Explanation:

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Solution 1

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$n\left(S\right)$ = Total number of ways of drawing 2 cards from 52 cards = ⁵²C₂

Let E = event of getting two Queens

We know that there are total 4 Queens in the 52 cards

$Hence, n\left(E\right)$ = Number of ways of drawing 2 Queens out of 4= ⁴C₂

$\text{P(E) = }\dfrac{\text{n(E)}}{\text{n(S)}} = \dfrac{4_{C_2}}{52_{C_2}}$

$= \dfrac{\left( \dfrac{4 \times 3}{2}\right)}{\left( \dfrac{52 \times 51}{2}\right)}$= $\dfrac{4\times 3}{ 52 \times 51}= \dfrac{3}{ 13 \times 51} = \dfrac{1}{ 13 \times 17} = \dfrac{1}{ 221}$

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Solution 2

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This problem can be solved using the concept of[Conditional Probability]

Let A be the event of getting a Queen in the first draw

Total number of Queens = 4

Total number of cards = 52

$\text{P(Queen in first draw) = }\dfrac{4}{52}$

Assume that the first event is happened. i.e., a Queen is already drawn in the first draw

and now B = event of getting a Queen in the second draw

Since 1 Queen is drawn in the first draw, Total number of Queens remaining = 3
Since 1 Queen is drawn in the first draw, Total number of cards = 52 - 1 = 51

$\text{P[Queen in second draw] = }\dfrac{3}{51}$

P[Queen in first draw and Queen in second draw] = P[Queen in first draw] × P[Queen in second draw]

$= \dfrac{4}{52} \times \dfrac{3}{51} = \dfrac{1}{13} \times \dfrac{1}{17} = \dfrac{1}{221}$