Q: What are chosen at random from a standard deck of cards with replacement what is the probability of getting to aces?

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hypergeometric distribution f(k;N,n,m) = f(3;52,4,3)

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 probability is (4/52) * (4/52) = 1/169 = 0.0059, approx.

The probability of a single point being chosen is 0.The probability of a single point being chosen is 0.The probability of a single point being chosen is 0.The probability of a single point being chosen is 0.

If the two marbles are drawn without replacement, the probability is 16/33.

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hypergeometric distribution f(k;N,n,m) = f(3;52,4,3)

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

For a single random choice from a standard deck, the probability is 1/13.For a single random choice from a standard deck, the probability is 1/13.For a single random choice from a standard deck, the probability is 1/13.For a single random choice from a standard deck, the probability is 1/13.

Q. A letter is chosen at random from the word STATistician.What is the probability that it is a vowel?What is the probability that it is T.

The probability is (4/52) * (4/52) = 1/169 = 0.0059, approx.

The probability of a single point being chosen is 0.The probability of a single point being chosen is 0.The probability of a single point being chosen is 0.The probability of a single point being chosen is 0.

If the two marbles are drawn without replacement, the probability is 16/33.

You have a 4/52 chance of your card being an ace the first time and a 4/52 chance of your card being an ace the second time. 4/52 * 4/52 or (Simplified) 1/13 * 1/13 = 1/169

1 out of 20 this is because there are 20 numbers in total, and there is only one 7 in there. (Assuming that there is the same probability for each number to be chosen, and that 17 is excluded as an affirmative outcome)

The probability is 0.3692

Probability that a girl is chosen = 23/45 = .511 So, the probability that a boy is chosen = 1 - .511 = .489

according to me its 1/2 as the possibility of getting an even is 1/2