## Friday, August 13, 2010

### Simple probability on cross-tab data problems.

I thought all of my ECON studes will be able to solve these problems but NO. Most of them expect canned packaged problems but are lead astray when they have to think more.

Consider the following frequency table of second year Engg students taking various subjects:

Subject\StudentMale Female
Calculus 525
History5 30
Physics15 6

1. What is the probability that a 2nd Year Engg. student chosen is taking Calculus or is a Male?

2. What is the probability that a 2nd Year Enng. student is a Female or is taking History?

3. What is the probability that 2nd Year Engg. student is a female taking History?

4. What is the probability that a 2nd Year Engg. student is not taking Caculus?

First form the full table with column and row sums.

Subject\StudentMale FemaleSum
Calculus 15 2540
History5 3035
Physics15 6 21
Total356196

1.Use the fact that $$P(A\cup B) = P(A) + P(B) - P(A\capB)$$

The answer is then $$P(Calulus \cup Male) = P(Calculus) + P(Male) - P(Calculus \cap Male)$$
or $$\frac{40 + 35 - 15}{96}= \frac{60}{96}= 0.625$$

2. Similar to #1,$$P(History \cup Female) = P(Calculus) + P(Female) - P(History \cap Female)$$
Solving, $$\frac{61+35-30}{96}=\frac{66}{96}=\frac{11}{96}= 0.114583$$.

3. This is the simplest problem. From the above table, it is $$\frac{30}{96} = 0.3125$$

4. The complement is $$1 - P(Calculus)$$. Therefore $$1- \frac{40}{96} = \frac{54}{96} = 0.5625$$