Results 1 to 3 of 3

Math Help - Mutually/ NonMutually Exclusive events

  1. #1
    Member
    Joined
    Jul 2010
    Posts
    85

    Mutually/ NonMutually Exclusive events

    I'm still confuse about these events. Also 180 + 160 is more than 300. Where have I made the mistake?

    The owners of a music store placed ads in the newspaper and on TV advertising their annual 40% sale. They did a survey of the first 300 customers, and discovered that 180 saw the ad on tv, 160 saw the ad in the newspaper and 75 saw the on TV and in the newspaper. What is the probability that a randomly select customer

    a. saw the advertisement in the newspaper and on Tv?

    75/100
    b. saw the advertisement in the newspaper or on TV?
    P(AandB) = P(A) + P(B) = 160/300 + 180/300 - (it is not a probability!!)
    c. did not see the ad?
    P(AorB) = P(A) + P(B) - P(AandB)= 160/300 + 180/300-??

    Please help
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor harish21's Avatar
    Joined
    Feb 2010
    From
    Dirty South
    Posts
    1,036
    Thanks
    10
    You seem to have messed up everything here.. First of all, your questions tells you that 300 people were interviewed. Then what made you write that the people saw the advertisement in the newspaper and on Tv is 75/100???

    also you seem to be unclear about union, intersection, and complement of sets..

    (a) find P(A \cap B) the denominator you have written for (a) is wrong!!

    (b) find P(A \cup B)

    (c) Do you know what complement of a set means? See the defintion in your book/notes
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,719
    Thanks
    634
    Hello, terminator!

    The owners of a music store placed ads in the newspaper and on TV
    advertising their annual sale.

    They did a survey of the first 300 customers and discovered that:
    180 saw the ad on TV,
    160 saw the ad in the paper,
    75 saw the on TV and in the paper.

    \begin{array}{ccc}P(\text{TV}) &=& \dfrac{180}{300} \\ \\[-2mm]<br />
P(\text{news}) &=& \dfrac{160}{300} \\ \\[-3mm]<br />
P(\text{TV} \wedge \text{news}) &=& \dfrac{75}{300} \end{array}




    What is the probability that a randomly selected customer

    a. saw the advertisement in TV and in the paper?

    P(\text{TV} \wedge\text{paper}) \;=\;\dfrac{75}{300} \;=\;\dfrac{1}{4}




    b. saw the advertisement in TV or in the paper?

    P(\text{TV} \vee\text{paper}) \;=\;P(\text{TV}) + P(\text{paper}) - P(\text{TV} \wedge\text{paper})

    . . . . . . . . . . . . =\;\dfrac{180}{300} + \dfrac{160}{300} - \dfrac{75}{300} \;=\;\dfrac{265}{300} \;=\;\dfrac{53}{60}




    c. did not see the ad?

    This is the opposite of part (b).

    P(\text{not TV} \wedge \text{not paper}) \;=\;1 - \dfrac{53}{60} \;=\;\dfrac{7}{60}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Mutually Exclusive Events
    Posted in the Advanced Statistics Forum
    Replies: 6
    Last Post: March 10th 2012, 02:20 PM
  2. Non-mutually/mutually exclusive events
    Posted in the Statistics Forum
    Replies: 1
    Last Post: September 29th 2010, 09:37 AM
  3. mutually exclusive events
    Posted in the Statistics Forum
    Replies: 3
    Last Post: July 24th 2010, 12:30 PM
  4. Mutually Exclusive Events
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: January 19th 2009, 07:10 PM
  5. Mutually exclusive events? need help for sure
    Posted in the Statistics Forum
    Replies: 4
    Last Post: August 8th 2006, 09:03 PM

Search Tags


/mathhelpforum @mathhelpforum