By definition of conditional probability,
P(X ≤ 0.5 | X ≤ 1) = P((X ≤ 0.5) and (X ≤ 1)) / P(X ≤ 1)
but if X ≤ 0.5, then it's automatic that X ≤ 1, so
P(X ≤ 0.5 | X ≤ 1) = P(X ≤ 0.5) / P(X ≤ 1)
Given the PDF of X,
[tex]f_X(x) = \begin{cases}2e^{-2x}&\text{if }x\ge0\\0&\text{otherwise}\end{cases}[/tex]
the CDF would be
[tex]P(X\le x) = F_X(x) = \displaystyle\int_{-\infty}^x f_X(t)\,\mathrm dt[/tex]
[tex]F_X(x) = \begin{cases}0&\text{if }x<0\\1-e^{-2x}&\text{if }x\ge0\end{cases}[/tex]
So we have
P(X ≤ 0.5 | X ≤ 1) = (1 - exp(-2 × 0.5)) / (1 - exp(-2 × 1))
… = (1 - exp(-1)) / (1 - exp(-2))
… = (1 - 1/e) / (1 - 1/e ²)
… = (e ² - e) / (e ² - 1)
… = e / (e + 1) ≈ 0.7312
