If X and Y are two random variables having density function

$f (x, y) = {\begin{matrix} \frac{(6 - x - y)}{8}, 0 < x < 2, 2 < y < 4 \\ 0, e l s e w h e r e \end{matrix}$

then P(X + Y < 3) is equal to:

Question

Question

This question was previously asked in

SSC CGL Tier-II (JSO) 2022 Official Paper (Held On: 4 March 2023)

Answer 1

The correct option is b.

Detailed Solution:

Concept:

The Joint cumulative function of two random variables X and Y is defined as

$F_{X Y} = P (X < x, Y < y) .$

The joint CDF satisfies the following properties:

$F_{X} (x) = F_{X Y} (X, \in f t y)$ , for any x (marginal CDF of X)
$F_{X Y} (\in f t y, \in f t y) = 1$
$F_{X Y} (x, - \in f t y) = F_{X Y} (- \in f t y, y) = 0$
P(x₁ < X < x₂, y₁ < Y < y₂) = F_XY(x₂, y₂) - FXY(x₁, y2) - FXY(x2, y₁) + FXY(x₁, y₁)
If X and Y are independent, then FXY(x, y) = FXY(x) FXY(y)

Two random variables X and Y are jointly continuous if there exists a nonnegative function f_XY : ℝ² → ℝ, such that for any set A ∈ ℝ², we have

$P ((X, Y) \in A) = \iint_{A} f_{X Y} (x, y) d x d y (5.15)$

The function f_XY(x, y) is called the joint probability density function (PDF) of X and Y.
Marginal PDFs

$f_{X} (x) = \int_{- \infty}^{\infty} f_{X Y} (x, y) d y, f o r a l l x$

$f_{Y} (y) = \int_{- \infty}^{\infty} f_{X Y} (x, y) d y, f o r a l l y$

Calculation:

f (x, y) = {\begin{matrix} \frac{(6 - x - y)}{8}, 0 < x < 2, 2 < y < 4 \\ 0, e l s e w h e r e \end{matrix}

P(X + Y < 3) =

\in t_{y = 2}^{y = 4} \in t_{x = 0}^{x = 3 - y} f (x, y) d x d y

P(X + Y < 3) =

\in t_{y = 2}^{y = 4} \in t_{x = 0}^{x = 3 - y} (\frac{(6 - x - y)}{8}) d x d y

P (X + Y < 3) = \int_{y = 2}^{y = 4} \int_{x = 0}^{x = 3 - y} (\frac{6 - x - y}{8}) d x d y

= \frac{1}{8} \int_{y = 2}^{y = 4} \int_{x = 0}^{x = 3 - y} (6 - x - y) d x d y

= \frac{1}{8} \int_{y = 2}^{y = 4} {(6 x - \frac{x^{2}}{2} - x y)}_{x = 0}^{x = 3 - y} d y

= \frac{1}{8} \int_{y = 2}^{y = 4} {6 (3 - y) - \frac{(3 - y)^{2}}{2} - (3 - y) y} d y

= \frac{1}{8} \int_{y = 2}^{y = 4} {(18 - 6 y) - (\frac{9}{2} - 3 y + \frac{y^{2}}{2}) - (3 y - y^{2})} d y

= \frac{1}{8} \int_{y = 2}^{y = 4} (\frac{27}{2} - 6 y + \frac{y^{2}}{2}) d y

= \frac{1}{8} {\frac{27}{2} y - 3 y^{2} + \frac{y^{3}}{6}}_{y = 2}^{y = 4}

= \frac{1}{8} {\frac{27}{2} (4 - 2) - 3 (16 - 4) + \frac{1}{6} (64 - 8)}

= \frac{1}{8} {27 - 36 + \frac{28}{3}}

= \frac{1}{24}