Question

5. 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 theprobability that a 2cm2sample of this zircon will reveal at most three tracks,thus leading to an underestimation of the age of the material

Answer 1

The probability that a 2cm2 sample of a zircon will reveal at most three tracks can be calculated using the Poisson distribution. The average number of tracks is five, so we can use the Poisson distribution formula to find the probability. The probability is approximately 0.1404, or 14.04%.

Step-by-step explanation:

To determine the probability of finding at most three tracks on a 2cm2 sample of the zircon, we need to use the Poisson distribution. The Poisson distribution is used to model the number of events that occur in a fixed interval of time or space when these events occur with a known average rate and independently of the time since the last event.

In this case, the average number of tracks per square centimeter is five. The probability of finding at most three tracks can be calculated using the Poisson distribution formula:

P(X ≤ 3) = e-λ(λ0/0! + λ1/1! + λ2/2! + λ3/3!)

Substituting λ = 5:

P(X ≤ 3) = e-5(50/0! + 51/1! + 52/2! + 53/3!)

Calculating this gives us a probability of approximately 0.1404, or 14.04%.

Answer 2

the probability that a 2cm² sample of the zircon will reveal at most three tracks is 0.0104

Step-by-step explanation:

Given that; 7

uranium fission tracks occur on average 5 per every square centimeter

i.e λ = 5

the required centimeter square s = 2

let X be the number of uranium fission tracks occur on the average per every square centimeter

therefore X follows poisson distribution

so the average number of tracks occur on 2-centimeter is;

k = λs

k = 5 × 2

k = 10

the required probability is;

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

= [ (e⁻¹⁰(10)⁰) / 0! ] + [ (e⁻¹⁰(10)¹) / 1! ] + [ (e⁻¹⁰(10)²) / 2! ] + [ (e⁻¹⁰(10)³) / 3! ]

= 0.00004 + 0.0005 + 0.0023 + 0.0076

= 0.0104

therefore the probability that a 2cm² sample of the zircon will reveal at most three tracks is 0.0104