Question #140631
Question
a. A radioactive source is emitting, on the average, one particle per minute. If counting continues for several hundred minutes, during which time the particles are emitted randomly, in what proportion of these minutes is to be expected that
i. There will be 2 or more particles emitted? (3 Marks)
ii. There will be 2 or more particles emitted in 2 minutes? (3 Marks)
1
Expert's answer
2020-10-27T19:56:46-0400

Let X=X= the number of particles emitted: XPo(λt)X\sim Po(\lambda t)


P(X=x)=eλt(λt)xx!,=0,1,2,...P(X=x)=\dfrac{e^{-\lambda t}(\lambda t) ^x}{x!}, =0, 1, 2, ...

i. λt=1\lambda t=1


P(X2)=1P(X=0)P(X=1)=P(X\geq2)=1-P(X=0)-P(X=1)==1e1100!e1111!=12e1=1-\dfrac{e^{-1}\cdot1^0}{0!}-\dfrac{e^{-1}\cdot1 ^1}{1!}=1-2e^{-1}

i. λt=1(2)=2\lambda t=1(2)=2


P(X2)=1P(X=0)P(X=1)=P(X\geq2)=1-P(X=0)-P(X=1)==1e2200!e2211!=13e2=1-\dfrac{e^{-2}\cdot2^0}{0!}-\dfrac{e^{-2}\cdot2 ^1}{1!}=1-3e^{-2}

12e113e2=e22ee230.445\dfrac{1-2e^{-1}}{1-3e^{-2}}=\dfrac{e^2-2e}{e^2-3}\approx0.445


13e212e1=e23e22e2.248\dfrac{1-3e^{-2}}{1-2e^{-1}}=\dfrac{e^2-3}{e^2-2e}\approx2.248



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