Q.A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.

Concept: Conditional Probability — we update the probability of the lost card being a diamond given the evidence that two diamonds were drawn from the remaining deck.

Step 1: Define events

Let $E$ = lost card is a diamond.

Let $A$ = two cards drawn (from remaining 51) are both diamonds.

Step 2: Find prior probabilities

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

$P(\bar{E}) = \frac{39}{52} = \frac{3}{4}$.

Step 3: Find likelihoods

If $E$ occurs: 12 diamonds remain in 51 cards.

$P(A \mid E) = \frac{\binom{12}{2}}{\binom{51}{2}} = \frac{66}{1275}$.

If $\bar{E}$ occurs: 13 diamonds remain in 51 cards.

$P(A \mid \bar{E}) = \frac{\binom{13}{2}}{\binom{51}{2}} = \frac{78}{1275}$.

Step 4: Apply Bayes' theorem …

Why conditional probability?

The problem gives us an outcome (two diamonds drawn from the remaining cards) and asks for the probability of a prior event (the lost card being a diamond) given that outcome. This is a classic Bayes' theorem situation — we need to reverse the conditioning.

Think of it this way: before any cards are drawn, the lost card could be any of the 52. After we see two diamonds drawn from the remaining 51, it becomes more likely that the lost card was not a diamond (because if it were, there would be fewer diamonds left to draw). The maths will quantify exactly how much more likely.

Let’s define:

• $E$ = event that the lost card is a diamond.
• $F$ = event that two cards drawn from the remaining 51 are both diamonds.

We want $P(E \mid F)$.

Step-by-step solution

1. Find the prior probability $P(E)$.

Before any draw, the lost card is equally likely to be any of the 52 cards. There are 13 diamonds.

2. Find $P(\text{not }E)$ — the lost card is not a diamond.

3. Compute $P(F \mid E)$ — probability of drawing two diamonds given the lost card was a diamond.

If the lost card was a diamond, then the remaining 51 cards contain $13 - 1 = 12$ diamonds.

Number of ways to draw 2 diamonds from these 12: $\binom{12}{2}$.

Total ways to draw any 2 cards from 51: $\binom{51}{2}$.

4. Compute $P(F \mid \text{not }E)$ — probability of two diamonds given the lost card was NOT a diamond.

If the lost card was not a diamond, all 13 diamonds remain in the 51 cards.

$P(F \mid \text{not }E) = \frac{\binom{13}{2}}{\binom{51}{2}} = \frac{13 \times 12}{51 \times 50} = \frac{156}{2550} = \frac{26}{425}$ …

Method: Bayes' theorem when the "cause" is an earlier unseen event

Use this for problems where something happened first but is unknown (a lost card, an unseen transfer), later evidence is observed, and you must update the probability of the hidden first event.

Steps

Step 1: Make the hidden event the partition.

Let the unknown first event and its complement be the two (or more) mutually exclusive causes, and write their priors — e.g. lost card is a diamond ($\tfrac{13}{52}$) or is not ($\tfrac{39}{52}$).

Step 2: Compute the likelihood of the evidence under each cause, using counting.

Under each hypothesis the remaining pool changes, so the probability of the observed draw changes. Use combinations for "both drawn are diamonds": …

Common Mistakes

Mistake 1: Ignoring the case where the lost card is NOT a diamond.

Why it's wrong: the evidence (two diamonds drawn) is more likely when all $13$ diamonds remain, so that branch must appear in the denominator. Correct approach: include both $P(A\mid E)$ and $P(A\mid E')$ in Bayes' total-probability denominator.

Mistake 2: Using $\tfrac{13}{52}$ directly as the answer.

Why it's wrong: that is only the prior; the observed diamonds lower it. Correct approach: update via Bayes to get $\tfrac{11}{50}$, which is less than the prior $\tfrac14$. …