CBSE Sample Paper 2023-24 · Section C · 3 marks

Q27: Probability distribution $k, 2k, 3k$ ⁠: find $k$ ⁠, $P(X < 2)$ and $P(X > 2)$

Sample 2023-24Ch 13 · Probability

The question

The random variable $X$ has a probability distribution $P(X)$ of the following form, where ' $k$ ' is some real number:

P(X) = \begin{cases} k, & \text{if } x = 0 \\ 2k, & \text{if } x = 1 \\ 3k, & \text{if } x = 2 \\ 0, & \text{otherwise} \end{cases}

(i) Determine the value of $k$ ⁠.

(ii) Find $P(X < 2)$ ⁠.

(iii) Find $P(X > 2)$ ⁠.

This question uses a random variable and its probability distribution. This topic is no longer in the CBSE syllabus (2025-26).

Exam answer

(i) The sum of the probabilities is 1:

\sum P(X_i) = 1 \Rightarrow k + 2k + 3k = 1

Why?

In any probability distribution the probabilities of all the values add up to 1. $X$ takes the values 0, 1 and 2 with probabilities $k$ ⁠, $2k$ and $3k$ ⁠, and every other value has probability 0.

\Rightarrow 6k = 1

Why?

$k + 2k + 3k = 6k$ ⁠. Dividing by 6 gives $k = \frac16$ ⁠.

\begin{aligned} k + 2k + 3k &= (1 + 2 + 3)k = 6k \\ 6k &= 1 \\ k &= \frac16 \quad (\text{divide by } 6) \end{aligned}

Check: the three probabilities $\frac16$ ⁠, $\frac26$ ⁠, $\frac36$ all lie between 0 and 1, as probabilities must.

$\therefore\ k = \frac16$

(ii)

\begin{aligned} &P(X < 2) \\ &= P(X = 0) + P(X = 1) \\ &= k + 2k = 3k \\ &= 3 \times \frac16 = \frac12 \end{aligned}

Why?

Among the values 0, 1 and 2, the ones less than 2 are 0 and 1. $X$ cannot be 0 and 1 at the same time, so these probabilities are added.

Watch out: " $X < 2$ " does not include $X = 2$ ⁠. That would be $X \le 2$ ⁠.

$\therefore\ P(X < 2) = \frac12$

(iii) $X$ takes only the values 0, 1 and 2, and $P(X) = 0$ otherwise.

Why?

$X > 2$ would need a value such as 3 or 4, and all of those have probability 0.

The same answer from the complement:

\begin{aligned} P(X > 2) &= 1 - P(X \le 2) \\ &= 1 - 6k = 0 \end{aligned}

Here $P(X \le 2) = k + 2k + 3k = 6k$ ⁠, and

6k = 6 \times \frac16 = 1, \quad 1 - 1 = 0

$\therefore\ P(X > 2) = 0$ ⁠.

Where this comes from

Chapter 13, Probability: Random variable: probability distribution.

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Q27: Probability distribution k,2k,3kk, 2k, 3kk,2k,3k⁠: find kkk⁠, P(X<2)P(X < 2)P(X<2) and P(X>2)P(X > 2)P(X>2)

Q27: Probability distribution $k, 2k, 3k$ ⁠: find $k$ ⁠, $P(X < 2)$ and $P(X > 2)$