Question

A high percentage of people who fracture or dislocate a bone see a doctor for that condition. Suppose the percentage is 99%. Consider a sample in which 300 people are randomly selected who have fractured or dislocated a bone. (i) What is the probability that exactly five of them did not see a doctor? (4 marks) (ii) What is the probability that fewer than four of them did not see a doctor? (5 marks) (iii) What is the expected number of people who would not see a doctor? (5 marks)

Answer 1

(i) 0.15708

(ii) 0.432488

(iii) 3

Explanation:

Given that, 99% of people who fracture or dislocate a bone see a doctor for that condition.

There is only two chance either the person having fracture or dislocation of bone will either see the doctor or not.

As per previous data, if one person got a fracture or dislocation of bone, the chance of seeing the doctor is 0.99. Assuming this chance is the same for every individual, so the total number of people having fractured or dislocated a bone can be considered as Bernoulli's population.

Let p be the probability of success represented by the chances of not seeing a doctor by any one individual having fractured or dislocated a bone.

So, p=1-0.99=0.01

According to Bernoulli's theorem, the probability of exactly r success among the total of n randomly selected from Bernoulli's population is

`P(r)=\binom{n}{r}p^r(1-p)^(n-r)\cdots(i)`

(i) The total number of persons randomly selected, n=400.

The probability that exactly 5 of them did not see a doctor

So, r=5 , p=0.01

Using equation (i),

`P(r=5)=\binom{400}{5}(0.01)^5(1-0.01)^(400-5)`

`=(400!)/((400-5)!* 5!)(0.01)^5(0.99)^(395)`

=0.15708

(ii) The probability that fewer than four of them did not see a doctor

`=P(r<4)`

`=P(r=0)+P(r=1)+P(r=2)+P(r=3)`

`=\binom{400}{0}(0.01)^0(0.99)^(400)+\binom{400}{1}(0.01)^1(0.99)^(399)+\binom{400}{2}(0.01)^2(0.99)^(398)+\binom{400}{3}(0.01)^3(0.99)^(397)`

`=0.017951+0.072527+0.146154+0.195856`

`=0.432488`

(iii) The expected number of people who would not see a doctor

`=np`

`=300* 0.01`

=3