Statistics for CS Problem

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  • #1
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Homework Statement



Suppose P(A) = .6, P([tex]\overline{A}[/tex] | B) = .4. Check whether events A and B are independent.

Homework Equations



Two events A and B are said to be independent if P(A|B) = P(A). This is equivalent to stating that P(A [tex]\cap[/tex] B) = P(A)P(B)

If A and B are any two events, then the conditional probability of A given B, denoted by P(A | B), is P(A | B) = [tex]\frac{P(A \cap B)}{P(B)}[/tex] provided P(B) > 0.

The Attempt at a Solution



I know that P([tex]\overline{A}[/tex]) = 1 - P(A) = .4

However, I'm not sure how to use the information given to check for independence. The professor says the solution should be very brief, but it's not coming to me.

Thinking on it a little more and looking it up online, it would seem that if P(A|B) = P(A) means A and B are independent, but P([tex]\overline{A}[/tex]) != P(A), then A and B are dependent. But I'm not sure if this logic is correct.
 
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Answers and Replies

  • #2
statdad
Homework Helper
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You have P(A-complement | B) - it should be very simple to obtain P(A | B) from it.

Note: couldn't get the \bar{A} construct to work in Latex.
 
  • #3
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But that's my problem, how do I use that information to get P(A | B)?
 
  • #4
statdad
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But that's my problem, how do I use that information to get P(A | B)?

What do you get when you add P(A) to P(A-complement)? the same result works with P(A |B) and P(A-complement | B).
 
  • #5
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I get 1. So P(A | B) + P(A-complement | B) = P(A|B) + .4 = 1... P(A | B) = .6?
 

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