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**Topic**

From the draw set, there are 52 normal cards being face-down, 4 cards are randomly drawn at the same time. Calculate the probability of the following events:

a) A: “Draw 4 cards of the same value”

b) B: “Draw 4 cards of the same suit”

c) C: “Out of 4 cards drawn, only 2 Aces”

**Solution method – See details**

The probability of event A being a number, symbol \(P\left( A \right)\) is determined by the formula: \(P\left( A \right) = \frac{{n\left( A \right)}}{{n\left( \Omega \right)}}\), where \(n\left( A \right)\) and \(n\left( \Omega \right)\) denote the number of elements of set A and \(\Omega \) respectively.

**Detailed explanation**

+ Draw 4 cards (unordered) from 52 cards \( \Rightarrow n\left( \Omega \right) = C_{52}^4\)

a) A: “Draw 4 cards of the same value”

In a deck of 52 cards, there are 13 groups of 4 cards of the same value.

\( \Rightarrow n\left( A \right) = 13\)

\( \Rightarrow P\left( A \right) = \frac{{n\left( A \right)}}{{n\left( \Omega \right)}} = \frac{{13}}{{ C_{52}^4}} = \frac{1}{{20825}}\)

b) B: “Draw 4 cards of the same suit”

There are 4 ways to choose the suit of the deck. Each suit has 13 cards.

The number of ways to choose 4 cards in each suit is the number of combinations of 4 of 13.

\( \Rightarrow n\left( B \right) = 4.C_{13}^4\)

\( \Rightarrow P\left( B \right) = \frac{{n\left( B \right)}}{{n\left( \Omega \right)}} = \frac{{4.C_{13 }^4}}{{C_{52}^4}} = \frac{{44}}{{4165}}\)

c) C: “Out of 4 cards drawn, only 2 Aces”

The number of Aces in the deck is 4. After choosing 2 Aces (from 4 Aces), the remaining 2 cards are chosen from 48 non-Aces cards.

\( \Rightarrow n\left( C \right) = C_4^2.C_{48}^2\)

\( \Rightarrow P\left( C \right) = \frac{{n\left( C \right)}}{{n\left( \Omega \right)}} = \frac{{C_4^2.C_ {48}^2}}{{C_{52}^4}} = \frac{{6768}}{{270725}}\)

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