Let X be a random variable with density function () = 2^−2 Calculate ( ≤ 0.5| ≤ 1)

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,

the CDF would be

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

6 equilateral triangles

answer is  

a.

93.98 sq. in.

hope you got it chief