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I have seen the post here, but this doesn't seem to answer my problem. Say this as an example(from OpenIntro Stats Book page 112):

A Professor select a student at random to answer a question. Each student has an equal chance. There are 15 people in total. If the professor asks 3 questions, what is the probability that you will not be selected? Assume that she will not pick the same person twice in a given lecture.

The answer is 

14/15 * 13/14 * 12/13

Looking at the answer, this is clearly an independent events. But I don't understand how it is an independent event. The events in question are :

  • Probability of NOT getting picked for the first question
  • Probability of NOT getting picked for second question
  • Probability of NOT getting picked for third question

Definition of Independence is Knowing event 2, doesn't change the probability of event 1. E.g. 1 die 2 rolls. Knowing the first roll does not change the probability of the second roll.

In this case, the probability of the first event(not getting picked for the first question) is 14/15. However, because no same person will be picked twice, this CHANGES the probability of not getting picked for the second time.

So, given the information of the first event, the probability of second event (not getting picked for the second question) is no longer 14/15. It is now 13/14.

Independent law states that:

P(B|A) = P(B)

If P(A) denotes Probability of getting picked for the first question,
P(B) denotes probability getting picked for the second question, and P(B|A) denotes probability not getting picked for the second question given a person is already picked,

the equality condition is NOT satisfied. So, why is this being solved with independent formula??

More importantly, Why is this not addition operation instead?

user2773013
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1 Answers1

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Recall that $$P(B|A)=\frac{P(A\cap B)}{P(A)}$$

Now you want to compute the probability of not be selected for the first two questions (for simplicity).

If $A$= "you have not been selected for the first question" and $B$="you have not been selected for the first question", you want to compute $P(A\cap B)$. We know $P(A)=14/15$, what is $P(B|A)$?

Well, in the hypotesis that you didn't get selected for the first questions, there are still 14 possible students and you are one of them, so 13 "positive" possibilities.

In fact, as you said, these events are not independent.

In general, you use addition when you are computing the union, e.g. if you wanted $P(A\cup B)$ and $A$ and $B$ were disjoint you would have to add the respective probabilities. But they are not disjoint (we proved above that $P(B|A)\neq 0$). Consider that in this case, adding would give you a number which is greater then 1.

karmalu
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  • AHA!! It isnt using independent rule, but it's using General Multiplication rule. Your answer helps me understand the problem more.THANK YOU!. Clarification maybe? According to the formula I saw in text book: P(A and B) = P(A|B) * P(B). However, P(A and B) = P(B and A). Therefore, P(A and B) = P(B and A) = P(B|A) * P(A). I assume this is how you get to the formula you mentioned? Also, to think from operation perspective, adding will make probability bigger, which is not the case in this problem. Probability of not getting picked will become smaller each time. (hence multiplication) – user2773013 Jul 10 '19 at 18:04
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    Yes, to be fair I wrote $A\cap B$ by habit, but it would have been slightly better to write $B\cap A$ even though as you said, the probability is the same. – karmalu Jul 10 '19 at 18:09
  • THANK YOU AGAIN! – user2773013 Jul 10 '19 at 18:15