## Related questions with answers

Let X denote the number of times a certain numerical control machine will malfunction: 1, 2, or 3 times on any given day. Let Y denote the number of times a technician is called on an emergency call. Their joint probability distribution is given as

$\begin{matrix} & & & x & \\ \hline & & 1 & 2 & 3 \\ & 1 & 0.05 & 0.05 & 0.10 \\ y & 3 & 0.05 & 0.10 & 0.35 \\ & 5 & 0.00 & 0.20 & 0.10 \end{matrix}$

(a) Evaluate the marginal distribution of X. (b) Evaluate the marginal distribution of Y. (c) Find P(Y=3|X=2).

Solution

Verified$\textbf{(a)}$ Since the joint probability distribution is given with the table, we will be finding the marginal distribution $g(x)$ of the random variable $X$ in each value by reading from the table and inserting those read values in the formula $g(x) = \sum_yf(x,y)$.

$\boxed{ g(1) }=\sum_{y}f(1,y)=f(1,1)+f(1,3)+f(1,5)=0.05+0.05+0.00 = \boxed{ 0.1 }$

$\boxed{ g(2) }=\sum_{y}f(2,y)=f(2,1)+f(2,3)+f(2,5)=0.05+0.10+0.20 = \boxed{ 0.35 }$

$\boxed{ g(3) }=\sum_{y}f(3,y)=f(3,1)+f(3,3)+f(3,5)=0.10+0.35+0.10 = \boxed{ 0.55 }$

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