One nanogram of  Plutonium-239 will have an average of 2.3 radioactive decays per second. What is the probability that:(i)   In a 2 second period there are exactly 3 radioactive decays?  (Give your answer correct to 3 decimal places)   (ii)  There are more than 3 radioactive decays in a particular second?  (Give your answer correct to 3 decimal places)

The problem can be solved using the Poisson distribution, which is a probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space.

(i) In a 2 second period there are exactly 3 radioactive decays:

The average rate of decay is 2.3 decays per second, so in 2 seconds, we would expect 2*2.3 = 4.6 decays on average. We can use the formula for the Poisson distribution to find the probability of exactly 3 decays in 2 seconds:

P(X=k) = (λ^k * e^-λ) / k!

where λ is the average rate (4.6), k is the number of events we are interested in (3), and e is the base of the natural logarithm (approximately 2.71828).

P(X=3) = (4.6^3 * e^-4.6) / 3! = (97.336 * 0.010052) / 6 = 0.163

So, the probability of exactly 3 decays in 2 seconds is approximately 0.163 or 16.3%.

(ii) There are more than 3 radioactive decays in a particular second:

For this part, we need to find the probability of 4 or more decays in 1 second. This is the same as 1 minus the probability of 3 or fewer decays. We can use the Poisson distribution formula again, but this time we need to sum the probabilities for 0, 1, 2, and 3 decays:

P(X>3) = 1 - P(X≤3) = 1 - [P(X=0) + P(X=1) + P(X=2) + P(X=3)] = 1 - [(2.3^0 * e^-2.3) / 0! + (2.3^1 * e^-2.3) / 1! + (2.3^2 * e^-2.3) / 2! + (2.3^3 * e^-2.3) / 3!] = 1 - [(1 * 0.10026) + (2.3 * 0.10026) + (2.652 * 0.10026) + (2.039 * 0.10026)] = 1 - [0.10026 + 0.2306 + 0.2658 + 0.2044] = 1 - 0.801

So, the probability of more than 3 decays in a particular second is approximately 0.199 or 19.9%.