48/52 or 92% chance of NOT getting an ace.
The probability of drawing a Ace from a standard deck of 52 cards if one Ace is missing is 3 in 51, or about 0.05882. If the missing card is not an Ace, then the probability is 4 in 51, or about 0.07843.
Let A= getting an ace the first timeand B= getting an ace the secondWe are looking to for the probaliity of getting A and B that is P(A and B)We know P(A and B) = P(A) . P(B|A)= (4/52) . (3/51) = 1/122 = .00452NOTE that P(B|A) is the conditional probability of getting an ace the second time given that you got an ace the first time.
Probability not ace is 1 minus probability of an ace which is 4/52. So, 1 - 4/52 is 48/52 or 12/13.
Probability of not drawing an ace equals one minus the probability of drawing an ace. The probability of drawing an ace is 4/52 or 1/13. So the probability of not drawing an ace on one draw is 1 - 1/13 or 12/13 or 0.9231 (92.31%).
The probability of drawing the first ace is 4 in 52. The probability of getting the second ace is 3 in 51. The probability of getting the third ace is 2 in 50. The probability, then, of drawing three aces is (4 in 52) times (3 in 51) times (2 in 50), which is 24 in 132600, or 1 in 5525, or about 0.0001810
Assuming it is a standard poker deck with 52 cards and 4 aces The probability of getting your first card an ace is = 4/52 Over here you need to reread your questions. There are something you need to know before i continue - when you draw your second card, did you return the first card back into the pile? lets say if you draw an ace of spades, would you be able to redraw it again for the second card? If the first ace is return to the pile: probability of getting a second ace is also 4/52 so the total probability of getting both cards an ace is (4/52 x 4/52) If the first ace is not return to the pile: probability of getting a second ace is now 3/51 note that removing one ace also removes one card from the pile. total card is now 51 with 3 aces only so the total probability of getting both cards an ace is (4/52 x 3/51) hope i help.
the answer is 4
48/52 or 92% chance of NOT getting an ace.
you have a 1/13 chance of getting an ace in a pack of cards
In a standard 52 card deck, the probability of drawing an ace is 1/13, and the probability of drawing a diamond is 1/4. The probability of drawing both an ace and a diamond is 1/52.Thus the probability of drawing an ace or a diamond is1/13 + 1/4 - 1/52 = 4/13 or about .308.
The probability of drawing a Ace from a standard deck of 52 cards if one Ace is missing is 3 in 51, or about 0.05882. If the missing card is not an Ace, then the probability is 4 in 51, or about 0.07843.
Let A= getting an ace the first timeand B= getting an ace the secondWe are looking to for the probaliity of getting A and B that is P(A and B)We know P(A and B) = P(A) . P(B|A)= (4/52) . (3/51) = 1/122 = .00452NOTE that P(B|A) is the conditional probability of getting an ace the second time given that you got an ace the first time.
Probability not ace is 1 minus probability of an ace which is 4/52. So, 1 - 4/52 is 48/52 or 12/13.
Probability of not drawing an ace equals one minus the probability of drawing an ace. The probability of drawing an ace is 4/52 or 1/13. So the probability of not drawing an ace on one draw is 1 - 1/13 or 12/13 or 0.9231 (92.31%).
Since there are 4 aces is a normal deck of 52 cards, the probability of drawing an ace is 4 in 52, or 1 in 13.
There are 4 aces in a standard deck of 52 cards. What is the probability of NOT getting an ace if you select one card at random? Write it as a percentage:) Move decimal 2 times to the right and round !