CBSE Board Examination 2022 (Term 2) · Section A · 2 marks

Q5: Number of red balls in two draws without replacement

Board 2022 T2Ch 13 · Probability

The question

Two balls are drawn at random from a bag containing 2 red balls and 3 blue balls, without replacement. Let the variable X denotes the number of red balls. Find the probability distribution of X.

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

Exam answer

X can take the values 0, 1, 2.

Why?

Only 2 balls are drawn, so the number of red balls among them can be none, one or both.

P(X = 0) = \frac35 \times \frac24 = \frac{3}{10}

Why?

$X = 0$ means both balls are blue. The first draw is blue with probability $\frac35$ (3 blue out of 5 balls).

The ball is not put back, so 4 balls are left and only 2 of them are blue: $\frac24$ ⁠. Multiply, because both draws must happen.

\begin{aligned} \frac35 \times \frac24 &= \frac{3 \times 2}{5 \times 4} = \frac{6}{20} \\ &= \frac{6 \div 2}{20 \div 2} = \frac{3}{10} \end{aligned}

One red: red then blue, or blue then red.

P(X = 1) = 2\left(\frac25 \times \frac34\right) = \frac{6}{10}

Why?

The two orders are separate cases, so their probabilities add:

\begin{aligned} RB&:\ \frac25 \times \frac34 = \frac{6}{20} \\ BR&:\ \frac35 \times \frac24 = \frac{6}{20} \end{aligned}

Red first: 2 red out of 5. Then 4 balls are left, and all 3 blue ones are still there. Blue first: 3 blue out of 5. Then both red balls are among the 4 left.

They are equal, so the total is $2 \times \frac{6}{20} = \frac{6}{10}$ ⁠.

\begin{aligned} \frac{6}{20} + \frac{6}{20} &= \frac{12}{20} \\ &= \frac{12 \div 2}{20 \div 2} = \frac{6}{10} \end{aligned}

P(X = 2) = \frac25 \times \frac14 = \frac{1}{10}

Why?

Both red: 2 red out of 5, then 1 red left out of 4.

\begin{aligned} \frac25 \times \frac14 &= \frac{2 \times 1}{5 \times 4} = \frac{2}{20} \\ &= \frac{2 \div 2}{20 \div 2} = \frac{1}{10} \end{aligned}

Check: $\frac{3}{10} + \frac{6}{10} + \frac{1}{10} = \frac{3 + 6 + 1}{10} = \frac{10}{10} = 1$ ⁠, as the probabilities of all the values of X must add up to 1.

\begin{array}{c|c|c|c} X & 0 & 1 & 2 \\ \hline P(X) & \frac{3}{10} & \frac{6}{10} & \frac{1}{10} \end{array}

$\therefore$ 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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