Question

Consider an experimenter that consists of counting the number of ∝- particles given off in a 1 second interval by 1 gram of radioactive material. If we know from past experience 3 on the average 3.2 such ∝- particles are given off, what is a good approximation to the probability that no more than 2 ∝- particles will appear?

Answer 1

This question is based on Pure Poisson distribution as radio active decay is based on this only..Also this is a discrete probability instance..

So according to question ,

Δt   =   1s

λ   =  α Δt   =   3.2 [Given mean value]

So

P(X <= 2)  =  e^-λ . [ (λ⁰ / 0!)  +  (λ / 1!)  + (λ² / 2!)]

                =   e^-3.2. [ ((3.2)⁰ / 0!)  +  ((3.2) / 1!) + ((3.2)² / 2)]

                =   e^-3.2 .[  1 + 3.2  + 5.12]

                =   0.380 [Correct to 3 decimal places]

Therefore P(no more than 2 particles emitted)  =  0.380

Comments

what is meant by this line then ? "If we know from past experience 3 on the average 3.2 such ∝- particles are given off, ".i thought the probability is 3/3.2

and is there any reason that this queston follow poissn distribution??as in  can you explain briefly why radio active decay is based on pission distribution??because as far as i know poissn distribution is used whenever value of 'n' is quiet large and probability is low.

and also can you tell how will we know that in a question,poission distribution has to be used if nothing is mentioned??

Thankyou

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U can assume that on an average 3.2 particles are emitted in 1s time interval..

And Pure Poisson distribution is for rate based systems as  given in the question..

Hence my assumption should be correct..

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but what is meant by 3 in " 3 on average 3.2" here ??

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I think '3' is a typo..It makes no sense simply..:)