Calculate Probability Using Random Variable? I found the following question in my text-book:
A bus comes in a delay which is uniform between $0$ and $10$ minutes.
Daniel comes $3$ minutes before the bus is supposed to arrive the earliest.

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*Calculate the possibility that Daniel waits more than $6$ min.

My solution:
Let $X$ be a random variable that describes the delay of the bus, we know that $X\thicksim\operatorname{Uni}(0,10)$ and that for sure he will wait at least $3$ min so we calculate $P(X>3)$:
$$P(X>3)=\int_3^\infty f(x)\,dx.$$
Now we can replace $\infty$ with $10$ since $f(x)=0$ for $x>10$ (Probability density function) thus getting $7/10$.


*In a specific day Daniel woke up late and arrived to the bus station in a delay of $4$ minutes.
What's the possibility he missed the bus?

Daniel will miss the bus if it arrived in the first $4$ minutes so:
At first I though this is the same as
$$\frac{P(x<4\mid x=4)}{P(x=4)}=\frac{0}{P(x=4)}=0.$$
But It's clearly wrong and doesn't make sense.
 A: You established $X$ to be a random variable that describes the time Daniel waits for the bus, which is not the best choice here, because it does not follow a uniform distribution between $0$ and $10$.
For example, in the original settings, Daniel comes $3$ minutes before the bus is supposed to arrive the earliest, so the probability for him waiting more than $6$ min should $P(X>6)$ instead of $P(X>3)$.
However, in the second question, he indeed came to station of $4$ min late, so what does the event $X=4$ mean? It means the bus came $4$ min later than Daniel, i.e., the bus came in totally $4+4=8$ min late. Of course this does not make sense at all. In fact, in your setting, the probability should read $P(X\geq 0)$. If Daniel did not miss the bus, then the waiting time for him should be non-negative.
You should set $X$ to be the time for such bus come in a delay, which ranges uniformly from $[0,10]$. Then the first one is $P(X>3)$ and the second one is $P(X<4)$. If $X\geq 4$, then the bus came later than Daniel, so he would not miss the bus.
A: For 1) you need to calculate $P(X>6\mid X>3)$ since we know he will wait at least 3 minutes.
For 2) We can safely assume that the probability of him arriving 4 minutes late is independent of the arrival time of the bus, so we get $P(X<4\mid \text{4 min late})=P(X<4)$
