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Bunuel
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P(2 jack drawn w/o replacing)= 4/52 * 3/51
simplify - 1/221

IMO (A)
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MarmikUpadhyay
Given: A pack of 52 cards, 2 cards are drawn one by one without replacement.
Probability that both cards are Jack = ?


Probability that first card is Jack = \(\frac{4}{52}\)
After the first cared is removed,
Probability that second card is Jack = \(\frac{3}{51}\) [3 Jacks remaining as first card was Jack]

As both the events are dependent on each other, i.e. second card will be drawn after the first one, the probabilities will get multiplied.
=> Probability that both cards are Jack = \(\frac{4}{52}\) * \(\frac{3}{51}\) = \(\frac{1}{221}\).

So correct answer is option A.

Hi Marmik..why have you taken 4? 4/52?
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MarmikUpadhyay
Given: A pack of 52 cards, 2 cards are drawn one by one without replacement.
Probability that both cards are Jack = ?


Probability that first card is Jack = \(\frac{4}{52}\)
After the first cared is removed,
Probability that second card is Jack = \(\frac{3}{51}\) [3 Jacks remaining as first card was Jack]

As both the events are dependent on each other, i.e. second card will be drawn after the first one, the probabilities will get multiplied.
=> Probability that both cards are Jack = \(\frac{4}{52}\) * \(\frac{3}{51}\) = \(\frac{1}{221}\).

So correct answer is option A.

Hi Marmik..why have you taken 4? 4/52?

Because in a deck of 52 cards, there are four cards of Jack.
I hope it helps!
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There are 52 cards in a pack of cards. Of these 52 cards, there are a total of 4 jacks, one from each suite (Hearts, Clubs, Spades & Diamonds)

Sample Space when two cards are drawn sequentially without replacement = 52 * 51

Number of ways in which 2 jacks can be drawn = 4 * 3 (since we are not replacing the cards).

Therefore, Probability (Both are Jacks) = \(\frac{4 * 3 }{ 52 * 51}\).

Simplifying, we obtain 1/221 as the final value.
The correct answer option is A.

Hope that helps!
Aravind BT
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