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Jeffrey works as a DJ at a local radio station.  On occasion, he chooses some of the...

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lbawa | Student, Undergraduate | (Level 1) Honors

Posted February 15, 2012 at 9:26 AM via web

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Jeffrey works as a DJ at a local radio station.  On occasion, he chooses some of the songs he will play based on the phone-in requests received by the

switchboard the previous day.  Jeffrey's list of 200 possible selections includes:

.  All the songs in the top 100

.  134 hard rock songs

.  80 phone-in requests

.  45 hard rock songs in the top 100

.  20 phone-in request in the top 100

.  24 phone-in requests for hard rock songs

How many phone-in requests were for hard rock songs in the top 100 and how many of the songs in the top 100 were neither phone-in requests nor hard rock selections?

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted February 15, 2012 at 9:22 PM (Answer #1)

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The DJ at a local radio station on occasion chooses some of the songs he will play based on the phone-in requests received by the switchboard the previous day.

The songs played by the DJ can be divided into 3 sets:

songs from the top 100 - set A

hard rock songs - set B

phone-in requests - set C

From the information given set A has 100 elements, set B has 134 elements and set C has 80 elements. There are 45 elements that belong to both A and B, 20 songs that belong to both A and C and 24 songs that belong to both B and C.

=> `P(AnnB)` = 45, `P(AnnC)` = 20 and `P(BnnC)` = 24

`P(AuuBuuC) = P(A) + P(B) + P(C) - P(AuuB) - P(BuuC) - P(AuuC) + P(AnnBnnC)`

=> 200 = 100 + 134 + 80 - 45 - 20 - 24 + `P(AnnBnnC)`

It is seen that `P(AnnBnnC)` is a negative number. This indicates an error in the information provided, either the total number of songs played is greater or the number of songs in each set smaller or the number of common elements between the sets is greater.

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted February 15, 2012 at 9:27 PM (Answer #2)

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Looks like the expression is not displayed correctly.

`P(AuuBuuC) = P(A) + P(B) + P(C) - P(AuuB) - P(BuuC) - P(AuuC) `

`+ P(AnnBnnC)`

=> `200 = 100 + 134 + 80 - 45 - 20 - 24 + P(AnnBnnC)`

=> `P(AnnBnnC) = -24`

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