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I have this exercise that I'm having trouble with, specifically on the conditional probability and intersection use. It goes like this:

A binary communication channel, constituted by a transmitter and a receiver, only transmits 0 or 1.

-The probability of a 1 being transmitted is 0.4

-The probability that a transmitted 0 is correctly received is equal to 0.9

-The probability that a transmitted 1 is correctly received is equal to 0.95

Determine the probability of an error.

To me, what makes sense is using the conditional, since you know that in order for an error to occur you have to either have transmitted one of the two possible signal and received the opposite. However, the solution uses the intersection probability to get the result.

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  • $\begingroup$ What is the problem? $\endgroup$
    – kludg
    Jun 9, 2017 at 18:31
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    $\begingroup$ Conditional probability sounds right: $P(error) = P(error | 1)P(1) + P(error|0)P(0)$. Can you square this with what your solution says? $\endgroup$
    – Jeff
    Jun 9, 2017 at 18:35
  • $\begingroup$ Instead of describing your work and the solution, please provide them here (in every post). Use MathJax. Formatting tips here. $\endgroup$
    – Em.
    Jun 9, 2017 at 18:41

2 Answers 2

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$P(E) = 1 - P(C)$

$P(C) = (.4)(.95) + (.6)(.9)$

So:

$P(E) = 1 - (.38 + .54) = 0.08$

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The conditional probability and the probability of the intersection are connected as

$P(A\cap B)=P(A)\cdot P(B|A)=P(B)\cdot P(A|B)$

In your case let $X=1$ be the event that the transmitter sent a 1.

$X=0$ is the event that transmitter sent a $0$.

$Y=0:$ received number is $0$.

$Y=1:$ received number is $1$

The given probabilities are $P(X=1)=0.4, P(Y=0|X=0)=0.9, P(Y=1|X=1)=0.95$

And it is asked for $P(X=1\cap Y=0)+P(X=0\cap Y=1)$

$P(X=1\cap Y=0)=P(X=1)\cdot P(Y=0|X=1)$, with $P(Y=0|X=1)=1-P(Y=1|X=1)$

$P(X=0\cap Y=1)=P(X=0)\cdot P(Y=1|X=0)=(1-P(X=1))\cdot P(Y=1|X=0)$, with $P(Y=1|X=0)=1-P(Y=0|X=0)$

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