A pack of cards has one card missing. Two cards are drawn randomly and are found to be spades. The p

Q: Wait, why is the denominator $51C2$ and not $52C2$?

Great catch! The problem states that one card is already missing from the deck. Therefore, when we draw two cards, we are drawing from a pool of $51$ cards, not $52$. Always account for the state of the system *after* the initial condition (the missing card) has occurred.

Q: Why do we use Bayes' Theorem instead of just simple probability?

Simple probability works when you know the cause and want to predict the effect. Here, we know the effect (we drew two spades) and we are working backward to find the probability of the cause (what card was missing). This 'inverse' nature is exactly what Bayes' Theorem is designed to handle.

Q: Can I just use the ratio of the number of favorable outcomes?

You can, but you must be careful. You need to weight the outcomes by the prior probability of each hypothesis ($E_1$ and $E_2$). Bayes' Theorem provides the formal structure to ensure you don't miss these weights, which is crucial for getting the correct answer.

Question

A pack of cards has one card missing. Two cards are drawn randomly and are found to be spades. The probability that the missing card is not a spade, is :

Options:

(A) $3/4$

(B) $52/867$

(C) $39/50$

(D) $22/425$

Sigma Percentile · Accepted Answer

### The Mystery of the Missing Card: A Bayesian Journey

Imagine you are standing at a table with a standard deck of $52$ cards. Suddenly, one card vanishes. You don't know what it is, but you know the deck is now incomplete.

From this mysterious, reduced deck of $51$ cards, you draw two cards at random. To your surprise, both are spades.

The question that now haunts you is: what is the probability that the missing card was not a spade? This is a classic problem of inverse probability and a perfect gateway into the elegant world of **Bayes' Theorem**.

### Phase 1: Defining the Hypotheses

In any probability problem, the first step is to define your universe. We have two mutually exclusive hypotheses regarding the missing card:

*   $E_1$: The missing card is a spade.
*   $E_2$: The missing card is not a spade.

A full deck has $13$ spades and $39$ non-spades. Thus, the prior probabilities are:

$$P(E_1) = \frac{13}{52} = \frac{1}{4}$$

$$P(E_2) = \frac{39}{52} = \frac{3}{4}$$

These represent our starting beliefs before observing the drawn cards.

### Phase 2: The Evidence

Now, we introduce the evidence: the event $A$, which is drawing two spades from the remaining $51$ cards. We must calculate the conditional probability of $A$ occurring under each hypothesis.

If $E_1$ is true (a spade is missing), there are only $12$ spades left in the $51$-card deck. The probability of drawing two spades is:

$$P(A|E_1) = \frac{^{12}C_2}{^{51}C_2} = \frac{66}{1275}$$

Conversely, if $E_2$ is true (a non-spade is missing), all $13$ spades remain. The probability of drawing two spades is:

$$P(A|E_2) = \frac{^{13}C_2}{^{51}C_2} = \frac{78}{1275}$$

### Phase 3: The Bayesian Engine

We now invoke **Bayes' Theorem** to find $P(E_2|A)$, the probability that the missing card was not a spade given that we drew two spades:

$$P(E_2|A) = \frac{P(E_2)P(A|E_2)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}$$

Substituting our values, we get:

$$P(E_2|A) = \frac{\frac{3}{4} 	imes \frac{78}{1275}}{\frac{1}{4} 	imes \frac{66}{1275} + \frac{3}{4} 	imes \frac{78}{1275}}$$

Notice the beauty of the algebra here: the term $\frac{1}{4 	imes 1275}$ is common to all parts of the fraction and cancels out entirely. This leaves us with:

$$P(E_2|A) = \frac{3 	imes 78}{66 + 3 	imes 78} = \frac{234}{66 + 234} = \frac{234}{300}$$

Simplifying this by dividing both the numerator and denominator by $6$, we arrive at our final answer:

$$\mathbf{P(E_2|A) = \frac{39}{50}}$$

### Conclusion

Probability is often counter-intuitive, but **Bayes' Theorem** provides a rigorous path through the fog. By starting with our prior beliefs and updating them with the evidence of the two spades, we have successfully solved the mystery of the missing card.

Remember, in JEE, the math is just the language; the real skill is in visualizing the scenario and choosing the right theorem to unlock the solution.

Animated Solution for Mathematics - Probability: A pack of cards has one card missing. Two cards are drawn randomly and are found to be spades. The probability that the missing card is not a spade, is :

Select Answer:

Visualized Solution

Understanding the Scenario

Defining the Hypotheses $E_{1}$ and $E_{2}$

Prior Probabilities $P (E_{1})$ and $P (E_{2})$

Defining the Observed Event $A$

Conditional Probability $P (A ∣ E_{1})$

Calculating $P (A ∣ E_{1})$

Conditional Probability $P (A ∣ E_{2})$

Calculating $P (A ∣ E_{2})$

Applying Bayes' Theorem

Substitution of Values

Simplifying the Expression

Final Calculation

Conclusion and Key Takeaway

The Sigma Insight: Bayes' Theorem

The Mystery of the Missing Card

Phase 1

Phase 2

Phase 3

Conclusion

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