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A system sends 1s and 0s in sequence with equal probability. Errors are made, so the probability of receiving a digit $r_j$ (equal to 0 or 1) given that a certain digit $s_i$ (equal to a 0 or 1) was sent as follows:

sent / received 0     1

0              .90   .1

1              .005  .995

Compute the conditional probability P($s_i =0| r_j=1$) that a digit 0 was sent given that a 1 was received?

My proposed solution is that this is in the Bayes theorem problem domain.

So the P(A/B) = $\frac {P(B/A).P(A)}{P(B)}$

$\equiv \frac{(.1 * .91)}{1.05}$ = .0087

Please could someone confirm if this is correct?

Thanks in advance

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Your table seems off. Do you want $P(r=1|s=0)=.1$? – David Mitra Jan 17 '13 at 16:09
up vote 2 down vote accepted

It seems off.

First note the upper right entry of your table, $P(r=1|s=0)$, seems off. I believe it should be $.1$ (if a $0$ is sent, then either a $0$ or a $1$ is recieved. So $P(r=1| s=0)+P(r=0|s=0)=1$).

So, using this value for $P(r=1|s=0)$:

You are using the correct formula, with $A$ being the event that the sent digit was $0$ and $B$ being the event that the recieved digit was $1$.

You calculated $P(B|A)=.1$ correctly; its value given in the (corrected) table.

But $P(A)$ is given as $1/2$; $P(A)$ is just the (unconditional) probability that a $0$ was sent.

You also calculated $P(B)$ incorrectly. $P(B)$ is the (unconditional) probability that a $1$ was received. It can be calculated as $$\eqalign{ P(B)&=P(B\cap A)+P(B\cap A^C)\cr &=P(B|A)P(A)+P(B|A^C)P(A^C)\cr &=.1\cdot\textstyle{1\over2}+.995\cdot{1\over2}.} $$

So $$ P(A|B)={P(B|A)P(A)\over P(B)}{.1\cdot{1\over2}\over.1\cdot{1\over2}+.995\cdot{1\over2} } ={.1\over .1+.995}\approx.09132. $$

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Thanks for the quick response David – bosra Jan 17 '13 at 16:12
@bosra You're welcome. – David Mitra Jan 17 '13 at 16:32

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