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 0.004^{1} x 0.996^{4}

= 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
= 0.003^{8}

Probability that none of the persons from NT is Covid positive =
0.002^{5}

Probability that none of the persons from VIC is Covid positive
= 0.003^{7}

Probability that none of the 20 persons is Covid positive =
0.003^{8} x 0.002^{5} x0.003^{7}

Probability that at elast one of them is Covid positive = 1 -
0.003^{8} x 0.002^{5} x0.003^{7} = 1 -
4.5961 x 10^{-52}

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