• July 26th 2009, 11:20 AM
mj.alawami
Question:
The table gives the classification of all the employees of a company by gender and college degree.If one of these employees is selected at random for a member ship on the employee management committee ,determine the following.

a)P(F intersection G)
(Table)

Female-------3---------7--------------10
male----------6--------14--------------20
----------------9---------21-------------30

Attempt:
10/30I*3/30=1/30

Am I correct?
Thank you
• July 26th 2009, 12:23 PM
Plato
There are three female graduates out of thirty employees.
• July 26th 2009, 12:47 PM
mj.alawami
Quote:

Originally Posted by Plato
There are three female graduates out of thirty employees.

Since they are joint probability we have to use the multiplication rule:
P(F intersection G)=P(F) * P(G/F)

P(F) = 10/30
P(G/F)=3/30
10/30 * 3/30 =1/30
• July 26th 2009, 12:57 PM
Plato
You are asked to find $P(G \cap F)$.
But that is $P(G \cap F)=P(G|F)P(F)=\left(\frac{3}{10}\right)\left(\frac {10}{30}\right)=\left(\frac{3}{30}\right)$.

But also $P(F \cap G)=P(F|G)P(G)=\left(\frac{3}{9}\right)\left(\frac{ 9}{30}\right)=\left(\frac{3}{30}\right)$.
• July 26th 2009, 01:31 PM
mj.alawami
if $P(M \cap G)$ was asked then

$P(G) * P(m/g)$

$9/30 * 6/9 =1/5$

Am i correct?
• July 26th 2009, 01:39 PM
Plato
Yes it is correct.
Out of thirty, how many male graduates are there?
Simply look at the table. It is quite clear.