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Required probability = P(Club and non queen, non queen) + P(club and queen, non queen)

$= \frac{12}{52} \times \frac{47}{51} + \frac{1}{52} \times \frac {48}{51}$

$=0.2307$
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since,it is given that first card has to be the club.so we can choose it in 13 ways.(it can also be a queen of club because there is no restriction on first card to be a queen)

now,for second card,it is mentioned that it should not be queen(but it can be a club).

1)when second card is non-queen of club-there are 12 non queen cards from 13 club cards.

2) when second card is non-queen of any suit but not club -there are 52 -13(for clubs) - 3(queens) =36 cards to choose from.

hence,probability = 13C1 * 12C1 /52C2    + 13C1 * 36C1/ 52C2.

@vaishali,am i correct??please do let me know the answer

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