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Since no information is given about the first two draws, they do not matter.

The odds against drawing a 6 of clubs from a standard deck of 52 cards is 51 in 52, or about 0.9808.

Q: What are the odds against drawing a 6 of clubs from a deck of cards on the third draw?

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The probability of drawing the first face card is 12 in 52. The probability of drawing the second is 11 in 51. The probability of drawing the third is 10 in 50. Thus, the probability of drawing three face cards is (12 in 52) times (11 in 51) times (10 in 50) or (1320 in 132600) or about 0.009955.

The probability of drawing three aces from a deck of cards is 1 in 5525. The probability of the first ace is 4 in 52, or 1 in 13. The second ace is 3 in 51, or 1 in 17. The third ace is 2 in 50, or 1 in 25. Multiply these three probabilities together and you get 1 in 5525.

In an ordinary deck of cards, there are 52 cards out of which 13 are spades. This means that the chances of drawing the first spade is 13 out of 52. The probability of the second spade is 12 out of 51 because one spade and, incidentally, one card are now missing. The third spade comes out as 11 out of 50 and the last one as 10 out of 49. Total probability of events that don't affect each other is the product of the individual probabilities. Thus, the chances of drawing four spades is (13 / 52) * (12 / 51) * (11 / 50) * (10 / 49), which is about 0.00264 -- in other words, one to 379.

It is 0.077, approx.

try drawing a circle and shade in 2 thirds

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The probability of drawing the first face card is 12 in 52. The probability of drawing the second is 11 in 51. The probability of drawing the third is 10 in 50. Thus, the probability of drawing three face cards is (12 in 52) times (11 in 51) times (10 in 50) or (1320 in 132600) or about 0.009955.

The probability of drawing three aces from a deck of cards is 1 in 5525. The probability of the first ace is 4 in 52, or 1 in 13. The second ace is 3 in 51, or 1 in 17. The third ace is 2 in 50, or 1 in 25. Multiply these three probabilities together and you get 1 in 5525.

No there is not a third crest in phantom hourglass.

I assume we do not replace the 8's after we pick one of them. This is sampling without replacement. Since there are four 8s, the odd of picking one of them out of 52 cards is 4/52 the first time, then we have three 8s left and only 51 cards so, 3/51 and then 2/50 the third time. The odd of drawing three consecutive eights is the product of those three probabilities. This is 4/52x3/51x2/50

Aprox. 0.018%There are 4 queens in a regular deck of 52 cards.The probability of drawing a queen on the first draw is: P(Q1) = 4/52.The probability of drawing a queen on the second draw given that the first card wasa queen is: P(Q2â”‚Q1) = 3/51.The probability of drawing a queen on the third draw given that the first two cardswere queens is: P(Q3â”‚(Q2UQ1)) = 2/50.The probability of drawing 3 queens on the first 3 cards drawn from a deck of cardsis: P(Q1UQ2UQ3) = (4/52)âˆ™(3/51)âˆ™(2/50) = 1.80995... x 10-4 â‰ˆ 0.00018 â‰ˆ 0.018%

cards, king of diamonds, king of hearts, king of spades, and the king of clubs.

yes along with golf clubs

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

In an ordinary deck of cards, there are 52 cards out of which 13 are spades. This means that the chances of drawing the first spade is 13 out of 52. The probability of the second spade is 12 out of 51 because one spade and, incidentally, one card are now missing. The third spade comes out as 11 out of 50 and the last one as 10 out of 49. Total probability of events that don't affect each other is the product of the individual probabilities. Thus, the chances of drawing four spades is (13 / 52) * (12 / 51) * (11 / 50) * (10 / 49), which is about 0.00264 -- in other words, one to 379.

It is 0.077, approx.

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