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Math Help - 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 \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 P(A|B)=\frac{P(A\cap B)}{P(B)}

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

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

    So how can we work out that 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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    \begin{array}{l}<br />
 P(X \cap Y) = P(X|Y)P(Y) \\ <br />
 P(A \cap B) = P(B)P(A|B) = \left( {\frac{{19}}{{20}}} \right)\frac{{18}}{{19}} = \frac{{18}}{{20}} \\ <br />
 \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 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 P(A\cap B)?
    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 P(A\cap B).

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