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Question
From a pack of 52 cards, 4 are drawn one by one without replacement. Find the probability that all are aces(or kings).
Solution
Consider the given events.
A = An ace in the first draw
B = An ace in the second draw
C = An ace in the third draw
D = An ace in the fourth draw
\[\text{ Now } , \]
\[P\left( A \right) = \frac{4}{52} = \frac{1}{13}\]
\[P\left( B/A \right) = \frac{3}{51} = \frac{1}{17}\]
\[P\left( C/A \cap B \right) = \frac{2}{50} = \frac{1}{25}\]
\[P\left( D/A \cap B \cap C \right) = \frac{1}{49}\]
\[ \therefore \text{ Required probability } = P\left( A \cap B \cap C \cap D \right) = P\left( A \right) \times P\left( B/A \right) \times P\left( C/A \cap B \right) \times P\left( D/A \cap B \cap C \right)\]
\[ = \frac{1}{13} \times \frac{1}{17} \times \frac{1}{25} \times \frac{1}{49}\]
\[ = \frac{1}{270725}\]
In case of kings, the required probablity will be = \[\frac{1}{270725}\]
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