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

3: 2

step-by-step explanation:

shorter piece is 2: 3, so just reverse it, and you get 3: 2