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

d is the only accurate one

step-by-step explanation:

тgє qυєѕтισи αѕкє∂: ιи тнє ѕєяιєѕ σя яєα¢тισиѕ тgαт σ¢¢υя ιи ρнσтσѕуитнєѕιѕ, ωнι¢н fσямѕ ωнєи ℓιgнт ѕтяιкєѕ α мσℓє¢υℓє σf ¢нℓσяσρнуℓℓ.

α. мσℓє¢υℓє σf ωαтєя

в.тнєє-¢αявσи ѕυgαя

¢. "єχ¢ιтє∂" єℓє¢тяσи

∂. иα∂ρн

му αиѕωєя: ιт ιѕ ¢. "єχ¢ιтє∂" єℓє¢тяσи

нσρє ι ¢συℓ∂ нєℓρ υ