The probability of not getting on ace is 48 in 52, or 12 in 13, or about 0.9231. The probability of not getting a heart is 39 in 52, or 3 in 4, or 0.75.
These two events, however, are not exclusive, so you can not just multiply them together. You need to look at the big picture. But look at the summary below.
There are 4 aces in the deck, one of which is a heart. There are 13 hearts in the deck, one of which is an ace. The inclusion set is 16 cards that are either an ace or a heart. Flip this over by subtracting from 52 and you get 36 cards that are not an ace nor a heart.
So, the probability of getting neither an ace nor a heart is 36 in 52, or 18 in 26, or 9 in 13, or about 0.6923.
Now, it turns out that you could have just multiplied the probabilities and obtained the same result. This worked because you were looking at inverse probabilities. It would not have worked if the question was to find the probability of getting an ace or a heart.
In summary, I went ahead and showed the thought process involved to illustrate a point - that you need to understand the set of possible outcomes in order to make a correct calculation.
3/52*13/51+1/52*12/51= .01923
you have a 1/13 chance of getting an ace in a pack of cards
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
The chance of picking a heart is 1/4 as there are four suits. The chance of picking an ace is 1/13, as there are 4 aces in a deck of 52. If I want the probability of drawing a heart of an ace, I can add these probabilities together, but I have double counted on possibility, the ace of hearts. So: Probability = 4/52 + 13/52 - 1/52 = 16/52 or 4/13 Probability = 0.308 expressed as a fraction.
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.
3/52*13/51+1/52*12/51= .01923
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
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
The chance of picking a heart is 1/4 as there are four suits. The chance of picking an ace is 1/13, as there are 4 aces in a deck of 52. If I want the probability of drawing a heart of an ace, I can add these probabilities together, but I have double counted on possibility, the ace of hearts. So: Probability = 4/52 + 13/52 - 1/52 = 16/52 or 4/13 Probability = 0.308 expressed as a fraction.
1/13 (one thirteenth
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.
48/52 or 12/13 or .92307
1/13
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.
With a regular deck (2-Ace with ♥,♦,♣, and ♠), the probability of drawing a three is 1 in 13. The probability for a heart is 1 in 4, a.k.a. 25% or 1/4 . The probability for a 3 of hearts is 1 in 52.Improve:And to be more specific:because a heart is 25%, and a three is 4/52 % and the h of hearts is 1/52 getting a 3 OR heart is 25% (heart) + 7.69% (a 3) - 1.923% (3 of hearts can't be counted twice!) = .30767
Probability that it is one of these eight cards is 8/52. Hence the probability of not getting these eight cards is 44/52