Probability question with a deck of cards

Just started to learn maths, so I'm sorry if this an elementary question.

The question is: here is a deck with 40 cards ; 10 cards, each 10 with one of the 4 shapes (hearts, diamonds, clubs and spades). All the cards are numbered from 1 to 10. Every card has the same probability to be drawn. We start drawing cards from the deck. If the 11th card which was drawn is the first one with number 10 on it, what is the probability that the next card is a clubs?

What I have tried: We need to find P(A|B), where A is the probability that the 12's card is a clubs, and B is the probability that the 11's card is a first ten.

I chose the sample space to be picking 12 cards from 40. $$P$$($$\Omega$$) = $$\binom{40}{12}$$

$$P(B)$$= $$\binom{36}{10}\binom{4}{1}\binom{29}{1}$$. Where in $$P(B)$$, i chose 10 cards from 36 cards(without 10's), the picking 1 card from the the 10, the choosing from the rest.

$$P(A\cap B)$$ = $$\binom{9}{9}\binom{26}{1}(\binom{3}{1}\binom{1}{1} +\binom{1}{0})$$ $$+$$ $$\binom{9}{8}\binom{26}{2}(\binom{3}{1}\binom{2}{1}+\binom{1}{1})$$ $$+$$ $$\binom{9}{7}\binom{26}{3}(\binom{3}{1}\binom{3}{1} + \binom{2}{1})$$ $$+$$ $$\binom{9}{6}\binom{26}{4}(\binom{3}{1}\binom{4}{1}+\binom{3}{1})$$ $$+$$ $$\binom{9}{5}\binom{26}{5}(\binom{3}{1}\binom{5}{1}+\binom{4}{1})$$ $$+$$ $$\binom{9}{4}\binom{26}{6}(\binom{3}{1}\binom{6}{1}+\binom{5}{1})+$$ $$\binom{9}{3}\binom{26}{7}(\binom{3}{1}\binom{7}{1}+\binom{6}{1})+$$ $$\binom{9}{2}\binom{26}{8}(\binom{3}{1}\binom{8}{1}+\binom{7}{1})+$$ $$\binom{9}{1}\binom{26}{9}(\binom{3}{1}\binom{9}{1}+\binom{8}{1})+$$ $$\binom{9}{0}\binom{26}{10}(\binom{3}{1}\binom{10}{1}+\binom{9}{1})$$ . Here, $$\binom{9}{8}$$ is the possibility that we choose 8 from 9 clubs ( we dont want to pick 10), $$\binom{26}{2}$$ is completing the picking of the rest 10 cards, $$\binom{3}{1}\binom{3}{1}$$ is picking 1 10 from 3 and picking a club from the 3 that left, $$\binom{1}{1}$$ is picking the ten of clubs and then picking the only left clubs. The rest of the calculations are the same.

So, $$P(A|B) = \frac {P(A\cap B)}{P(B)} = \frac {5271351254}{29485675300}$$ $$\approx$$ 0.178

The correct answer is $$\frac 14$$. My questions are: am I wrong, and, if so, is there a "nicer" way to compute $$P(A\cap B)$$(or generally solving this exercise)?

*Sorry if there are mistakes in grammar.

• The probability is the same for the 12th card whatever the suit specified, so it should be clear what the probabability is for the 12th card being a club Dec 7 '21 at 8:45
• What do you mean by "If the 11's card who was drawn is the first ten" Dec 7 '21 at 8:55
• @C.Koca We pull out 10 cards from the deck ( we didn't pick a card with a number of 10), and only in the 11's card we pull a card with a number of 10. Dec 7 '21 at 8:58
• @Arthur Vause If i understand, beacuse every card has the same possibility to be drawn and there are an even amount of card ( ten from each shape) , then we can say the probability is 1/4? Dec 7 '21 at 8:59
• @Yoxbox my argument is the same as C. Koca has described in his answer Dec 7 '21 at 9:15