(3c to the 5th power d 1/2)3
1 answer:
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Step-by-step explanation:
Given that,
a)
X ~ Bernoulli and Y ~ Bernoulli
X + Y = Z
The possible value for Z are Z = 0 when X = 0 and Y = 0
and Z = 1 when X = 0 and Y = 1 or when X = 1 and Y = 0
If X and Y can not be both equal to 1 , then the probability mass function of the random variable Z takes on the value of 0 for any value of Z other than 0 and 1,
Therefore Z is a Bernoulli random variable
b)
If X and Y can not be both equal to 1
then,
or
and
c)
If both X = 1 and Y = 1 then Z = 2
The possible values of the random variable Z are 0, 1 and 2.
since a Bernoulli variable should be take on only values 0 and 1 the random variable Z does not have Bernoulli distribution