Question

Let B ∼ Bin(20,0.2). Compute the following probabilities. I would suggest computing these with a hand calculator using the formula provided in class (you will not have R on the exams!), but you can check your answers using R if you wish.
(a) P(B=4).

(b) P(B≤1).

(c) P(B>1).

Answer 1

Answer:

a) 0.2182

b) 0.0691

c) 0.9309

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly b successes on n repeated trials, and B can only have two outcomes.

$P(B = b) = C_{n,b}.p^{b}.(1-p)^{n-b}$

In which $C_{n,b}$ is the number of different combinations of b objects from a set of n elements, given by the following formula.

$C_{n,b} = \frac{n!}{x!(n-b)!}$

And p is the probability of B happening.

In this problem we have that:

Bin(20,0.2).

This means that $n = 20, p = 0.2$

(a) P(B=4).

$P(B = b) = C_{n,b}.p^{b}.(1-p)^{n-b}$

$P(B = 4) = C_{20,4}.(0.2)^{4}.(0.8)^{16} = 0.2182$

(b) P(B≤1).

$P(B \leq 1) = P(B = 0) + P(B = 1)$

$P(B = b) = C_{n,b}.p^{b}.(1-p)^{n-b}$

$P(B = 0) = C_{20,0}.(0.2)^{0}.(0.8)^{20} = 0.0115$

$P(B = 1) = C_{20,1}.(0.2)^{1}.(0.8)^{19} = 0.0576$

$P(B \leq 1) = P(B = 0) + P(B = 1) = 0.0115 + 0.0576 = 0.0691$

(c) P(B>1).

Either B is less than or equal to 1, or B is larger than 1. The sum of the probabilities of these events is decimal 1. So

$P(B \leq 1) + P(B > 1) = 1$

We have that, from b), $P(B \leq 1) = 0.0691$

So

$0.0691 + P(B > 1) = 1$

$P(B > 1) = 0.9309$