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5 June, 06:26

In a certain Algebra 2 class of 28 students, 5 of them play basketball and 21 of them play baseball. There are 5 students who play neither sport. What is the probability that a student chosen randomly from the class plays both basketball and baseball? Does 'neither' mean 'not basketball AND not baseball'? Or 'not basketball OR not baseball'?

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  1. 5 June, 06:44
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    Answer:3/28; Neither means NOT basketball AND not baseball

    Step-by-step explanation:

    a) Using set notation to answer the question,

    Let U be universal set (total number of student)

    B be those that play basketball

    C be those that play baseball

    BUC be those that plays either of the games

    (BUC) ' will them be those that play neither

    BnC be those that plays both games

    (U) = 28

    n (B) = 5

    n (C) = 21

    Before we get the probability that a student chosen randomly from the class plays both basketball and baseball, we need to get BnC first

    Using the formula

    n (BUC) = n (B) + n (C) - n (BnC) ... (1) and;

    n (U) = n (BUC) + n (BUC) ' ... (2)

    n (BUC) = n (U) - n (BUC) '

    n (BUC) = 28-5 = 23

    Therefore using eqn 1,

    23 = 5+21-n (BnC)

    23=26-n (BnC)

    n (BnC) = 3

    P{n (BnC) } = n (BnC) / n (U) = 3/28

    b) Neither means NOT basketball AND not baseball i. e those that played NONE of the games.
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