Question

If 20% of the people in a community use the emergency room at a hospital in one year, find

Answer 1

Answer:

a) 87.91% probability that at most three used the emergency room

b) 20.13% probability that exactly three used the emergency room.

c) 3.28% probability that at least five used the emergency room​

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they use the emergency room, or they do not. The probability of a person using the emergency room is independent of any other person. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

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

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

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

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

And p is the probability of X happening.

Sample of 10 people:

This means that $n = 10$

20% of the people in a community use the emergency room at a hospital in one year

This means that $p = 0.2$

a) At most three used the emergency room

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.2)^{0}.(0.8)^{10} = 0.1074$

$P(X = 1) = C_{10,1}.(0.2)^{1}.(0.8)^{9} = 0.2684$

$P(X = 2) = C_{10,2}.(0.2)^{2}.(0.8)^{8} = 0.3020$

$P(X = 3) = C_{10,3}.(0.2)^{3}.(0.8)^{7} = 0.2013$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.1074 + 0.2684 + 0.3020 + 0.2013 = 0.8791$

87.91% probability that at most three used the emergency room

b) Exactly three used the emergency room

$P(X = 3) = C_{10,3}.(0.2)^{3}.(0.8)^{7} = 0.2013$

20.13% probability that exactly three used the emergency room.

c) At least five used the emergency room​

$P(X \geq 5) = 1 - P(X < 5)$

In which

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

From 0 to 3, we already have in a).

$P(X = 4) = C_{10,4}.(0.2)^{4}.(0.8)^{6} = 0.0881$

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.1074 + 0.2684 + 0.3020 + 0.2013 + 0.0881 = 0.9672$

$P(X \geq 5) = 1 - P(X < 5) = 1 - 0.9672 = 0.0328$

3.28% probability that at least five used the emergency room​