Probability of $y \leq x$ given $x \leq \frac 12$

Question

Please help me understand solve this problem.

Let the sample space be the unit square, $\Omega =[0,1]^2$ , and let the probability of a set be the area of the set. Let $A$ be the set of points $(x,y) \in [0,1]^2$ for which $y \leq x$ . Let $B$ be the set of points for which $x \leq \frac 12$. Find $\mathbf P(A∣B)$.

Is the region of $\mathbf P(A∣B)$ not the dark shaded triangle region in the following image?

If so, isn't the area $\frac 12$ base height. In which case, I get $\frac 12 \cdot (\frac 12 \cdot \frac 12) = \frac 18$. But my answer is wrong. How do I work this problem out? Thank you.

Answer 1

Don't forget $$P(A|B)=\frac{P(A\cap B)}{P(B)}$$You worked out the numerator to be $\frac18$. What about the denominator?

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Another way to see it visually is this: when asked for a conditional probability, you are given some information. So you should take this information as something you know for sure. In this example, you should assume you already know $B$ is true. So you are definitely in this region to the left of the line $x=1/2$. I.e. you are essentially now looking at a new, adjusted probability distribution, a uniform one $\tilde\Omega=[0,\tfrac12]\times[0,1]$. Now what's the probability $x\le y$? It is whatever proportion of your region has $x\le y$. From your image, its easy to see that the shaded region is $1/4$ of the total region to the left of $x=1/2$. So your answer is $1/4$ again.

Answer 2

I believe I understand where you made a mistake.

There are 2 ways to solve this problem.

1. By the definition of Conditional Probability. As noted in the other answer, you have: $P(A|B)=\frac{P(A\cap B)}{P(B)}$. Thus, $P(A\cap B)=1/8$ (area of the triangle) and $P(B) = 1/2$ (area of B, the vertical rectangle in your picture). Thus, $P(A|B)=1/4$. If you had followed this idea, you would be okay.

2. However, you probably followed the "area" idea but forgot to "revise" your model. "Revised" comes from the idea that now you have knowledge that B has occurred, so you are now looking at a "revised probability model". Initially, you had a "Probability Law" behind your model. Now you have a revised one, and you can call it the "Conditional Probability Law". Thus, you can't apply the initial "probability law", where you were calculating probabilities based on the unit square area as the point of reference (aka the sample space $\Omega$), because now there is 0 probability to outcomes outside B. The revised model gives you a new law: you will look at B as the new sample space (hence the new point of reference for the areas) and calculate probabilities in this space in relation to the area of B and not of $\Omega$. So what is the probability of A happening given that you know B occurred? $\frac{Area-of-the-triangle}{Area-of-B}=\frac{1/8}{1/2}=1/4$.