# Probability problem

• Jan 30th 2011, 06:21 PM
Glitch
Probability problem
The question:

Employment data at a large company reveals that 72% of the workers are married, that 44% are university graduates and that half of the university graduates are married. What is the probability that a randomly chosen worker...

a) is neither married nor a university graduate?
b) is married but not a university graduate?
c) is married or is a university graduate?

My attempt:
I tried to convert the wording into math as follows,

Let M = "people married"

$P(M) = 0.72$
$P(G) = 0.44$
$P(M \cap G) = P(G)/2 = 0.22$

a) We want to find $P(M^c \cap G^c)$, so:

$P(M^c) = 1 - P(M) = 0.28$
$P(G^c) = 1 - P(G) = 0.56$

Thus we get 0.28 x 0.56 = 0.1568

However, this is incorrect. I'm sure I've made a wrong assumption somewhere. I'm yet to attempt part b) and c) since I'm not sure where I've gone wrong with a). Any assistance would be great.
• Jan 30th 2011, 06:34 PM
dwsmith
Quote:

Originally Posted by Glitch
The question:

Employment data at a large company reveals that 72% of the workers are married, that 44% are university graduates and that half of the university graduates are married. What is the probability that a randomly chosen worker...

a) is neither married nor a university graduate?
b) is married but not a university graduate?
c) is married or is a university graduate?

My attempt:
I tried to convert the wording into math as follows,

Let M = "people married"

$P(M) = 0.72$
$P(G) = 0.44$
$P(M \cap G) = P(G)/2 = 0.22$

a) We want to find $P(M^c \cap G^c)$, so:

$P(M^c) = 1 - P(M) = 0.28$
$P(G^c) = 1 - P(G) = 0.56$

Thus we get 0.28 x 0.56 = 0.1568

However, this is incorrect. I'm sure I've made a wrong assumption somewhere. I'm yet to attempt part b) and c) since I'm not sure where I've gone wrong with a). Any assistance would be great.

Have you drawn a Venn Diagram?
• Jan 30th 2011, 06:35 PM
Glitch
No I haven't. Probably a good idea. Will report back.
• Jan 30th 2011, 07:11 PM
Soroban
Hello, Glitch!

Quote:

Employment data at a large company reveals that 72% of the workers are married,
44% are university graduates, and half of the university graduates are married.

What is the probability that a randomly chosen worker:

a) is neither married nor a university graduate?
b) is married but not a university graduate?
c) is married or is a university graduate?

Did you consider entering all this data into a chart?

. . $\begin{array}{c||c|c||c|}
& \text{Grad} & \text{non-G} & \text{Total} \\ \hline \hline
\text{Married} & 22\% & 50\% & 72\% \\ \hline
\text{Single} & 22\% & 6\% & 28\% \\ \hline \hline
\text{Total} & 44\% & 56\% & 100\% \\ \hline \end{array}$

• Jan 30th 2011, 07:16 PM
Glitch
Thanks guys. I used a Venn diagram and worked it out. Cheers.