Back to Probability Theory

### Problem

- Among 1,000 students 400 have cell phones (C), 300 daypacks (D), and 350 neither. What are the following probabilities for a randomly selected student:

- P(C&D)
- P(~C&~D)
- P(C)
- P(D)

- P(CvD)
- P(~C)
- P(D|C)
- P(C|D)

### Solution

**Probability Table**

- The probabilities can be calculated from the probabilities of the true-false combinations of C and D:
- C and D true
- C and D false
- C true and D false
- C false and D true.

- Of 1,000 students we know 350 are neither C nor D, as shown in this
*Probability Table*(T means*true*, F means*false*):

- Since the sum of C’s (400), D’s (300) and neither-C-nor-D’s (350) exceeds the total students (1,000), the overage (50) equals the students that are both C and D. The number of C’s not D is therefore 400 – 50 = 350; and the number of D’s not C is 300 – 50 = 250. Thus:

**Venn Diagram**

- The probabilities can also be represented by a Venn Diagram:

- All but two probabilities are calculated as follows:
- P(C&D) = 5/100, represented by the first row of the table; also by the region common to the C and D circles in the Venn diagram.
- P(~C&~D) = 35/100, represented by the last row of the table; also by the region outside the circles.
- P(C) = 40/100, represented by the first two rows of the table; also by the two regions of the C circle. That is, P(C) = P(C&~D) + P(C&D) = 35/100 + 5/100 = 40/100.
- P(D) = 30/100, represented by the first and third rows of the table; also by the two regions of the D circle. That is, P(D) = P(C&D) + P(~C&D) = 5/100 + 25/100 = 30/100.
- P(CvD) = 65/100, represented by the first three rows of the table; also by the three regions inside the circles. That is, P(CvD) = P(C&~D) + P(C&D) + P(~C&D) = 35/100 + 5/100 + 25/100 = 65/100.
- P(~C) is 60/100, represented by the last two rows of the table; also by the regions outside the C circle. That is, P(~C) = P(~C&D) + P(~C&~D) = 25/100 + 35/100 = 60/100.

- For P(D|C), if a student is among the 400 C’s, the probability she’s a D, of which there are 50, is 50/400 or 5/40. The
*General Conjunction Rule*produces the same result:**P(D&C) = P(D|C) x P(C).**- Dividing the sides of the equation by P(C) and rearranging:
**P(D|C) = P(D&C) / P(C)**

- P(D|C) thus equals (5/100) / (40/100), i.e. 5/40.

- P(C|D) is calculated similarly. If a randomly selected student is one of the 300 D’s, the probability she’s among the 50 C’s is 50/300 or 5/30.
- Here’s how to look at this. If a student is a D, only what’s inside the D circle is relevant to calculating P(C|D); nothing outside matters. Given that P(C&D) = 5/100 and P(~C&D) = 25/100, the contents of D, probability-wise, are 5 parts C and 25 parts not-C. Thus P(C|D) is 5 parts of 30.