Error in applying Chebyshev's inequality

Question

I'm trying to solve a problem using Chebychev^' s Inequality:

"Suppose that X is a random variable with mean and variance both equal to 20. What can be said about P(0<X<40)?"

P(|X-μ|≥kσ)≤1/k^2

Restating the original equation:

P(0<X<40)=P(|X-20|≤20)

Rewriting to match Chebychev's Inequality

P(|X-20|≤20)≤1- 1/k^2 ;k>0

Find k: P(|X-20|≤(kσ=20))≤1-1/k^2

P(|X-20|≤(k√20=20))≤1- 1/k^2

P(|X-20|≤(k=√20))≤1- 1/k^2

P(|X-20|≤20)≤1-1/20

P(|X-20|≤20)≤19/20

The probability that X falls between 0 and 40 is less than or equal to 0.95.

Now, I know that my answer is wrong. I am either sloppy at inequalities, or I don't understand how to use them correctly.

In this case, the correct answer should be

P(|X-20|≤20)≥19/20.

Can anyone help me understand where my misunderstanding is?

Here's an image that may be easier to read:

Answer 1

This line is false: $P(|X-20|≤20)≤1- 1/k^2 ;k>0$.

You started with: $A=P(|X-20|\geq20)≤ 1/k^2=B$.

To get to the next line, you need to change the direction of the inequality because you multiply both side by a negative constant ($-1$):

$$A\leq B \Rightarrow 1-A \geq 1-B$$