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Thread: just wondering

  1. #1
    Senior Member slevvio's Avatar
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    just wondering

    A student moving into a new flat buys a box of 20 light bulbs. Unknown to the student, exactly one of the bulbs is defective. Find:

    b) The conditional probability that the second bulb used is not defective, given that the first bulb used is not defective;

    If the first bulb is not defective then you have 19 bulbs from which you can choose, and one of these is defective so the probability is $\displaystyle \frac{18}{19}$.

    This is the easy way to work it out but I am having trouble using the actual formula which may come in handy for other examples:

    Let A be the event that the 2nd bulb is not defective.
    Let B be the event that the 1st bulb is not defective.

    Then $\displaystyle P(A|B)=\frac{P(A\cap B)}{P(B)}$

    $\displaystyle P(B) = \frac{19}{20}$ but how can you work out $\displaystyle P(A\cap B))$?

    $\displaystyle P(A\cap B) = P(A|B)\cdot P(B)$
    $\displaystyle P(A\cap B) = \frac{18}{19}\cdot \frac{19}{20}=\frac{18}{20}$

    So how can we work out that $\displaystyle P(A\cap B) = \frac{18}{20}$ from what we are given?

    I apologise for any offence my pedantry has caused
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  2. #2
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    $\displaystyle \begin{array}{l}
    P(X \cap Y) = P(X|Y)P(Y) \\
    P(A \cap B) = P(B)P(A|B) = \left( {\frac{{19}}{{20}}} \right)\frac{{18}}{{19}} = \frac{{18}}{{20}} \\
    \end{array}$
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  3. #3
    Senior Member slevvio's Avatar
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    Let me rephrase my question:

    A student moving into a new flat buys a box of 20 light bulbs. Unknown to the student, exactly one of the bulbs is defective. Let A be the event the 2nd bulb is defective, and let B be the event the 1st bulb is defective. What is $\displaystyle P(A\cap B)$?
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  4. #4
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    Quote Originally Posted by slevvio View Post
    Let me rephrase my question:

    A student moving into a new flat buys a box of 20 light bulbs. Unknown to the student, exactly one of the bulbs is defective. Let A be the event the 2nd bulb is defective, and let B be the event the 1st bulb is defective. What is $\displaystyle P(A\cap B)$?
    $\displaystyle P(A \cap B) = \left( {\frac{2}{{20}}} \right)\left( {\frac{1}{{19}}} \right)$
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  5. #5
    Senior Member slevvio's Avatar
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    Thank you but once again I have asked the wrong thing.

    A student moving into a new flat buys a box of 20 light bulbs. He takes a number of bulbs from the box without replacement. Unknown to the student, exactly one of the bulbs is defective. Find:

    Let A be the event the 2nd bulb is NOT DEFECTIVE and B the event that the 1st bulb is NOT DEFECTIVE. Find $\displaystyle P(A\cap B)$.

    Thanks and sorry about this the stress is making me do silly things.
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