CBSE Board Examination 2022 (Term 2) · Section C · 4 marks

Q12: A lost card and two aces drawn: was the lost card an ace?

Board 2022 T2Ch 13 · Probability

The question

A card from a pack of 52 playing cards is lost. From the remaining cards, 2 cards are drawn at random without replacement, and are found to be both aces. Find the probability that lost card being an ace.

Exam answer

Let $E_1$ ⁠: the lost card is an ace, $E_2$ ⁠: the lost card is not an ace, and $A$ ⁠: the 2 cards drawn are both aces.

Why?

We know what happened afterwards (two aces were drawn) and ask about a cause before it (the lost card). That is a job for Bayes' theorem.

$E_1$ and $E_2$ cannot both happen, and one of them must happen, since the lost card either is or is not an ace. So they are mutually exclusive and exhaustive.

P(E_1) = \frac{4}{52} = \frac{1}{13},\quad P(E_2) = \frac{12}{13}

Why?

A pack has 4 aces among 52 cards. Then $P(E_2) = 1 - \frac{1}{13} = \frac{12}{13}$ ⁠, the 48 non-aces out of 52.

\begin{aligned} \frac{4}{52} &= \frac{4 \div 4}{52 \div 4} = \frac{1}{13} \\[4pt] 1 - \frac{1}{13} &= \frac{13}{13} - \frac{1}{13} = \frac{12}{13} \end{aligned}

P(A/E_1) = \frac{^3C_2}{^{51}C_2} = \frac{3}{^{51}C_2}

Why?

If the lost card was an ace, 51 cards remain and only 3 of them are aces. The number of ways to pick 2 aces is $^3C_2 = 3$ ⁠, out of $^{51}C_2$ ways to pick any 2 cards.

^3C_2 = \frac{3 \times 2}{2 \times 1} = \frac62 = 3

P(A/E_2) = \frac{^4C_2}{^{51}C_2} = \frac{6}{^{51}C_2}

Why?

If the lost card was not an ace, all 4 aces are still among the 51 cards, and

^4C_2 = \frac{4 \times 3}{2 \times 1} = \frac{12}{2} = 6

By Bayes' theorem,

P(E_1/A) = \frac{P(E_1)P(A/E_1)}{\displaystyle\sum_{i = 1}^{2} P(E_i)P(A/E_i)}

Why?

The denominator is $P(A)$ ⁠, found by the theorem of total probability:

\begin{aligned} P(A) &= P(E_1)P(A/E_1) \\ &\quad + P(E_2)P(A/E_2) \end{aligned}

The numerator is the part of $P(A)$ that comes through $E_1$ ⁠.

\begin{aligned} &= \frac{\frac{1}{13}\cdot\frac{3}{^{51}C_2}}{\frac{1}{13}\cdot\frac{3}{^{51}C_2} + \frac{12}{13}\cdot\frac{6}{^{51}C_2}} \\ &= \frac{3}{3 + 12 \times 6} = \frac{3}{75} \end{aligned}

Why?

Every term on the top and the bottom has the common factor

\frac{1}{13 \times {^{51}C_2}}

so it cancels. There is no need to work out $^{51}C_2 = 1275$ ⁠.

Write each product with this factor shown:

\begin{aligned} &\frac{1}{13} \cdot \frac{3}{^{51}C_2} \\ &= 3 \times \frac{1}{13 \times {^{51}C_2}} \\[4pt] &\frac{12}{13} \cdot \frac{6}{^{51}C_2} \\ &= 12 \times 6 \times \frac{1}{13 \times {^{51}C_2}} \end{aligned}

Cancel the factor from the top and every term of the bottom: the top leaves 3, the bottom leaves $3 + 12 \times 6$ ⁠.

Then $12 \times 6 = 72$ ⁠, $3 + 72 = 75$ and $\frac{3}{75} = \frac{1}{25}$ ⁠.

\begin{aligned} \frac{3}{3 + 72} &= \frac{3}{75} \\ &= \frac{3 \div 3}{75 \div 3} = \frac{1}{25} \end{aligned}

$\therefore$ The probability that the lost card is an ace $= \dfrac{1}{25}$ ⁠.

Where this comes from

Chapter 13, Probability: Bayes' theorem.

Question type: Bayes' theorem

How to spot it: click to see

Type 8. Bayes' theorem: from the result back to the cause

Spot it: the result has already happened (a red ball, a faulty item, an A grade), and you are asked for the chance that it came from one cause. It often comes as a case study or a 5-mark question.

Here: Two aces were drawn (the result), and you are asked about the lost card (the cause). With $E_1$ ⁠: the lost card is an ace, only 3 aces are left, so $P(A/E_1)$ uses $^3C_2$ ⁠; Bayes then gives $\frac{3}{3 + 72}$ ⁠.

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