Answer:
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
