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24 December, 10:53

Suppose odd numbers 1, 3, 5 with probability C and on each of the even numbers with probability 2C.

(a) Find C.

(b) Suppose that the die is tossed. Let X equal 1 if the result is an even number, and let it be 0 otherwise. Also, let Y equal 1 if the result is a number greater than three and let it be 0 otherwise. Find the joint probability mass function of X and Y. Suppose now that 12 independent tosses of the die are made.

(c) Find the probability that each of the six outcomes occurs exactly twice.

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  1. 24 December, 11:45
    0
    Step-by-step explanation:

    a) Odd numbers if n then even numbers also n

    Total prob = 1 gives 3c=1

    c=1/3

    b) X=1, if even and

    0, if odd

    Y = 1, if x >3 and

    =0 otherwise

    xy can take values as 0 or 1

    xy = 1 if x even and y is greater than 3, or favorable outcomes are

    4, 6.

    xy = 1 if die shows 4 or 6

    =0 if otherwise

    Prob (xy) = 1/3, for xy = 1 and

    = 2/3 for xy = 0

    This is binomial since each trial is independent.

    If 12 tosses are made then if we have binomial proba

    with n = 12 and p = 1/3 for success
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