Statistics, Probability

Question

In a certain country, 40% of the population has covid-19. If one person at a time is selected at random, what is the probability that

1. The 1st person found with covid-19 is the 5th person selected?
2. The 5th person not having covid-19 is the 7th person selected?
3. Let Y be the number of people sampled in order to get 5 people with covid-19. Find V(Y).

If a random sample of 20 people is taken, find the probability that

(1) At most 5 have covid-19

(2) Either 3, 4, 5, 6 or 7 have covid-19

(3) At most 15 will not have covid-19

Answer 1

p = .4 q = 1 - .4 = .6

1. This is the geometric distribution: n failures before a success = qⁿp

1st person with Covid is the 5the person means 4 non-Covids before Covid has probability .6⁴*.4 = .05184

2 This is the negative binomial. P(2 successes before fifth failure) is C(6,2)p²q⁵ = 15 * .4²*.6⁵ = 0.186624

3 Again, we use the negative binomial. Here, we have the fifth with Covid, not the fifth without. 5q/p^2 = 5 * .6/.4² = 18.75

For the next three problems, we use the binomial distribution.

The sample is size 20, so n = 20

P(5 or less successes) = from Excel (and most calculators have similar functions, and you will also see this in tables), binom.dist(5, 20, .4, 1) - in the last position, 1 means cumulative -

.1256

P( X = 3, 4, 5, 6, 7) = P(X <= 7) - P(X <= 2) - notice that it is 2 in the second expression, not 3.

binom.dist(7, 20, .4, 1) - binom.dist(2, 20, .4, 1)

.4123

P(at most 15 will not have Covid) - we can solve this in 2 ways; we can find the probability that at least 5 will have Covid, which equals 1 - P(4 or less have Covid), or we can find the probability as in the first two parts by just switching p and q

1 - binom.dist(4, 20, .4, 1) = .9490

binom.dist(15, 20, .6, 1) = .9490