Two cards are drawn without replacement from a standard deck of 52 playing cards. What is the probability of choosing a king for the second card drawn, if the first card, drawn without replacement, was a king? Express your answer as a fraction or a decimal number rounded to four decimal places.

the probability of choosing a king for the second card drawn is 3/51, if the first card, drawn without replacement, was a king

Step-by-step explanation:

defining the variable F = choosing a king in the first drawn, then the probability is

P (F) = 4/52

then using the theorem of Bayes for conditional probability and denoting the event S = choosing a king in the second drawn, then

P (S/F) = P (S∩F) / P (F)

where

P (S∩F) = probability of choosing a king in the first drawn and second drawn = 4/52 * 3/51

P (S∩F) = probability of choosing a king in the second drawn given that a king was chosen in the first drawn

then

P (S/F) = P (S∩F) / P (F) = 4/52 * 3/51 / 4/52 = 3/51

The probability of choosing a king for the second card drawn, if the first card, drawn without replacement, was a king

P (K2|K1) = 3/51

Step-by-step explanation:

In a standard deck of 52 playing cards consists of 4 kings.

Let P (K1) represent the probability of selecting a king as the first card and P (K2) the probability of selecting a king as the second card.

P (K1) = N (K1) / N (total) = 4/52

Since it's without replacement

N (total) and N (K) are both reduced by 1.

P (K2) = N (K2) / N (total)

P (K2) = 3/51

the probability of choosing a king for the second card drawn, if the first card, drawn without replacement, was a king can be given as;

P (K2|K1) = P (K2) P (K1) / [P (K1) P (K2) + P (K1) P (K2')

P (K2') = 1 - P (K2) = 1 - 3/51 = 48/51

P (K2|K1) = (3/51*4/52) / [4/52*3/51 + 4/52*48/51]

P (K2|K1) = (3/51*4/52) / [4/52*51/51] = (3/51*4/52) / 4/52

P (K2|K1) = 3/51