hypergeometric distribution f(k;N,n,m) = f(1;51,3,1)
or binominal distribution f(k;n,p) = f(1;1,3/51) would result in same probability
"Playing cards" are chosen at random.
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
If you answer randomly, 1 in 8.
There are infinitely many numbers and so the probability of the second event is 0. As a result the overall probability is 0.
The answer depends on how many cards are drawn.
"Playing cards" are chosen at random.
The probability of getting an 8 on a standard six-sided die is zero.
If the events can be considered independent then the probability is (0.7)4 = 0.24 approx.
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
If you answer randomly, 1 in 8.
There are infinitely many numbers and so the probability of the second event is 0. As a result the overall probability is 0.
The probability is 0 if you pick the the card from one end of a mint pack (2 of clubs) and 1 if you pick it from the other end (A spades). Also, if you pick 49 cards without replacement, the probability is 1. So, the answer depends on how many cards are drawn, and whether or not they are drawn from a well shuffled pack. The probability of getting an ace when one card is randomly picked from a pack is 4/52 = 1/13.
The answer depends on how many cards are drawn.
hypergeometric distribution f(k;N,n,m) = f(3;52,4,3)
If you pick enough cards, without replacement, the probability is 1. The probability for a single random draw is 1/26.
The probability of getting the first answer correct is 1/2 The probability of getting the first two correct is 1/2 * 1/2 = 1/(22) The probability of getting all 9 correct is 1/(29) = 1/512 which is just under 0.2%
The probability is 0.5