Problem Statement
\(p\) and \(q\) are two points chosen at random between 0 \(\&\) 1. What is the probability that the ratio \(p/q\) lies between 1 \(\&\) 2?
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Solution
We have 2 random variables (RVs) \(p \& q\) here. Let’s try to fix one for simplicity, and then use that result to develop a general solution where both are RVs.
Step 1: Fixing q
Say we fix \(q\) at some value between 0 and 1. Now, the given problem reduces to:
“q is a fixed value between 0 and 1. p ~ Uniform(0,1). What is the probability that 1 < p/q < 2?”
This question is fairly straightforward.
Step 2: Rearranging the Inequality
Let’s rearrange the inequality:
\[1 < \frac{p}{q} < 2\] \[q < p < 2q\]Also, note that since p is restricted in (0,1), we must say:
\[\max(0,q) < p < \min(2q, 1)\]But since q is also restricted in (0,1), we have \(\max(0,q) = q\)
Thus, the final expression is:
\[q < p < \min(2q,1)\]Step 3: Computing the Probability
Now, let’s compute the probability that p will lie within this range.
As \(p \sim \text{Uniform}(0,1)\), the required probability:
\[P = \frac{\min(1,2q) - q}{1} = \min(2q,1) - q\](Note: The probability is simply the length of the required range divided by the length of the total range. This is also visible by intuition)
Step 4: Considering q as a Random Variable
Now, note that this is the case when q is fixed. But q itself will vary from 0 to 1.
We can write the above probability as “conditioned on” a particular q:
\[P(E|q) = \min(2q,1) - q\](Where E is our required event of \(1 < \frac{p}{q} < 2\))
Step 5: Using the Law of Total Probability
Using fundamental laws, we have:
\[P(E) = \sum P(E|q)P(q)\]What is P(q)? Essentially the probability of choosing a particular value from 0 to 1, an infinite set! Clearly very small! Let’s use calculus to denote this.
\[P(q) = \frac{dq}{1} = dq\]Thus,
\[P(E) = \int_0^1 (\min(2q,1) - q) \, dq\]Step 6: Solving the Integral
To solve this integral, we need to split it into two parts:
- From 0 to 1/2, where \(\min(2q,1) = 2q\)
- From 1/2 to 1, where \(\min(2q,1) = 1\)
Now,
\[P(E) = \int_0^{1/2} (2q - q) \, dq + \int_{1/2}^1 (1 - q) \,\] \[= \int_0^{1/2} q \, dq + \int_{1/2}^1 (1 - q) \, dq\] \[= [\frac{q^2}{2}]_0^{1/2} + [q - \frac{q^2}{2}]_{1/2}^1\] \[= (\frac{1}{8} - 0) + ((1 - \frac{1}{2}) - (\frac{1}{2} - \frac{1}{8}))\] \[= \frac{1}{8} + \frac{1}{2} - \frac{3}{8} = \frac{1}{4}\]Therefore, the probability that the ratio \(p/q\) lies between 1 and 2 is \(\frac{1}{4}\) or \(0.25\) or \(25\%\).
PS - A similar integration is involved in the problem Random-Angle-II. Check it out for a more detailed explaination incase of any confusions!.
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