CBSE Sample Paper 2021-22 (Term 2) · Section A · 2 marks

Q5: Number of red balls in two draws from 1 red and 3 white balls

Sample 2021-22 T2Ch 13 · Probability

The question

A bag contains 1 red and 3 white balls. Find the probability distribution of the number of red balls if 2 balls are drawn at random from the bag one-by-one without replacement.

This topic is no longer in the CBSE syllabus (2025-26).

Exam answer

Let X be the random variable defined as the number of red balls. Then X can take the values 0, 1.

Why?

The bag has only one red ball, so two draws give at most one red ball. That is why $X = 2$ is impossible.

P(X = 0) = \frac34 \times \frac23 = \frac{6}{12} = \frac12

Why?

$X = 0$ means both balls are white. First draw: 3 white balls out of 4. The ball is not put back, so 3 balls remain and 2 of them are white. Multiply:

P(W_1W_2) = P(W_1)\,P(W_2/W_1)

\begin{aligned} \frac34 \times \frac23 &= \frac{3 \times 2}{4 \times 3} = \frac{6}{12} \\ &= \frac{6 \div 6}{12 \div 6} = \frac12 \end{aligned}

One red ball: red then white, or white then red.

\begin{aligned} P(X = 1) &= \frac14 \times \frac33 + \frac34 \times \frac13 \\ &= \frac{6}{12} = \frac12 \end{aligned}

Why?

Red first has probability $\frac14$ ⁠; then all 3 balls left are white, giving $\frac33$ ⁠. White first has probability $\frac34$ ⁠; then 1 of the 3 balls left is red, giving $\frac13$ ⁠.

The two orders cannot both happen, so their probabilities are added.

\begin{aligned} \frac14 \times \frac33 &= \frac{3}{12} \\ \frac34 \times \frac13 &= \frac{3}{12} \\[4pt] \frac{3}{12} + \frac{3}{12} &= \frac{6}{12} = \frac12 \end{aligned}

\begin{array}{c|c|c} X & 0 & 1 \\ \hline P(X) & \frac12 & \frac12 \end{array}

Why?

The table lists each value of X with its probability. Check: $\frac12 + \frac12 = 1$ ⁠, as the probabilities of a distribution must add to 1.

$\therefore$ The probability distribution of X is given by the table above.

Where this comes from

Chapter 13, Probability: Probability distribution of a random variable.

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