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12 April, 16:37

Let p (z) = 0.54, p (y) = 0.22, and p (z u y) = 0.62. Find each probability.

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  1. 12 April, 17:47
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    Answer : p (only z) = 0.40 and p (only u) = 0.08

    Explanation:

    We have given that,

    p (u) = 0.22,

    p (z∪u) = 0.62,

    p (z) = 0.54

    By using probability rule of union of 2 events, we have,

    p (z) + p (u) - p (z∩u) = p (z∪u)

    ⇒0.54+0.22-p (z∩u) = 0.62

    ⇒0.76-p (z∩u) = 0.62

    ⇒ (-) p (z∩u) = 0.62-0.76

    ⇒ (-) p (z∩u) = (-) 0.14

    ⇒p (z∩u) = 0.14

    Now,

    p (only z) = p (z) - p (z∩u)

    =0.54-0.14

    =0.40

    and,

    p (only u) = p (u) - p (z∩u)

    =0.22-0.14

    =0.08

    ∴ Each probability will be

    p (only z) = 0.40

    p (only u) = 0.08
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