Can ANYONE fathom this statistics question?

Factual Questions
1

A stats question, exactly as it is worded. Is it just me or is something missing?

“On a given circuit some packets are sniffed. The probability of a given packet being TCP is P(T)=.25. The probability of a given packet being to or from a certain server is P(S)=.33. The probability of P(T or S)=.43. What is the probability of a packet being TCP GIVEN THAT it is to or from the server?”

That is the entire question.

Man, I thought I was good at this, but I’m baffled (and so is MRs. RickJay.) .43 bears no mathematical relationship to the other two probabilities, so they must be statistically dependent. But how? How does P(T or S) - which I believe means P(T or S) but not P(TS) - tell me the probability of one given the other?

If these are statistically dependent events, then the probability of T or S but not both is .415. If it’s actually .43, dependency must exist, so the probability of either T AND S or Neither T nor S, or both, must be less than if they were statistically independent. But to determine T IF S… don’t I need more information?

2

Sorry, the probability of T OR S but not both is .58.

3

No it’s not. Sorry; they are not mutually exclusive.

4

I’m taking your “no, it’s not” to mean that the value of the or is .43, not .58. In statistics (the way I learned it, anyway), (A or B) includes all cases of (A and B).
P(T or S) = .43. This is the total of P(S) + P(T) - P(S and T).

.43 = .33 + .25 - X. X is therefore .15.

P(T|S) = P(T and S) / P(S)

Y = .15 / .33 = .4545454545…

So the probability of T given S = .45.

5

Unless there’s something you’re not telling us, I think you’re incorrect assuming P(T or S) is an XOR. I worked it a different way than BraheSilver but got the same answer:



      |  S  |  !S | Total
----------------------
   T  | .15 |  .1 | .25
----------------------
  !T  | .18 | .57 | .75
----------------------
Total | .33 | .67

You’re given the Total row and Column (they’re P(S) and P(T)) and the sum of all but the bottom left entry. So, you know the bottom left P(!T) AND P(!S) is 1 - P(S OR T), so you can fill out the rest of the table with basic algebra. Then the question is P(T) given S, so that’s the left column, which you normalize to get the answer.

Hal

6

We overanalyzed it. You’re both right. Thanks.