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

42.4

step-by-step explanation:

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