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    • Thread Starter

    An apple grower classifies his apples by mass, x grams, as small, medium or large as follows.

    Small: x<90
    Med: 90=<x<120
    Large: x=<120

    Assume the masses are normally distributed with mean 110 and standard dev. 15 to the nearest integer.
    (might be relevant) In a large crop of apples ti was found that 9% were small and 25% were large.

    iii) Two apples were chosen at random from the crop. Find the probability that at least one of them has a mass of 100g or more.

    ... my maths teacher couldn't do this.
    I thought of doing P(X>99.5) + P(X>99.5)^2 (mass of one or mass of both >100) but it's not right.

    Any tips?

    P( at least one has mass >= 100g) = 1 - P( both have mass < 100g)

    = 1- [P(X < 100)]^2

    How does that look to you?
    • Thread Starter

    I don't understand how P of at least 1 led to 1 - P of both <100 but I plugged in the numbers anyway. I got 0.7454+0.0019 (whatever that is) which I think is that I got from my method (probability of one OR probability of both). The textbook says 0.9363 though

    In probability, for an event A, P(A) = 1-P(A').
    So P( at least one has mass >= 100g) = 1 - P( both have mass < 100g).

    Alternatively: P( at least one has mass >= 100g) = P(first apple has mass >= 100 AND second apple mass<100) + P(second apple has mass >= 100 AND first apple mass<100) + P(both apples have mass>=100)

    =P(X>100)*P(X<100) + P(X<100)*P(X>100) + P(X>100)*P(X>100)= ....0.936 also.
    • Thread Starter

    I see it now, thanks!
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