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 answer depends on how many cards are drawn.
hypergeometric distribution f(k;N,n,m) = f(3;52,4,3)
The probability of getting two prime numbers when two numbers are selected at random and without replacement, from 1 to 10 is 2/15.
The probability is 7,893,600/311,875,200 = 0.0253
24 out of 6497400 = 1 out of 270725.
O.25
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 two prime numbers when two numbers are selected at random and without replacement, from 1 to 10 is 2/15.
completely useless.
The probability is 7,893,600/311,875,200 = 0.0253
24 out of 6497400 = 1 out of 270725.
If you draw more than 24 cards from a standard pack, without replacement, the probability is 1. That is, it is a certainty. The probability of the outcome for a single, randomly drawn card from a standard pack, is 7/13.
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
When you pick an object and do not return it, in probability it is termed "without replacement".
A card is drawn from a standard deck of playing cards. what is the probability that a spade and a heart is selected?