A question of Bayes' theorem

Let $A$ be the event that the lost card is a spade and let $B$ be the event that the two cards drawn are spades. Then

$P(A | B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{(\frac{1}{4})*\binom{12}{2}}{(\frac{1}{4})*\binom{12}{2} + (\frac{3}{4})*\binom{13}{2}} = \dfrac{66}{66 + 3*78} = \dfrac{11}{50} = \boxed{0.22}$ .

Comments: The probability that the lost card is a spade is $\frac{13}{52} = \frac{1}{4}$ , and so the probability that the lost card is not a spade is $\frac{3}{4}$ . If the lost card is a spade then we have only $12$ spades left in the deck, and thus $\binom{12}{2}$ possible combinations of two spades. If the lost card is not a spade then there are $\binom{13}{2}$ possible combinations of two spades.

Consider the card has not been lost, so 11 spades and 50 cards remaining after drawing two spades. Then if a card is lost, the probability of it being a spade=11/50=0.22

As total no. Of cards are 52 and the probability of. Spades is 13/52. but two spades are drawn so no. of spades left to be =13-2=11, and total. No. Of Cards left to be equal to,,,,, total no. Of cards-no. Of cards drawn=52-3=49.....
so probability =11/49=0.22

A deck of cards has 13 cards of each suite for a total of 52 cards. Taking 2 spades out leaves 11 spades in the deck. 52 cards minus the 2 spades and the lost card leaves 49 cards in the deck, so there is a 11/49 chance of the lost card being a spade. 11/49=0.22