If the interval of A has been uniformly chosen as [0,1] and B as [0,6] then what is the probability of A being a lower number than B?
I'm completely lost here, do I somehow calculate the uniform distribution probabilities of both and then compare them? Not sure how to get the probability of one being lower than the other. Any help would be appreciated
One way:
\begin{align} P(A\lt B) &= \int_{x=0}^1 P(A\lt B\mid A=x)f_X(x)\;dx \qquad\qquad\text{by Law of Total Probability} \\ &= \int_{x=0}^1 P(x\lt B)\cdot 1\;dx \\ &= \int_{x=0}^1 \left(1-\dfrac{x}{6}\right)\;dx \\ &= \left[ x-\dfrac{x^2}{12}\right]_{x=0}^1 \\ &= \dfrac{11}{12}. \end{align}
Another way:
\begin{align} P(A\lt B) &= P(A\lt B\mid B\lt 1)P(B\lt 1) + P(A\lt B\mid B\gt 1)P(B\gt 1) \\ & \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\text{by Law of Total Probability} \\ &= \dfrac{1}{2}\cdot\dfrac{1}{6} + 1\cdot\dfrac{5}{6} \\ & \qquad\text{since, given $B\lt 1$, events $A\lt B$ and $B\lt A$ are equally likely} \\ &= \dfrac{11}{12}. \end{align}