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Probability question where an object is chosen randomly out of two objects

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    Dear friends,

    I am stuck at the following probability problem. Will appreciate your help.

    Two opera singers, Mario and Clarissa both perform on the same night, in separate recitals. The independent probabilities that two newspapers X and Y publish reviews of their recitals are given below:

    Probability of review in newspaper X
    ================================ ====
    Mario's recital - 1/2
    Clarissa's recital - 2/3

    Probability of review in newspaper Y
    ================================ ====
    Mario's recital - 1/4
    Clarissa's recital - 2/5

    Mario buys one of the newspapers at random. What is the probability that it has reviewed "both" recitals?


    I did this way: P(reviewed both recitals) = P(buys paper X)*P(X reviews Mario)*P(X reviews Clarissa) = (1/2)*(1/2)*(2/3) = 1/6

    But 1/6 is not the correct answer. Let me know where I am wrong and why.

    Thanks in advance.
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    (Original post by tantrik)
    Dear friends,

    I am stuck at the following probability problem. Will appreciate your help.

    Two opera singers, Mario and Clarissa both perform on the same night, in separate recitals. The independent probabilities that two newspapers X and Y publish reviews of their recitals are given below:

    Probability of review in newspaper X
    ================================ ====
    Mario's recital - 1/2
    Clarissa's recital - 2/3

    Probability of review in newspaper Y
    ================================ ====
    Mario's recital - 1/4
    Clarissa's recital - 2/5

    Mario buys one of the newspapers at random. What is the probability that it has reviewed "both" recitals?

    I did this way: P(reviewed both recitals) = P(buys paper X)*P(X reviews Mario)*P(X reviews Clarissa) = (1/2)*(1/2)*(2/3) = 1/6

    But 1/6 is not the correct answer. Let me know where I am wrong and why.

    Thanks in advance.
    You also need to account for the possibility that Y publishes both. Hence the total is:
    1/2*1/2*2/3 + 1/2*1/4*2/5 = 1/6 + 1/20 = 13/60.
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    (Original post by tantrik)

    But 1/6 is not the correct answer. Let me know where I am wrong and why.

    Thanks in advance.
    So you've worked out the probability if he buys paper X, what about paper Y? You don't seem to include that.

    You can either do it the way above this post, or a more straight forward way to think about it is this:

    For paper X, the probability of both reviews is \frac{1}{2}\cdot \frac{2}{3}= \frac{1}{3}

    For paper Y, the probability of both reviews is \frac{1}{4} \cdot \frac{2}{5}=\frac{1}{10}

    Overall probability of both reviews: \frac{1}{3}+\frac{1}{10}=\frac{1  3}{30}

    In a SINGLE given paper it would be half of this probability (ie probability of choosing a single paper):
    \frac{1}{2}(\frac{13}{30})=\frac  {13}{60}
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    (Original post by RDKGames)
    So you've worked out the probability if he buys paper X, what about paper Y? You don't seem to include that.

    You can either do it the way above this post, or a more straight forward way to think about it is this:

    For paper X, the probability of both reviews is \frac{1}{2}\cdot \frac{2}{3}= \frac{1}{3}

    For paper Y, the probability of both reviews is \frac{1}{4} \cdot \frac{2}{5}=\frac{1}{10}

    Overall probability of both reviews: \frac{1}{3}+\frac{1}{10}=\frac{1  3}{30}

    In a SINGLE given paper it would be half of this probability (ie probability of choosing a single one):
    \frac{1}{2}(\frac{13}{30})=\frac  {13}{60}
    Thanks for the reply. I now understand where the mistake was made.
 
 
 
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