Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|cite|improve this question
Your table seems off. Do you want $P(r=1|s=0)=.1$? – David Mitra Jan 17 '13 at 16:09

1 Answer 1

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.