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:

A standard pack has 52 cards with 13 spades. One card is missing. We draw two cards and both are spades. We need to find the probability that the missing card is not a spade.

Let $$S$$ be the event that the missing card is a spade and $$S'$$ that it is not. Let $$E$$ be the event that two drawn cards are both spades.

Using Bayes' theorem: $$P(S' | E) = \frac{P(E | S') \cdot P(S')}{P(E | S') \cdot P(S') + P(E | S) \cdot P(S)}$$.

We have $$P(S) = \frac{13}{52} = \frac{1}{4}$$ and $$P(S') = \frac{39}{52} = \frac{3}{4}$$.

If the missing card is a spade, the remaining deck has 51 cards with 12 spades: $$P(E | S) = \frac{\binom{12}{2}}{\binom{51}{2}} = \frac{66}{1275}$$.

If the missing card is not a spade, the remaining deck has 51 cards with 13 spades: $$P(E | S') = \frac{\binom{13}{2}}{\binom{51}{2}} = \frac{78}{1275}$$.

Substituting: $$P(S' | E) = \frac{\frac{78}{1275} \cdot \frac{3}{4}}{\frac{78}{1275} \cdot \frac{3}{4} + \frac{66}{1275} \cdot \frac{1}{4}} = \frac{78 \times 3}{78 \times 3 + 66 \times 1} = \frac{234}{234 + 66} = \frac{234}{300} = \frac{39}{50}$$.

The required probability is $$\frac{39}{50}$$, which corresponds to option (3).