CBSE Board Examination 2023 · Section C · 3 marks

Q30: Find $k$ ⁠, $P(1 \le X < 3)$ and $E(X)$ ⁠, or $P(A)$ and $P(B)$ for independent events

Board 2023Ch 13 · Probability

The question

(a) The probability distribution of a random variable X is given below :

\begin{array}{|c|c|c|c|} \hline X & 1 & 2 & 3 \\ \hline P(X) & \dfrac{k}{2} & \dfrac{k}{3} & \dfrac{k}{6} \\[6pt] \hline \end{array}

(i) Find the value of k.

(ii) Find $P(1 \le X < 3)$ ⁠.

(iii) Find E(X), the mean of X.

(b) A and B are independent events such that $P(A \cap \overline{B}) = \dfrac14$ and $P(\overline{A} \cap B) = \dfrac16$ ⁠. Find P(A) and P(B).

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

Exam answer

(a)

(i) The sum of the probabilities is 1:

\frac{k}{2} + \frac{k}{3} + \frac{k}{6} = 1 \Rightarrow k = 1

Why?

In a probability distribution, the probabilities of all the values add up to 1.

Use the common denominator 6:

\begin{aligned} \frac{k}{2} &= \frac{3k}{6},\quad \frac{k}{3} = \frac{2k}{6} \\[4pt] \frac{3k + 2k + k}{6} &= \frac{6k}{6} = k \end{aligned}

The sum must be 1, so $k = 1$ ⁠.

(ii)

\begin{aligned} &P(1 \le X < 3) \\ &= P(X = 1) + P(X = 2) \\ &= \frac{k}{2} + \frac{k}{3} = \frac{5k}{6} = \frac56 \end{aligned}

Why?

$1 \le X < 3$ includes $X = 1$ and $X = 2$ ⁠, but not $X = 3$ ⁠, because the second sign is strict.

With denominator 6, $\frac{3k}{6} + \frac{2k}{6} = \frac{5k}{6}$ ⁠. Then put $k = 1$ ⁠.

\begin{aligned} \frac{k}{2} + \frac{k}{3} &= \frac{3k}{6} + \frac{2k}{6} \\ &= \frac{5k}{6} \\ &= \frac{5(1)}{6} = \frac56 \end{aligned}

(iii)

\begin{aligned} E(X) &= \sum p_ix_i \\ &= 1\cdot\frac{k}{2} + 2\cdot\frac{k}{3} + 3\cdot\frac{k}{6} \\ &= \frac{k}{2} + \frac{2k}{3} + \frac{k}{2} = \frac{5k}{3} \end{aligned}

Why?

The mean multiplies each value by its probability, and adds:

E(X) = \sum p_ix_i

The three products:

\begin{aligned} 1\cdot\frac{k}{2} &= \frac{k}{2} \\ 2\cdot\frac{k}{3} &= \frac{2k}{3} \\ 3\cdot\frac{k}{6} &= \frac{3k}{6} = \frac{k}{2} \end{aligned}

With denominator 6, $\frac{k}{2} = \frac{3k}{6}$ and $\frac{2k}{3} = \frac{4k}{6}$ ⁠:

\begin{aligned} \frac{3k + 4k + 3k}{6} &= \frac{10k}{6} \\ &= \frac{10k \div 2}{6 \div 2} = \frac{5k}{3} \end{aligned}

With $k = 1$ ⁠, this is $\frac53$ ⁠.

$\therefore\ k = 1$ ⁠, $P(1 \le X < 3) = \frac56$ and $E(X) = \frac53$ ⁠.

(b) A and B are independent, so $A$ ⁠, $\overline B$ and $\overline A$ ⁠, $B$ are also independent:

\begin{aligned} P(A)\,P(\overline B) &= \frac14 \\ P(\overline A)\,P(B) &= \frac16 \end{aligned}

Why?

If A and B are independent, so are A and $\overline B$ ⁠:

\begin{aligned} P(A \cap \overline B) &= P(A) - P(A \cap B) \\ &= P(A) - P(A)P(B) \\ &= P(A)\,P(\overline B) \end{aligned}

The same argument works for $\overline A$ and B. So each given probability becomes a product.

Let $P(A) = x$ and $P(B) = y$ ⁠. Then

x(1 - y) = \frac14,\quad (1 - x)y = \frac16

Why?

A probability and its complement add to 1, so $P(\overline B) = 1 - y$ and $P(\overline A) = 1 - x$ ⁠.

