Question

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 1

`\begin{equation*} 0.0588 \end{equation*}`

Step-by-step explanation

A standard deck of cards has 52 cards.

The number of king is 4

The probability of choosing the first king is;

`\begin{gathered} P(K_1)=(n(K))/(n(cards)) \\ =(4)/(52) \end{gathered}`

The probability of choosing king again without replacement i;

`P(K_2)=(3)/(51)`

Hence the probability of K2 AND K1 is;

`\begin{gathered} P((K_1)/(K_2))=(P(K_1)* P(K_2)=)/(P(K_1)) \\ =((4)/(52)*(3)/(51))/((4)/(52)) \\ =(3)/(51) \\ =0.0588 \end{gathered}`