Answer
Verified
438.3k+ views
Hint: 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
\[\begin{align}
& P\left( X=2 \right)=9p\left( X=4 \right)+90p\left( X=6 \right) \\
& \dfrac{{{\lambda }^{2}}{{e}^{-\lambda }}}{2!}=9\times \dfrac{{{\lambda }^{4}}{{e}^{-\lambda }}}{4!}+90\times \dfrac{{{\lambda }^{6}}{{e}^{-\lambda }}}{6!} \\
\end{align}\]
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:
\[\begin{align}
& 1=\dfrac{3{{\lambda }^{2}}}{4}+\dfrac{{{\lambda }^{4}}}{4} \\
& 4=3{{\lambda }^{2}}+{{\lambda }^{4}} \\
& {{\lambda }^{4}}+3{{\lambda }^{2}}-4=0 \\
\end{align}\]
Now, according to options:
\[\begin{align}
& \lambda =3,{{\left( 3 \right)}^{4}}+3{{\left( 3 \right)}^{2}}-4=50\left( \text{Wrong} \right) \\
& \lambda =2,{{\left( 2 \right)}^{4}}+3{{\left( 2 \right)}^{2}}-4=24\left( \text{Wrong} \right) \\
& \lambda =1,{{\left( 1 \right)}^{4}}+3{{\left( 1 \right)}^{2}}-4=0\left( \text{Right} \right) \\
& \lambda =0,{{\left( 0 \right)}^{4}}+3{{\left( 0 \right)}^{2}}-4=-4\left( \text{Wrong} \right) \\
\end{align}\]
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.
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
\[\begin{align}
& P\left( X=2 \right)=9p\left( X=4 \right)+90p\left( X=6 \right) \\
& \dfrac{{{\lambda }^{2}}{{e}^{-\lambda }}}{2!}=9\times \dfrac{{{\lambda }^{4}}{{e}^{-\lambda }}}{4!}+90\times \dfrac{{{\lambda }^{6}}{{e}^{-\lambda }}}{6!} \\
\end{align}\]
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:
\[\begin{align}
& 1=\dfrac{3{{\lambda }^{2}}}{4}+\dfrac{{{\lambda }^{4}}}{4} \\
& 4=3{{\lambda }^{2}}+{{\lambda }^{4}} \\
& {{\lambda }^{4}}+3{{\lambda }^{2}}-4=0 \\
\end{align}\]
Now, according to options:
\[\begin{align}
& \lambda =3,{{\left( 3 \right)}^{4}}+3{{\left( 3 \right)}^{2}}-4=50\left( \text{Wrong} \right) \\
& \lambda =2,{{\left( 2 \right)}^{4}}+3{{\left( 2 \right)}^{2}}-4=24\left( \text{Wrong} \right) \\
& \lambda =1,{{\left( 1 \right)}^{4}}+3{{\left( 1 \right)}^{2}}-4=0\left( \text{Right} \right) \\
& \lambda =0,{{\left( 0 \right)}^{4}}+3{{\left( 0 \right)}^{2}}-4=-4\left( \text{Wrong} \right) \\
\end{align}\]
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
10 examples of friction in our daily life
One cusec is equal to how many liters class 8 maths CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
What is the type of food and mode of feeding of the class 11 biology CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
At which age domestication of animals started A Neolithic class 11 social science CBSE
One cusec is equal to how many liters class 8 maths CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
What organs are located on the left side of your body class 11 biology CBSE
Can anyone list 10 advantages and disadvantages of friction
Difference between physical and chemical change class 11 chemistry CBSE
How do you graph the function fx 4x class 9 maths CBSE
What happen to the gravitational force between two class 11 physics CBSE