# urgent

• November 8th 2007, 05:14 AM
MS BATOOL
urgent
• The personnel department of a company has records which show the following analysis of its 200 engineers.

Age
Bachelor Degree
Master Degree
Total
Under 30
90
10
100
30 to 40
20
30
50
Over 40
40
10
20
Total
150
50
200

If one engineer is selected at random from the company, find

1. The probability that he has only a bachelor’s degree.
2. The probability that he has a master’s degree, given that he is over 40.
The probability that he is under 30, given that he has only a bache
• November 9th 2007, 05:23 AM
Soroban
Hello, Ms Batool!

Quote:

The personnel department of a company has records
which show the following analysis of its 200 engineers.

$\begin{array}{cccccccc}\text{Age} &|& \text{BS} &|& \text{MS} &|& \text{Total} & | \\ \hline \text{under 30} &|& 90 &|& 10 &|& 100 &| \\
\text{ 30-40} &|& 20 &|& 30 &|& 50 &| \\ \text{over 40} &|& 40 &|& 10 &|& 20 &| \\ \hline \text{Total} &|& 150 &|& 50 &|& 200 &| \end{array}$

If one engineer is selected at random from the company, find:

1. The probability that he has only a BS.
2. The probability that he has an MS, given that he is over 40.
3. The probability that he is under 30, given that he has only a BS.

$1.\;P(\text{BS}) \;=\;\frac{150}{200} \;=\;\frac{3}{4}$

$2.\;P(\text{ MS }|\text{ over 40}) \;=\;\frac{P(\text{MS }\wedge\text{ over 40})}{P(\text{over 40})} \;=\;\dfrac{\frac{10}{200}}{\frac{20}{200}} \;=\;\frac{1}{2}$

$3.\;P(\text{under 30 }|\text{ BS}) \;=\;\frac{P(\text{under 30} \wedge\text{ BS})}{P(\text{BS})} \;=\;\dfrac{\frac{90}{200}}{\frac{150}{200}} \;=\;\frac{3}{5}$

• November 9th 2007, 06:41 AM
MS BATOOL
Hello Sir,
I want to thank you for your help........thanks again for this usefull reply.