Solution of problem 4.

(a)
Yes. Reason: $P(A^c \cap B^c) = P((A \cup B)^c) = 1- P(AUB)= 1-[P(A)+P(B) -P(A \cap B)] = 1- [P(A)+P(B) -P(A)P(B)]= 1- P(A) -P(B) + P(A)P(B).$ Now

\begin{displaymath}P(A^c)P(B^c)= (1-P(A))(1-P(B)) = 1 -P(B)-P(A) + P(A)P(B).\end{displaymath}

Hence $P(A^c \cap B^c) =P(A^c)P(B^c).$ Therefore they are independent.

(b)
True. Reason: Since $X$ takes values $0$ or $1$, $X^2$ takes values $0$ or $1$. Let us try to calculate the probability mass function of $X^2$. Let us call it $pmf_{X^2}(\cdot).$

\begin{displaymath}pmf_{X^2}(1)= P(X^2=1) = P(X=1)= p\end{displaymath}


\begin{displaymath}pmf_{X^2}(0)= P(X^2=0) = P(X=0)= 1-p\end{displaymath}

As the probability mass function solely identifies the random variable, we can conclude that $X^2$ is a Bernoulli(p) random variable.

(c)
False. Reason:
First method: $X$ measures the number of trials required to obtain the first success in a sequence of independent Bernoulli Trials. So does $Y$ and independent of $X$. Therefore $X+Y$ measures the number of trials required to obtain the second success in a sequence of independent Bernoulli trials. Hence $X+Y$ is a negative binomial random variable with parameters (2,p) and NOT geometric.

Second method: $X$ is a Geometric(p) random variable and thereby takes values $1,2,3,4,5,...$. $Y$ is independent of $X$. $Y$ is also a Geometric(p) random variable and thereby takes values $1,2,3,4,5,...$. Now, $X+Y$ takes values $2,3,4,5,6...$. Let their probability mass function be denoted as $pmf_X$, $pmf_Y$ and $pmf_{X+Y}.$ We calculate the $pmf_{X+Y}.$ For $k \geq 2,$

\begin{eqnarray*}pmf_{X+Y}(k) &=& P(X+Y =k) \\ &=& \sum_{m=1}^{k-1}P(X=m, Y=k-m)... ...& (1-p)^{k-2}p^2\sum_{m=1}^{k-1} \\ &=& (1-p)^{k-2}p^2 (k-1)\\ \end{eqnarray*}



Hence $X+Y$ is not geometric(p). But one can realise that the $pmf_{X+Y}$ resembles that of negative binomial (2,p). (the question did not ask that).

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