A random variable X has the following probability mass function : | X | $-2$ | $3$ | $1$ | |---|---|---|---| | $P(X=x)$ | $\dfrac{\lambda}{6}$ | $\dfrac{\lambda}{4}$ | $\dfrac{\lambda}{12}$ | Then the value of $\lambda$ is : (a) $3$ (b) $1$ (c) $4$ (d) $2$

> [!TLDR] > Using $\sum P(X=x)=1$ for the given pmf gives $\lambda=2$. > > 1. The probability mass function must satisfy $\displaystyle\sum_x P(X=x)=1$. > 2. So $\dfrac{\lambda}{6}+\dfrac{\lambda}{4}+\dfrac{\lambda}{12}=1$. > 3. Take LCM of $6,4,12$, which is $12$: $\dfrac{2\lambda}{12}+\dfrac{3\lambda}{12}+\dfrac{\lambda}{12}=1$. > 4. Combine: $\dfrac{2\lambda+3\lambda+\lambda}{12}=\dfrac{6\lambda}{12}=1$. …

A random variable X has the following probability – Class 12 Mathematics

Question 70 of 105

A random variable X has the following probability mass function :

X	$-2$	$3$	$1$
$P(X=x)$	$\dfrac{\lambda}{6}$	$\dfrac{\lambda}{4}$	$\dfrac{\lambda}{12}$

Then the value of $\lambda$ is :

Puducherry TnboardTamil Nadu HSC (DGE) Board 2018MCQ· 1mImportance★★★★★

67% · 70/105 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Using $\sum P(X=x)=1$ for the given pmf gives $\lambda=2$ .

The probability mass function must satisfy $\displaystyle\sum_x P(X=x)=1$ .
So $\dfrac{\lambda}{6}+\dfrac{\lambda}{4}+\dfrac{\lambda}{12}=1$ .
Take LCM of $6,4,12$ , which is $12$ : $\dfrac{2\lambda}{12}+\dfrac{3\lambda}{12}+\dfrac{\lambda}{12}=1$ .
Combine: $\dfrac{2\lambda+3\lambda+\lambda}{12}=\dfrac{6\lambda}{12}=1$ . …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ14 — Verify f(x) = 30x⁴ e-6x^5 ; x > 0; 0 ; Otherwise for p.d.f. If f(x) is…→Next questionQ16 — If, in a Poisson distribution P(X=0) = k then the variance is : (a) eλ…

🎓Unlock everything free for 14 days

Unlock everything free for 14 days