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 Post subject: ProbabilityPosted: Tue, 6 Dec 2011 22:52:38 UTC
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Joined: Tue, 6 Dec 2011 22:37:32 UTC
Posts: 1
Hey
This is a fairly easy task and my answer dont seem to be correct according to the answersheet.

Let A and B be two events from a random experiment such that P(A)=0.4 and P(AUB)=0.7

Compute P(B) and P(AnB)
You know that A and B are independent events.

This is my solution:

Since A and B are independent we know that: P(AnB)=P(A)P(B)

P(AUB) = P(A) + P(B) - P(AnB) = P(A) + P(B) - P(A)P(B)
P(AUB) = P(A) + P(B) - P(A)P(B)
0.7 = 0.4 + P(B) - 0.4*P(B)
0.3 = P(B)*(1 - 0.4)
P(B) = 0.5

=> P(AnB) = 0.4*0.5 = 0.2

However in the answer it says that P(B) = 0.6 and P(AnB) = 0.3.
Am I just stupid and missed something or is it wrong in the answer?

Cheers.

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 Post subject: Re: ProbabilityPosted: Tue, 6 Dec 2011 22:54:19 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12172
Location: Austin, TX
Vandeley wrote:
Hey
This is a fairly easy task and my answer dont seem to be correct according to the answersheet.

Let A and B be two events from a random experiment such that P(A)=0.4 and P(AUB)=0.7

Compute P(B) and P(AnB)
You know that A and B are independent events.

This is my solution:

Since A and B are independent we know that: P(AnB)=P(A)P(B)

P(AUB) = P(A) + P(B) - P(AnB) = P(A) + P(B) - P(A)P(B)
P(AUB) = P(A) + P(B) - P(A)P(B)
0.7 = 0.4 + P(B) - 0.4*P(B)
0.3 = P(B)*(1 - 0.4)
P(B) = 0.5

=> P(AnB) = 0.4*0.5 = 0.2

However in the answer it says that P(B) = 0.6 and P(AnB) = 0.3.
Am I just stupid and missed something or is it wrong in the answer?

Cheers.

What you have looks right to me.

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 Post subject: Re: ProbabilityPosted: Thu, 8 Dec 2011 01:18:15 UTC
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Joined: Fri, 1 Jul 2011 01:17:26 UTC
Posts: 329
A simple check shows that the given answer is wrong. P(A)P(B)=.24, while the given answer has .3 for P(A∩B).

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