Geophysicists determine the age of a zircon by counting the number ofuranium fission tracks on a polished surface. A particular zircon is of such anage that the average number of tracks per square centimeter is five. What is the probability that a 2cm^2 sample of this zircon will reveal at most three tracks, thus leading to an underestimation of the age of the material?

p (x = 3, λ = 5) = 0.14044

Step-by-step explanation:

Given

λ = 5 (the average number of tracks per square centimeter)

ε = 2.718 (constant value)

x = 3 (the variable that denotes the number of successes that we want to occur)

p (x,λ) = probability of x successes, when the average number of occurrences of them is λ

We can use the equation

p (x,λ) = λˣ*ε∧ (-λ) / x!

⇒ p (x = 3, λ = 5) = (5) ³ * (2.718) ⁻⁵/3!

⇒ p (x = 3, λ = 5) = 0.14044

0.0108

Step-by-step explanation:

Let X denote the number of uranium fission tracks occurring on the average 5 per square centimetre. We need to find the probability that a 2cm² sample of this zircon will reveal at most three tracks. X follows Poisson distribution, λ = 5 and s = 2.

k = λs = 5*2 = 10

Since we need to reveal at most three tracks the required probability is:

P (X≤3) = P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3)

P (X≤3) = (((e^ -10) * (10) ⁰) / 0!) + (((e^ -10) * (10) ¹) / 1! + (((e^ -10) * (10) ²) / 2! + (((e^ -10) * (10) 3) / 3!

P (X≤3) = 0.0004 + 0.0005 + 0.0023 + 0.0076

P (X≤3) = 0.0108

Therefore, the probability that a 2cm² sample of this zircon will reveal at most three tracks is 0.0108