1. ## Conditional probability problem

Another problem I am working on:

A survey of consumers in a particular community showed that 10% were dissatisfied with plumbing jobs done in their homes. Half the complaints dealt with plumber A. If plumber A does 40% of the plumbing jobs in the
town, find the following probabilities:

(a) A consumer will obtain an unsatisfactory plumbing job, given that the plumber was A.
(b) Consumer will obtain a satisfactory plumbing job, given that the plumber was A.

Solution

a)

I defined 2 events A and B
A: obtain unsatisfactory plumbing
B: the plumber is A

I am not sure if I find the right P(A and B) for the conditional prob.
Thats what I got:
P(A)= 40% or 0.4 P(B)=10% or 0.1

P(A|B)= 0.05/0.4 = 12.5%

b)
100-12.5=87.5%

Am I on the right track? Thanks

2. Originally Posted by somestudent2
Another problem I am working on:

A survey of consumers in a particular community showed that 10% were dissatisfied with plumbing jobs done in their homes. Half the complaints dealt with plumber A. If plumber A does 40% of the plumbing jobs in the
town, find the following probabilities:

(a) A consumer will obtain an unsatisfactory plumbing job, given that the plumber was A.
(b) Consumer will obtain a satisfactory plumbing job, given that the plumber was A.

Solution

a)

I defined 2 events A and B
A: obtain unsatisfactory plumbing
B: the plumber is A

I am not sure if I find the right P(A and B) for the conditional prob.
Thats what I got:
P(A)= 40% or 0.4 P(B)=10% or 0.1

P(A|B)= 0.05/0.4 = 12.5%

b)
100-12.5=87.5%

Am I on the right track? Thanks
I always find it easiest to try and show these things visually using, say, a karnaugh table. I'll let A represent plumber A (that makes more sense to me than letting A represent something else and letting B represent plumber A!!)

I start off with: $\, \begin{tabular}{l | c | c | c} & A & A\, ' & \\ \hline Dissatisfied & a & b & a + b \\ \hline Satisfied & c & d & c + d \\ \hline & a + c & b + d & 1 \\ \end{tabular}$

From the given data it's known that:

a + c = 0.4 and therefore b + d = 0.6.

a + b = 0.1 and therefore c + d = 0.9

a = 0.05 (half of the complaints were about A. Half of 0.1 = 0.05).

a + b = 0.1 therefore 0.5 + b = 0.1 therefore b = 0.5.

Update the table: $\, \begin{tabular}{l | c | c | c} & A & A\, ' & \\ \hline Dissatisfied & 0.05 & 0.05 & 0.1 \\ \hline Satisfied & c & d & 0.9 \\ \hline & 0.4 & 0.6 & 1 \\ \end{tabular}$

It's not hard now to add in the other values:

0.05 + c = 0.4 therefore c = 0.35.

0.05 + d = 0.6 therefore d = 0.55.

Update the table: $\, \begin{tabular}{l | c | c | c} & A & A\, ' & \\ \hline Dissatisfied & 0.05 & 0.05 & 0.1 \\ \hline Satisfied & 0.35 & 0.55 & 0.9 \\ \hline & 0.4 & 0.6 & 1 \\ \end{tabular}$

As a check, note that 0.35 + 0.55 adds up to 0.9, which it should.

(a) From the table: Pr(Dissatisfied|A) = 0.05/0.4 = 1/8.

(b) From the table: Pr(Satisfied|A) = 0.35/0.4 = 7/8.

Counter-intuitive results, perhaps ......

Find the percentage of consumers that will obtain:

(a) an unsatisfactory plumbing job, given that the plumber was A.
(b) a satisfactory plumbing job, given that the plumber was A.

But you were NOT asked to find percentages, you were asked to find probability..... A probability is always a number between 0 and 1 inclusive ........ Capisce?

Here's another one for you: Find the probability that plumber A was NOT used, given that the consumer was satisfied.

3. Thank You very much for a detailed explaination. I did not know about Karnaugh tables before now.

Here's another one for you: Find the probability that plumber A was NOT used, given that the consumer was satisfied.
I guess from your Karnaugh's table it would be 0.55/0.6 = 0.92

4. Originally Posted by somestudent2
Thank You very much for a detailed explaination. Mr F says: You're welcome. And thankyou for your thanks - lovely manners.

I did not know about Karnaugh tables before now.

I guess from your Karnaugh's table it would be 0.55/0.6 = 0.92
Not quite.

Pr(A'|satisfied) = 0.55/0.9 = 11/18.

The given condition is 'customer satisfied', so you divide by the total 'customer satisfied' probability (the end number of the 'satisfied' row), that is, 0.9.