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A student come into a classroom where a teaching assistant and professor are helping students, then the student will leave once their question has been answered. $20\%$ of students saw only the teaching assistant, then leave the building. $40\%$ of students saw the professor, then leave the building. $40\%$ saw the teaching assistant, then see the professor, then finally leave the building. Answer these three questions:

1.Total probability that student seek teacher assistance.

My attempt: I just needed to solve total probability that student seek the teacher assistance. .40+.20 = .60. Then everything else is

-solved- the comments help me evaluated the problems.

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  • $\begingroup$ Given your definition of the events $A$ and $B,$ both $P(A)$ and $P(B)$ should be equal to $0.6$ rather than $0.2$ and $0.4$ respectively. $\endgroup$ – jvdhooft Jun 20 '18 at 20:47
  • $\begingroup$ "What is the probability that the student saw the teacher assistance then left?" This is ambiguous. It could mean the probability of the saw the teacher assistant and then immediately left and didn't see the professor (20%). Or it could mean the probability they saw the the assistant and then some time later they left but the may or may not have seen the professor as well. (60%). Which one is meant is not a matter of math but of language. ... IMO as the problem used the same language both the setup and the answer should make the same assumptions. $\endgroup$ – fleablood Jun 20 '18 at 21:33
  • $\begingroup$ the question is more about sequence of events. Let says 20 student come for help. 11 student can seek Teacher assistance, then leave the building. 5 student can seek teacher assistance, then seek the professor. 4 student can seek professor then leave. 0 student went to doctor then teacher assistance. The probability of each are different. Once a student is at teacher assistance, he can $\endgroup$ – Nakashiro Jun 20 '18 at 21:39
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I think the easiest way to do this is simply:

$A:$ Saw only the teaching assistant. $20\%$. $P(A) = .2$.

$B:$ Saw only the professor and then left. $40\%$. $P(B) = .4$

And $C:$ Saw the assistant then the professor. $40\%$. $P(C) = .4$.

These events are mutually exclusive.

So "What is the total probability that the student saw the teacher assistance?" = $A \cup C$ and $P(A\cup C) = P(A) + P(C) = .2 + .4 = .6$.

"What is the probability that the student saw the teacher assistance then left?" = $A$ so $P(A) = .2$

"What is the probability that the student asked for teacher assistance, then asked the professor for additionally help and finally the student leaves?" = $C$ and $P(C) = .4$.

....

If we want to do it as $A$ saw only the assistant $B$ as saw the professor only. Then we must have "Saw neither" = $\emptyset$. Saw both = $(A \cup B)^c$.

"What is the total probability that the student saw the teaching assistant?"= $B^c$ so $P(A) = 1- B$ is okay.

"What is the probability that the student saw the teaching assistant then left?"

$P(A) = .2$ is okay

What is the probability that the student asked for teaching assistant, then asked the professor for additional help, and then finally leaves?

$P(A\cap B) = P(\emptyset) = 0$ is not okay.

Instead $P((A\cup B)^c)=1 - (P(A) + P(B)) = 1 - (.2 + .4) = 1- .6 = .4$.

But why go to such trouble? Use what was given. You were told almost all of these.

That

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