Calculating Probabilities for a cumulative distribution function within a given inequality

Question

Given that K = 1/36, I require some help understanding (b)

• Pr(1/2 ≤ X ≤ 1)

Is re-written as such:

Pr(X ≤ 1) - Pr(X < 1/2)

I do not understand why!

Is it because

Pr(X ≤ 1) is solved as F(1) and

Pr(X ≤ 1) - Pr(X < 1/2) = Pr(X ≤ 1) + (1 - Pr(X < 1/2))

Answer 1

You should understand the difference between a probability density function and a cumulative distribution function.

The cumulative distribution function, which in your case is $F(x)$, always gives the value for $P(X \leq x)$

So, $F(1)$ would give you $P(X\leq1)$ and $F(\frac{1}{2})$ would give you $P(X\leq\frac{1}{2})$. In order to find $P(\frac{1}{2} < X \leq 1)$, you calculate, $F(1) - F(\frac{1}{2})$

• Edited as per suggestion..

Answer 2

$$\{X\leq 1\}=\{X<\frac{1}{2}\}\cup\{\frac{1}{2}\leq X\leq 1\}$$ and these sets are disjoint so that: $$P\{X\leq 1\}=P\{X<\frac{1}{2}\}+P\{\frac{1}{2}\leq X\leq 1\}$$ or equivalently: $$P\{\frac{1}{2}\leq X\leq 1\}=P\{X\leq 1\}-P\{X<\frac{1}{2}\}$$