Sample 2021-22 T2 Class 12 Maths Q6 Solution: Two cards without replacement: first red, second a Jack

The question

Two cards are drawn at random from a pack of 52 cards one-by-one without replacement. What is the probability of getting first card red and second card Jack?

Exam answer

The first card is red and the second is a Jack in two mutually exclusive ways: the first is a red Jack and the second a Jack, or the first is a red non-Jack card and the second a Jack.

Why?

The chance that the second card is a Jack depends on whether the first card was a Jack. So split by the first card: a red Jack, or a red card that is not a Jack.

A pack has 26 red cards, and 2 of them are Jacks (hearts and diamonds). So there are $26 - 2 = 24$ red non-Jack cards, and 4 Jacks in all.

\begin{aligned} &\text{Required probability} \\ &= \frac{2}{52} \times \frac{3}{51} + \frac{24}{52} \times \frac{4}{51} \end{aligned}

Why?

In each case multiply: first card, then second card from the 51 cards left.

Red Jack first: $\frac{2}{52}$ ⁠. Then 3 Jacks remain in 51 cards: $\frac{3}{51}$ ⁠.

Red non-Jack first: $\frac{24}{52}$ ⁠. Then all 4 Jacks remain in 51 cards: $\frac{4}{51}$ ⁠.

The two cases cannot both happen, so add them. Watch out: $\frac{26}{52} \times \frac{4}{51}$ is wrong, because after a red Jack only 3 Jacks are left.

= \frac{6 + 96}{52 \times 51} = \frac{102}{2652}

Why?

Here $2 \times 3 = 6$ ⁠, $24 \times 4 = 96$ and $52 \times 51 = 2652$ ⁠. Since $2652 = 102 \times 26$ ⁠, the fraction simplifies to $\frac{1}{26}$ ⁠.

\begin{aligned} \frac{2}{52} \times \frac{3}{51} &= \frac{6}{2652} \\ \frac{24}{52} \times \frac{4}{51} &= \frac{96}{2652} \\[4pt] \frac{6}{2652} + \frac{96}{2652} &= \frac{102}{2652} \\[4pt] \frac{102 \div 102}{2652 \div 102} &= \frac{1}{26} \end{aligned}

$\therefore$ Required probability $= \dfrac{1}{26}$ ⁠.

Where this comes from

Chapter 13, Probability: Multiplication theorem: without replacement.

Question type: One after the other

How to spot it: click to see

Type 3. One after the other: the multiplication rule

Spot it: draws "without replacement", or a second event that depends on the first.

Here: Two cards without replacement, and the chance of a Jack second depends on whether the red card was a Jack. So multiply in two cases, $\frac{2}{52}\times\frac{3}{51}$ and $\frac{24}{52}\times\frac{4}{51}$ ⁠, and add.

Quick lesson · Worked NCERT solutions · Interactive lesson

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Q6: Two cards without replacement: first red, second a Jack