# Cell Phones and Day Packs

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.