Two cards are drawn without replacement from a standard deck of 52 playing cards. What is the probability of choosing a king and then without replacement, another king? Express your answer as a fraction or a decimal number rounded to four decimal places



Answer :

ANSWER

[tex]\begin{equation*} 0.0588 \end{equation*}[/tex]

EXPLANATION

A standard deck of cards has 52 cards.

The number of king is 4

The probability of choosing the first king is;

[tex]\begin{gathered} P(K_1)=\frac{n(K)}{n(cards)} \\ =\frac{4}{52} \end{gathered}[/tex]

The probability of choosing king again without replacement i;

[tex]P(K_2)=\frac{3}{51}[/tex]

Hence the probability of K2 AND K1 is;

[tex]\begin{gathered} P(\frac{K_1}{K_2})=\frac{P(K_1)\times P(K_2)=}{P(K_1)} \\ =\frac{\frac{4}{52}\times\frac{3}{51}}{\frac{4}{52}} \\ =\frac{3}{51} \\ =0.0588 \end{gathered}[/tex]