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\Student	Male	Female
Calculus	5	25
History	5	30
Physics	15	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?

Answers.
First form the full table with column and row sums.

Subject\Student	Male	Female	Sum
Calculus	15	25	40
History	5	30	35
Physics	15	6	21
Total	35	61	96

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 $$1- \frac{40}{96} = \frac{54}{96} = 0.5625$$