Question

3.68. If the joint probability density of X, Y, and Z is given by \begin{equation} f(x, y, z) = \begin{cases} \frac{1}{3}(2x + 3y + z) & \text{for } 0 < x < 1, 0 < y < 1, 0 < z < 1 \\ 0 & \text{elsewhere} \end{cases} \end{equation} find (a) $P(X = \frac{1}{2}, Y = \frac{1}{2}, Z = \frac{1}{2})$; (b) $P(X < \frac{1}{2}, Y < \frac{1}{2}, Z < \frac{1}{2})$.

          3.68. If the joint probability density of X, Y, and Z is given by
\begin{equation*}
f(x, y, z) = \begin{cases}
\frac{1}{3}(2x + 3y + z) & \text{for } 0 < x < 1, 0 < y < 1, 0 < z < 1 \\
0 & \text{elsewhere}
\end{cases}
\end{equation*}
find
(a) $P(X = \frac{1}{2}, Y = \frac{1}{2}, Z = \frac{1}{2})$;
(b) $P(X < \frac{1}{2}, Y < \frac{1}{2}, Z < \frac{1}{2})$.

3.68. If the joint probability density of X, Y, and Z is given by

f(x, y, z) = (1)/(3)(2x + 3y + z) for 0 < x < 1, 0 < y < 1, 0 < z < 1

0 elsewhere

find
(a) P(X = (1)/(2), Y = (1)/(2), Z = (1)/(2));
(b) P(X < (1)/(2), Y < (1)/(2), Z < (1)/(2)).

Added by Nicole C.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert Christopher Stanley

06/14/2021

Step 1

Therefore, $P(X = \frac{1}{2}, Y = \frac{1}{2}, Z = \frac{1}{2}) = 0$. Show more…

Show all steps

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3.68. If the joint probability density of X, Y, and Z is given by $$f(x, y, z) = \begin{cases} \frac{1}{3}(2x + 3y + z) & \text{for } 0 < x < 1, 0 < y < 1, 0 < z < 1 \\ 0 & \text{elsewhere} \end{cases}$$ find (a) $P(X = \frac{1}{2}, Y = \frac{1}{2}, Z = \frac{1}{2})$; (b) $P(X < \frac{1}{2}, Y < \frac{1}{2}, Z < \frac{1}{2})$.