Conditional Probability Calculation

We are given that the joint pdf of $X$ and $Y$ is

f_{X,Y}(x,y)= \frac{1}{9}(3-x)(2-y),\quad 0\le x\le 3,\; 0\le y\le 2.

We wish to find

P(Y\ge 1\mid X<2)=\frac{P(Y\ge 1\text{ and }X<2)}{P(X<2)}.

The probability that $X<2$ is obtained by integrating the joint density over the region $0\le x<2$ and $0\le y\le 2$ :

P(X<2)=\int_{x=0}^{2}\int_{y=0}^{2}\frac{1}{9}(3-x)(2-y)\,dy\,dx.

Integrate with respect to $y$ :

\int_{y=0}^{2}(2-y)\,dy.

Compute:

\int_{y=0}^{2}(2-y)\,dy = \left[2y-\frac{y^2}{2}\right]_0^2 = \left(2(2)-\frac{2^2}{2}\right)-0 = (4-2)= 2.

Thus,

P(X<2)=\frac{1}{9}\int_{0}^{2}(3-x)\cdot 2\,dx = \frac{2}{9}\int_{0}^{2}(3-x)\,dx.

Now integrate with respect to $x$ :

\int_{0}^{2}(3-x)\,dx = \left[3x-\frac{x^2}{2}\right]_0^{2} = \left(3(2)-\frac{2^2}{2}\right)-0 = (6-2)=4.

So,

P(X<2) = \frac{2}{9}\cdot 4 = \frac{8}{9}.

This probability is given by:

P(Y\ge 1,\,X<2)=\int_{x=0}^{2}\int_{y=1}^{2}\frac{1}{9}(3-x)(2-y)\,dy\,dx.

Integrate with respect to $y$ :

\int_{y=1}^{2}(2-y)\,dy.

Compute:

\int_{y=1}^{2}(2-y)\,dy = \left[2y-\frac{y^2}{2}\right]_1^{2}.

Evaluate the antiderivative:

Thus, the result is:

2-\frac{3}{2}=\frac{1}{2}.

Therefore, we have:

P(Y\ge 1,\,X<2)=\frac{1}{9}\int_{0}^{2}(3-x)\cdot \frac{1}{2}\,dx=\frac{1}{18}\int_{0}^{2}(3-x)\,dx.

We already computed $\int_{0}^{2}(3-x)\,dx=4$ . Hence,

P(Y\ge 1,\,X<2)=\frac{1}{18}\cdot 4 = \frac{4}{18} = \frac{2}{9}.

Now, the conditional probability is:

P(Y\ge 1\mid X<2)=\frac{P(Y\ge 1,\,X<2)}{P(X<2)}=\frac{\frac{2}{9}}{\frac{8}{9}}=\frac{2}{8}=\frac{1}{4}.

\boxed{\frac{1}{4}}

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