability: en.wikipedia.org/wiki/Probability#Conditional_probability $\endgroup$ – user144410 Sep 3 '17 at 23:57
  • $\begingroup$ Yes, I think it's clearing up. Maybe a good night's rest will do the trick. Thanks again. $\endgroup$ – RawlinsCross Sep 4 '17 at 0:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.