Question

The table below shows the total number of COVID-19 tests conducted in Australia and in selected state and territory, since 22 January 2020. This figure represents the percent of all tests that have returned a positive result for COVID-19. The number of total tests conducted is updated daily but the rate of positive tests has been found to be invariant. Suppose that people independently conducted their tests following a binomial distribution and the percent of positive tests reported in each state is the corresponding probability of one person having COVID-19 (assume no false positives). Show all working out for answers.

Jurisdiction	Total Tests Conducted	Positive Tests (%)
Australia	2,713,435	0.3%
ACT	32,273	0.3%
NT	14,748	0.2%
TAS	52,934	0.4%
VIC	928,171	0.3%

Question 1 : Find the probability that we can meet one people in Tasmania who has a positive result for COVID-19 test.

Question 2 : Suppose that you just returned from a coffee shop in Launceston, where you met 5 random Tasmanians. Find the probability that one of them has COVID-19.

Question 3 : A domestic flight just landed at Launceston airport with 20 passengers. These passengers haven’t met each other before and independently took the flight. Eight (8) of them are from ACT, 5 from NT and 7 from VIC. On the day they arrived at the airport, find the probability that we can meet 1 passenger from this flight who has a positive result for COVID-19.

Answer 1

1. Probability that a person selected at random in Tasmania has a positive result for Covid-19 = 0.004

2. This can be solved as binomial random variable with p = 0.004, q = 1-0.004 = 0.996; n=5

Probability that exactly one out of the five people met has COVID-19 = C[5,1] x 0.004¹ x 0.996⁴

= 5 x 0.004 x 0.9841 = 0.0197

3. Probability that at elast one person is Covid positive = 1 - Probability that none of them is Covid positive

Probability that none of the persons from ACT is Covid positive = 0.003⁸

Probability that none of the persons from NT is Covid positive = 0.002⁵

Probability that none of the persons from VIC is Covid positive = 0.003⁷

Probability that none of the 20 persons is Covid positive = 0.003⁸ x 0.002⁵ x0.003⁷

Probability that at elast one of them is Covid positive = 1 - 0.003⁸ x 0.002⁵ x0.003⁷ = 1 - 4.5961 x 10^-52