VIPSolutions**a)** $c = 1$ --- **b)** $P(Y_1 \ge 3Y_2) = \frac{2}{3}$ --- **c)** $P(Y_1 > 1.5 \mid Y_2 = 0.5) = 0.5$ --- **d)** No, $Y_1$ and $Y_2$ are not independent --- **e)** * $E(Y_1) = \frac{4}{3}$ * $E(Y_2) = \frac{1}{3}$ * $V(Y_1) = \frac{2}{9}$ * $V(Y_2) = \frac{1}{18}$ --- **f)** * $E(Y_1 - Y_2) = 1$ * $V(Y_1 - Y_2) = \frac{1}{6}$ rewrite as it is
Question:
**a)** $c = 1$ --- **b)** $P(Y_1 \ge 3Y_2) = \frac{2}{3}$ --- **c)** $P(Y_1 > 1.5 \mid Y_2 = 0.5) = 0.5$ --- **d)** No, $Y_1$ and $Y_2$ are not independent --- **e)** * $E(Y_1) = \frac{4}{3}$ * $E(Y_2) = \frac{1}{3}$ * $V(Y_1) = \frac{2}{9}$ * $V(Y_2) = \frac{1}{18}$ --- **f)** * $E(Y_1 - Y_2) = 1$ * $V(Y_1 - Y_2) = \frac{1}{6}$ rewrite as it is
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Asked by: Umesh
Created at: 2025-08-02 01:21:00
Asked by: Umesh
Created at: 2025-08-02 01:21:00
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**a)**
$c = 1$
---
**b)**
$P(Y_1 \ge 3Y_2) = \frac{2}{3}$
---
**c)**
$P(Y_1 > 1.5 \mid Y_2 = 0.5) = 0.5$
---
**d)**
No, $Y_1$ and $Y_2$ are not independent
---
**e)**
* $E(Y_1) = \frac{4}{3}$
* $E(Y_2) = \frac{1}{3}$
* $V(Y_1) = \frac{2}{9}$
* $V(Y_2) = \frac{1}{18}$
---
**f)**
* $E(Y_1 - Y_2) = 1$
* $V(Y_1 - Y_2) = \frac{1}{6}$
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