Verified answer

I expect "binomial (4, .5)" refers to the binomial distribution with

n=4, p = 0.5 = q

Hence mean μ = np = 2

and standard deviation σ = npq = 1

Now Chebyshev's inequality is:

"Let X be a random variable with expected value μ and variance σ². Then for any real number k ≥ 1

P(| x- ( μ )| <kσ) ≥ 1 - (1/k²)"

which in this case translates to

P(|X - 2| < 2*1) ≥ 1 - 1/4 = 3/4

Now the event |X-2|≥2 is the complement of |X - 2| < 2

Hence

P(|X-2|≥2) = 1 - P(|X - 2| < 2)

Now since P(|X - 2| < 2) ≥ 3/4

therefore -P(|X - 2| < 2) ≤ -3/4

and so 1-P(|X - 2| < 2) ≤ 1-3/4

i.e. P(|X-2|≥2) ≤ 1/4, or 0.25

How does this compare with exact probability? i.e. the probability that X = 0 or 4

[ |X-2| ≥ 2 means either X-2 ≥ 2, so X ≥ 4; or X-2 ≤ -2, so X ≤ 0. Of the possible values of X (0, 1, 2, 3, 4), the only ones that satisfy this are 0 and 4]

The probabilities for the binomial distribution are given by the terms in the expansion of (p+q)^n,

hence we look at the expansion of (0.5 + 0.5)⁴, and so

The full list of probabilities for this distribution is

(0.5)⁴ + 4*(0.5)³(0.5) + 6*(0.5)²(0.5)² + 4*(0.5)(0.5)³ + (0.5)⁴

P(X=0) = (0.5)⁴

Also P(X=4) = (0.5)⁴

Therefore

P(X=0) + P(X=4) = 0.125

which is less than 0.25 and so agrees with Chebyshev's inequality.