One in 52 - because there are 52 cars in a deck, and only one Queen of Clubs.
The probability is 0. One card cannot be a club and a spade!
Probability not a club = 1 - probability it is a club = 1 - 13/52 = 1 - 1/4 = 3/4.
It is 3/4.
Red card- 1/2 CLUB-1/4
It is 10/52 = 5/26.
One quarter of the pack are CLUB cards. Three quarters of the pack are NOT CLUB cards. So the chance (probability) of picking a CLUB card is 1 out of 4 = 0.25 The chance (probability) of picking a NOT CLUB card is 3 out of 4 = 0.75 Adding the various probabilities the answer must always be 1.0, which is true here. If the probability of something happening is 1.0, that means the probability is "certainty". It is bound to happen.
The probability is 22/52 = 11/26.
In an ordinary 52 card deck, there are 4 three's, and 13 clubs, one of which is the 3 of clubs, so there are 16 cards that are a three or a club. The probability of drawing a three or a club is 16 out of 52, or 4 out or 13, or 0.308.
To determine the probability that one card drawn is a club and the other is a diamond from a standard deck of 52 cards, you can use the concept of combinations. There are 13 clubs and 13 diamonds in the deck. The probability of drawing one club and one diamond in two draws (without replacement) can be calculated as follows: the probability of drawing a club first and then a diamond is (13/52) * (13/51), and the probability of drawing a diamond first and then a club is (13/52) * (13/51). Adding these two probabilities gives you the total probability of one card being a club and the other a diamond. The final probability is approximately 0.25 or 25%.
"Pick A Club Card" is uncertain, in the context of the question. In all probability (no pun intended) the probability of picking a club from a random draw from a standard 52 card deck is 13 in 52, or 1 in 4, or 0.25.If the question means something else, please restate the question, giving more details.
In a standard deck of 52 cards - the probability of drawing any single card of two suits is 1:2 or 50%.
4/13