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Question

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

x12310.050.050.10y30.050.100.3550.000.200.10\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

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

g(1)=yf(1,y)=f(1,1)+f(1,3)+f(1,5)=0.05+0.05+0.00=0.1\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 }

g(2)=yf(2,y)=f(2,1)+f(2,3)+f(2,5)=0.05+0.10+0.20=0.35\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 }

g(3)=yf(3,y)=f(3,1)+f(3,3)+f(3,5)=0.10+0.35+0.10=0.55\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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