Answer: Option B

Answer : Option B

Explanation :

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Solution 1
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n(S) = 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(E) = Number of ways of drawing 2 Queens out of 4= ⁴C₂

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

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

----------------------------------------------------------------------------------------- Solution 2 ----------------------------------------------------------------------------------------- This problem can be solved using the concept of
Let A be the event of getting a Queen in the first draw
Total number of Queens = 4
Total number of cards = 52

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

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

$MF#%\text{P(Queen in second draw) = }\dfrac{3}{51}$MF#%

P(Queen in first draw and Queen in second draw) = P(Queen in first draw) × P(Queen in second draw)

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