If a real number x is chosen at random in the interval $[0, 3]$, and a real number $y$ is chosen at random in the interval $[0, 4]$, what is the probability that $x < y$ ?
Is this an application of Normal Distribution ?
what is the probability that $x < y$?
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probability
probability-theory
probability-distributions
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0Use independence to obtain the joint pdf of $x$ and $y$ then integrate it over the region $x
3 Answers
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Break up the range of $y $ into two parts: $y>3$ and $0\geq y \geq 3$. We then consider the probability of $y>x $ in each case and add them up.
In case $1$, $y $ is always $> x$, so the probability that we choose $y $ in this interval is $\frac {1}{4} $.
In case $2$, as both $x $ and $y $ are in the same range, we can logically conclude that $y>x$ half the time. As the probability of choosing a $y $ in this range is $\frac {3}{4} $, the total probability is $\frac {1}{2}\times \frac {3}{4} = \frac {3}{8} $. Hence, $$\boxed {P_{\text {tot}} = P_1 +P_2 = \frac {1}{4} + \frac {3}{8} = \frac{5}{8}}$$ Hope it helps.
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No; presumably these are uniformly random variables.
Use the Law of Total Probability:
$$P(x You should be able to work out these conditional probabilities.
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Assuming $X\sim\text{Uniform}(0,3)$ and $Y\sim\text{Uniform}(0,4)$:
$$P(X Then using independence: $$f_{X,Y}(x,y)=f_X(x)f_Y(y)=\frac{1}{3}\cdot\frac{1}{4}=\frac{1}{12}$$ Thus: $$P(X