Answer:
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
Answer:
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
