# Conditional probability question and solution

Please could someone review my solutions for the problems below..thanks in advance

An e-mail message can travel through one of three server routes. The probability of transmission of error in each of the servers and the proportion of messages that travel each route are shown in the following table. Assume that the servers are independent.

                  % messages % errors
Server 1     40         1%
Server 2     25         2%
Server 3     35         1.5%


1) What is the probability of receiving an email containing an error? Solution: this would be .4*.01 + .25*.02 + .35*0.15 = 0.615

2) What is the probability a msg will arrive without error? Solution: .4*.99 + .25*.98 +.35*.95= .9735

3) If a msg arrives without an error, what is the probability that it was sent through server 2? Solution: Let event E = Sent through server 2. Let event F = arrives without an error. We are looking for P(E/F) or the conditional probability. We can use the formula $\frac {P(E \cap F) }{P(F)}$. P(EnF) = .25*.98 = .245. While the P(F) = 1 - (.01) - (.02) - (.15) = .92. Therefore P(E\F) = .245/.92 = .266 or 26.6%

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## 1 Answer

The procedure used in 1) was correct. Unfortunately, there was a numerical slip. The answer is $$(0.4)(0.01)+(0.25)(0.02)+(0.35)(0.015).$$ The slip was writing that $1.5\%$ is $0.15$.

One should always glance at the answer one gets to check for plausibility: surely these "good" servers cannot produce an error with probability $0.615$!

2) Again, the setup was right, though not optimal. There was a slip: $100\%$ minus $1.5\%$ is not $95\%$.

A more efficient way to solve 2) is to note that the event in 2) is the complement of the event in 1). So to find the answer for 2), it is simplest to find $1-a$, where $a$ is the answer to 1).

3) The procedure used began along correct lines: the conditional probability setup is good. However, $\Pr(F)$ was not computed correctly. The right number is the answer to 2).

There were several problems with the computation of $\Pr(F)$: (i) The wrong idea was used; (ii) There was the error of 1), writing $0.15$ for $1.5\%$; (iii) Even if we assume the expression is right, and $0.15$ is right, the subtraction is incorrect.

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Thanks @andre Nicolas. You mention in 3) that the wrong formula was used. Is this not a conditional probability problem or is it just my interpretation of the formula that is wrong? – bosra Dec 3 '12 at 20:10
I am sorry for possible ambiguity. Have changed "formula" to "idea." Your conditional probability setup was good. It is in the calculation of $\Pr(F)$ that you used, implicitly, an incorrect formula/method. The probability of arrival without error has nothing to do with $1-0.01-0.02-0.015$. – André Nicolas Dec 3 '12 at 20:26
OK, thanks again @Andre Nicolas. I really appreciate your comments and help – bosra Dec 3 '12 at 20:41
You are welcome. Maybe in other questions you ask, numerical stuff could be checked more carefully, so that one can concentrate on mathematical issues. – André Nicolas Dec 3 '12 at 20:51