Independent events in probabilities

  • #1

Homework Statement


Let S be the sample space for rolling a single die. Let A={1,2,3,4}, B={2,3,4}, and C={3,4,5}. Which of the pairs (A,B), (A,C), and (B,C) is independent?


Homework Equations


P(A|B)=P(A)
P(A|B)=P(A&B)/P(B)
P(A&B)=P(A)*P(B)

The Attempt at a Solution


P(A)=2/3 P(B)=1/2 P(C)=1/2
P(A&B)=P(A)*P(B)=(2/3)(1/2)=1/3 P(A&B)/P(B)=(1/3)/(1/2)=1/6
P(A&C)=1/3 P(A&C)/P(C)=(1/3)/(1/2)=1/6
P(B&C)=1/4 P(B&C)/P(C)=(1/4)/(1/2)=1/8

So based on my calculations, there is none of the pairs which match the independence rule. But the book says that (A,C) is independent.
 

Answers and Replies

  • #2
vela
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Homework Statement


Let S be the sample space for rolling a single die. Let A={1,2,3,4}, B={2,3,4}, and C={3,4,5}. Which of the pairs (A,B), (A,C), and (B,C) is independent?


Homework Equations


P(A|B)=P(A)
P(A|B)=P(A&B)/P(B)
P(A&B)=P(A)*P(B)
The first and third equations only hold when A and B are independent. The second equation holds generally.

The Attempt at a Solution


P(A)=2/3 P(B)=1/2 P(C)=1/2
P(A&B)=P(A)*P(B)=(2/3)(1/2)=1/3 P(A&B)/P(B)=(1/3)/(1/2)=1/6
P(A&C)=1/3 P(A&C)/P(C)=(1/3)/(1/2)=1/6
P(B&C)=1/4 P(B&C)/P(C)=(1/4)/(1/2)=1/8

So based on my calculations, there is none of the pairs which match the independence rule. But the book says that (A,C) is independent.
You can't ignore the outcomes included in each set. For example, if events A and B are both true, that means that you rolled a 2, 3, or 4, so P(A&B)=1/2.
 

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