Subtracting, $x - y = \frac14 - \frac16 = \frac{1}{12}$ ⁠, so $y = x - \frac{1}{12}$ ⁠.

Why?

Expand the left sides: $x - xy$ and $y - xy$ ⁠. On subtracting, the $xy$ terms cancel and leave $x - y$ ⁠.

\begin{aligned} &(x - xy) - (y - xy) \\ &= x - xy - y + xy \\ &= x - y \end{aligned}

With denominator 12, $\frac{3}{12} - \frac{2}{12} = \frac{1}{12}$ ⁠.

\begin{aligned} &x\left(\frac{13}{12} - x\right) = \frac14 \\ \Rightarrow\ &12x^2 - 13x + 3 = 0 \\ \Rightarrow\ &(3x - 1)(4x - 3) = 0 \\ \Rightarrow\ &x = \frac13 \text{ or } x = \frac34 \end{aligned}

Why?

Put $y = x - \frac{1}{12}$ in $x(1 - y) = \frac14$ ⁠. Then $1 - y = \frac{13}{12} - x$ ⁠.

\begin{aligned} 1 - y &= 1 - \left(x - \frac{1}{12}\right) \\ &= 1 - x + \frac{1}{12} \\ &= \frac{12}{12} + \frac{1}{12} - x \\ &= \frac{13}{12} - x \end{aligned}

Multiply by 12: $13x - 12x^2 = 3$ ⁠, which rearranges to the quadratic.

\begin{aligned} \frac{13x}{12} - x^2 &= \frac14 \\ 12 \times \frac{13x}{12} - 12x^2 &= 12 \times \frac14 \\ 13x - 12x^2 &= 3 \\ 0 &= 12x^2 - 13x + 3 \end{aligned}

To factorise, split the middle term. $12 \times 3 = 36 = 4 \times 9$ and $4 + 9 = 13$ ⁠, so

\begin{aligned} &12x^2 - 4x - 9x + 3 \\ &= 4x(3x - 1) - 3(3x - 1) \\ &= (3x - 1)(4x - 3) \end{aligned}

A product is 0 when one factor is 0:

\begin{aligned} 3x - 1 = 0 &\Rightarrow x = \frac13 \\ 4x - 3 = 0 &\Rightarrow x = \frac34 \end{aligned}

$P(A) = \frac13 \Rightarrow P(B) = \frac13 - \frac{1}{12} = \frac14$ ⁠; $P(A) = \frac34 \Rightarrow P(B) = \frac34 - \frac{1}{12} = \frac23$ ⁠.

Why?

Use $y = x - \frac{1}{12}$ ⁠, with denominator 12:

\begin{aligned} \frac13 - \frac{1}{12} &= \frac{4}{12} - \frac{1}{12} \\ &= \frac{3}{12} = \frac14 \\[4pt] \frac34 - \frac{1}{12} &= \frac{9}{12} - \frac{1}{12} \\ &= \frac{8}{12} = \frac23 \end{aligned}

Both pairs are probabilities between 0 and 1, so both are kept. Check the first pair in the two equations:

\frac13\cdot\frac34 = \frac14,\quad \frac23\cdot\frac14 = \frac16

The second pair checks the same way: $\frac34\cdot\frac13 = \frac14$ and $\frac14\cdot\frac23 = \frac16$ ⁠.

Watch out: the question has two answers, so give both.

$\therefore\ P(A) = \dfrac13$ ⁠, $P(B) = \dfrac14$ or $P(A) = \dfrac34$ ⁠, $P(B) = \dfrac23$ ⁠.

Where this comes from

Chapter 13, Probability: Random variable and mean; independent events.

Question type: (b) At least one, or exactly one

How to spot it: click to see

(b) Type 6. At least one, or exactly one: think "not" first

Spot it: independent events, with "at least one", "the problem is solved", "neither" or "exactly one".

Here: $P(A \cap \overline B)$ and $P(\overline A \cap B)$ are the two "one but not the other" pieces. $A$ and $B$ are independent, so write them as $x(1 - y) = \frac14$ and $(1 - x)y = \frac16$ ⁠, and solve.

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Q30: Find kkk⁠, P(1≤X<3)P(1 \le X < 3)P(1≤X<3) and E(X)E(X)E(X)⁠, or P(A)P(A)P(A) and P(B)P(B)P(B) for independent events

Q30: Find $k$ ⁠, $P(1 \le X < 3)$ and $E(X)$ ⁠, or $P(A)$ and $P(B)$ for independent events