110k views
1 vote
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).

User Kevin Cook
by
5.1k points

1 Answer

3 votes

Answer:

a) 0.2182

b) 0.0691

c) 0.9309

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) = (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

User Intcreator
by
4.8k points