Question

If X is a Poisson variate such that $P\left( X=2 \right)=9p\left( X=4 \right)+90p\left( X=6 \right)$ then the mean of x is

& \text{A}.\text{ 3} \\\ & \text{B}.\text{ 2} \\\ & \text{C}.\text{ 1} \\\ & \text{D}.\text{ }0 \\\ \end{aligned}$$

Answer 1

The concept in the question is 'Poisson distribution' concept. Basically, the probability of an event's for a Poisson distribution. Put the formula of $\text{P}\left( \text{k events in interval} \right)=\dfrac{{{\lambda }^{k}}{{e}^{-\lambda }}}{k!}$ in the given condition of the question and then solve the calculation part and finally we will get polynomial in $\lambda$ (which is average number of events per interval or mean value). According to the options given in the question put the value of $\lambda$ in the polynomial. Those which satisfies the polynomial will be the answer. We can also solve the problem (polynomial by factorization also).

Complete step-by-step solution:
Now, look over the given condition we have:
$P\left( X=2 \right)=9p\left( X=4 \right)+90p\left( X=6 \right)\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)}$
The probability of observing K events in an interval is given by the equation:
$\text{P}\left( \text{k events in interval} \right)=\dfrac{{{\lambda }^{k}}{{e}^{-\lambda }}}{k!}$
Where, $\lambda$ = Average number of events per interval or mean value.
e = Euler’s number (2.71626...)
k takes value 0, 1, 2 ...
$\text{k}!=\text{k}\times \left( k-1 \right)\times \left( k-2 \right)\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. 2}\times \text{1}$is factorization of K.
From equation (i) we have:
Here k = X

& P\left( X=2 \right)=9p\left( X=4 \right)+90p\left( X=6 \right) \\\ & \dfrac{{{\lambda }^{2}}{{e}^{-\lambda }}}{2!}=9\times \dfrac{{{\lambda }^{4}}{{e}^{-\lambda }}}{4!}+90\times \dfrac{{{\lambda }^{6}}{{e}^{-\lambda }}}{6!} \\\ \end{aligned}$$ Cancelling ${{e}^{-\lambda }}$ from LHS and RHS, we get: $$\dfrac{{{\lambda }^{2}}}{2!}=9\times \dfrac{{{\lambda }^{4}}}{4!}+90\times \dfrac{{{\lambda }^{6}}}{6!}$$ Now, $$\dfrac{{{\lambda }^{2}}}{2!}={{\lambda }^{2}}\left( \dfrac{9\times {{\lambda }^{2}}}{4\times 3\times 2!}+\dfrac{90\times {{\lambda }^{4}}}{6\times 5\times 4\times 3\times 2!} \right)$$ Cancelling ${{\lambda }^{2}}\text{ and }2!$ from both sides of equation we get: $$1=\dfrac{9\times {{\lambda }^{2}}}{4\times 3}+\dfrac{90\times {{\lambda }^{4}}}{90\times 4}$$ Dividing $\dfrac{9}{3}$ and cancelling 90, we get: $$\begin{aligned} & 1=\dfrac{3{{\lambda }^{2}}}{4}+\dfrac{{{\lambda }^{4}}}{4} \\\ & 4=3{{\lambda }^{2}}+{{\lambda }^{4}} \\\ & {{\lambda }^{4}}+3{{\lambda }^{2}}-4=0 \\\ \end{aligned}$$ Now, according to options: $$\begin{aligned} & \lambda =3,{{\left( 3 \right)}^{4}}+3{{\left( 3 \right)}^{2}}-4=50\left( \text{Wrong} \right) \\\ & \lambda =2,{{\left( 2 \right)}^{4}}+3{{\left( 2 \right)}^{2}}-4=24\left( \text{Wrong} \right) \\\ & \lambda =1,{{\left( 1 \right)}^{4}}+3{{\left( 1 \right)}^{2}}-4=0\left( \text{Right} \right) \\\ & \lambda =0,{{\left( 0 \right)}^{4}}+3{{\left( 0 \right)}^{2}}-4=-4\left( \text{Wrong} \right) \\\ \end{aligned}$$ **Hence, option C is the correct answer.** **Note:** The Poisson distribution works properly if the following assumptions are true: 1) k is the number of times an event occurs in an interval and k can take values 0, 1, 2,... 2) The occurrence of one event does not affect the probability that a second event will occur. That is, events occur independently. 3) The average rate at which events occur is independent of any occurrences. For simplicity, this is usually assumed to be constant, but may in practice vary with time. 4) Two events cannot occur at exactly the same instant; instead, at each very small sub-interval, exactly one event either occurs or does not occur. But, if we don't know the terms and variables used in the formula properly then, there will be some confusion and we will get the wrong answer corresponding to some other variables.