Suppose the probability of contracting a certain disease is 1 in 27,124 for a new case in a given year. Approximate the probability that in a town of 5,756 people there will be at least one new case of the disease next year.

The probability that there is at least one new case of the disease is approximately 0.19121

Step-by-step explanation:

Given that the probability of contracting a disease is p = 1/27124.

In an experiment n = 5756, we want to find the probability that there will be at least one new case of the disease. That is P(X ≥ 1).

If x is approximated Binomial(n, p)

Then

P(X = x) = (nCx)(p^x)q^(n - x)

Where q = 1 - p

Here, q = 1 - (1/27124) = 27123/27124

And nCx, read as "n combination x"

is given as n!/(n - x)! x!

Also note that

P(X ≥ a) = 1 - P(X < a)

So, it is sufficient to find P(X < 1) for this problem.

P(X < 1) = P(X = 0)

= (5756C0)(1/27124)^0(27123/27124)^(5756 - 0)

= 1 × 1 × 0.80879

≈ 0.80879

Now,

P(X ≥ 1) = 1 - 0.80879 = 0.19121

ian gonna lie that is way to much

the expression (1.26)2t reveals the approximate rate of increases in the number of plants semiannually.

hope this : )