Anil placed a deck of 52 cards if you pick three cards what is the probability that all are queens

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Ex: Find the probability of drawing 3 aces at random from a deck of 52 ordinary cards if the cards are not replaced.

Here's what I did:

The probability of choosing the first ace = $\dfrac{^4C_1}{^{52}C_1}$

The probability of choosing the second ace = $\dfrac{^3C_1}{^{51}C_1}$

The probability of choosing the third ace = $\dfrac{^2C_1}{^{50}C_1}$

The probability is = $\dfrac{^4C_1}{^{52}C_1}\times\dfrac{^3C_1}{^{51}C_1}\times\dfrac{^2C_1}{^{50}C_1}=\dfrac{1}{5525}$

But the answer is given as $\dfrac{1}{17,576}$

Please tell me where did I go wrong?

All questions on Maths

use probability with combinations to solve the problem

Assuming the deck is entirely random, then the probability of getting 3 queens is given by: This comes from the following, in order of the calculation: When we start dealing we have 52 cards, of which four are queens. We want any one of the four queens to be our first card. Assume that any card is equally likely to be dealt. Then there are 52 distinct cards that we can be given, and 4 of them are queens, so the probability of getting a queen is 4/52 = 1/13. Proceed to the second card. Now we have only 3 queens left, and also only 51 cards left. Using the same reasoning above, the probability of getting a queen is 3/51 = (13)/(173) = 1/17. On the third and final card we are interested in, now there are only two queens left and 50 cards in total - so we get P(queen dealt) = 2/50 = 1/25. Why multiply? Because each of the third and second cards being queens is only relevant if the first card was a quen, and then if the second cards was also a queen. This means that the overall probability is the product of all other probabilities. Finally, since there are 4 queens and we want any three of them, the suit of the queen does not matter to an individual probability, but does increase the number of ways we can get free queens. Called the Queen of Hearts H, etc., and (C = clubs, D = diamonds, S = Spades), these are the ways of getting 3 queens, assuming that the order does not matter (which it doesn't here): CHD CHS CDS DHS i.e. 4 distinct ways of getting 3 queens out of the deck. So we multiply our earlier calculation by 4. This can be written 4C3 where 4 is the number of queens available to choose from, and we choose 3. (Usually the 4 is written to the top left of the C, and the 3 to the bottom-right, but I don't know how to format this properly.) You should be told how to calculate this in classes, but it's written: with: and so on Hope this helps.