A random variable $X$ has the following probability mass function. $$\begin{array}{c|ccccc} x & 1 & 2 & 3 & 4 & 5\\\hline f(x) & k^2 & 2k^2 & 3k^2 & 2k & 3k\end{array}$$ Find (i) the value of $k$ (ii) $P(2\le X<5)$ (iii) $P(3<X)$.

A random variable X has the following probability mass function. c|ccccc x 1 2 3 4 5; hline f(x) k² 2k² 3k² 2k

Exercise 11.2 · Q6

X

(i) the value of

k

(ii)

P(2\le X<5)

(iii)

P(3<X)

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Summing the table to $1$ gives a quadratic in $k$ ; the root that keeps every probability non-negative is chosen, and the two probability questions are then sums of the resulting pmf.

Step 1. Sum the pmf to $1$ . $k^2+2k^2+3k^2+2k+3k=6k^2+5k=1$ , i.e. $6k^2+5k-1=0$ .

Step 2. Solve the quadratic. $k=\dfrac{-5\pm\sqrt{25+24}}{12}=\dfrac{-5\pm7}{12}$ , giving $k=\dfrac2{12}=\dfrac16$ or $k=-1$ .

Step 3. Reject the invalid root. $f(4)=2k$ and $f(5)=3k$ must be $\ge0$ (Theorem 11.1(i)), which forces $k\ge0$ ; so $k=-1$ is rejected and $k=\dfrac16$ . …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2025Set ANNUAL1 markMCQ
$P(X=i)=k\,P(X=i-1)$
(a) $3$
(b) $1$
(c) $4$
(d) $2$
›Reveal solutionSolution
Building all three probabilities in terms of $k$ from the given recursive relation and using $\sum P(X=i)=1$ gives a quadratic in $k$ , of which only the positive root is valid.

Given $P(X=0)=\dfrac17$ .

For $i=1$ : $P(X=1)=k\,P(X=0)=\dfrac k7$ .

For $i=2$ : $P(X=2)=k\,P(X=1)=k\cdot\dfrac k7=\dfrac{k^2}7$ .

Since $X$ only takes values $0,1,2$ , the probabilities must sum to $1$ : $\dfrac17+\dfrac k7+\dfrac{k^2}7=1$ . …
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CBSE 2023Set ANNUAL1 markMCQ
$0, 1, 2$
(a) $3$
(b) $1$
(c) $4$
(d) $2$
›Reveal solutionSolution
Building up $P(1)$ and $P(2)$ from $P(0)=1/7$ via the recurrence $P(i)=kP(i-1)$ , then using $\sum P(x)=1$ to solve for $k$ .

Given $P(X=0)=\dfrac17$ and $P(X=i)=kP(X=i-1)$ for $i=1,2$ .

$P(X=1)=kP(X=0)=\dfrac{k}{7}$ .

$P(X=2)=kP(X=1)=k\cdot\dfrac{k}{7}=\dfrac{k^2}{7}$ . …
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CBSE 2018Set ANNUAL1 markMCQ
$-2$
(a) $3$
(b) $1$
(c) $4$
(d) $2$
›Reveal solutionSolution
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: $\dfrac{2\lambda+3\lambda+\lambda}{12}=\dfrac{6\lambda}{12}=1$ . …
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CBSE 2017Set ANNUAL1 markMCQ
$\dfrac{1}{4}$
(a) $\dfrac{10}{21}$
(b) $\dfrac{2}{7}$
(c) $\dfrac{1}{14}$
(d) $\dfrac{1}{2}$
›Reveal solutionSolution
Using $\sum P(X=x)=1$ to find $a=\dfrac{1}{28}$ , then $P(1\le X\le4)=14a=\dfrac12$ .

Sum of all probabilities equals 1: $\dfrac14+2a+3a+4a+5a+\dfrac14=1$ .

Simplify: $\dfrac12+14a=1\Rightarrow14a=\dfrac12\Rightarrow a=\dfrac{1}{28}$ . …
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CBSE 2016Set ANNUAL1 markMCQ
$k^2$
(a) $\dfrac{1}{8}$
(b) $\dfrac{1}{10}$
(c) $0$
(d) $-1$ or $\dfrac{1}{10}$
›Reveal solutionSolution
Solving $\sum P(X=x)=1$ gives $k=\dfrac1{10}$ (the negative root is rejected as it makes some probabilities negative).

Sum all the probabilities and set the total equal to $1$ : $0+k+2k+2k+3k+k^2+2k^2+(7k^2+k)=1$ .

Collect linear terms: $k+2k+2k+3k+k=9k$ . Collect quadratic terms: $k^2+2k^2+7k^2=10k^2$ .

So $10k^2+9k=1$ , i.e. $10k^2+9k-1=0$ .

By the quadratic formula: $k=\dfrac{-9\pm\sqrt{81+40}}{20}=\dfrac{-9\pm\sqrt{121}}{20}=\dfrac{-9\pm11}{20}$ .

This gives $k=\dfrac{2}{20}=\dfrac1{10}$ or $k=\dfrac{-20}{20}=-1$ . …
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X	$-2$	$3$	$1$
$P(X=x)$	$\dfrac{\lambda}{6}$	$\dfrac{\lambda}{4}$	$\dfrac{\lambda}{12}$

X	0	1	2	3	4	5
P(X=x)	$\dfrac{1}{4}$	$2a$	$3a$	$4a$	$5a$	$\dfrac{1}{4}$

X	0	1	2	3	4	5	6	7
P(X=x)	0	k	2k	2k	3k	$k^2$	$2k^2$	$7k^2+k$

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Q.A random variable $X$ has the following probability mass function.
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Find

Concept understanding — Probability Mass Function

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Q.A random variable $X$ has the following probability mass function.
[!FORMULA] $\begin{array}{c|ccccc} x & 1 & 2 & 3 & 4 & 5\\\hline f(x) & k^2 & 2k^2 & 3k^2 & 2k & 3k\end{array}$
Find

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