Probability — Question 3

Question 3 – Solution

Step 1: Identify the two events. Let $$A=\{X>7\}, \qquad B=\{X>3\}.$$ Because $A\subset B$, we have $A\cap B=A$.

Step 2: Apply conditional probability. $$P(A\mid B)=\frac{P(A\cap B)}{P(B)} =\frac{P(A)}{P(B)}.$$

Step 3: Use uniform interval lengths. On $[0,10]$, probability equals interval length divided by $10$: $$P(X>7)=\frac{10-7}{10}=\frac3{10}, \qquad P(X>3)=\frac{10-3}{10}=\frac7{10}.$$

Step 4: Form the ratio. $$P(X>7\mid X>3) =\frac{\frac3{10}}{\frac7{10}} =\frac37.$$ $$\boxed{P(X>7\mid X>3)=\frac37}$$ Equivalently, after conditioning on $X>3$, the available interval is $(3,10]$, and the favorable portion $(7,10]$ has length $3$ out of $7$.