Problem: Out of a pack of 52 playing cards, 3 cards are drawn one after the other without replacement. If a fourth card is drawn, what is the probability that it is a king?
One of my friends has given this as the solution.
`The probability that the 4th card is a king when the cards are drawn one by one without replacement is same as the probability that the fourth card obtained is a king when the 4 cards are drawn simultaneously.
This is the same as the probability that the first card drawn is a king when the four cards are drawn simultaneously, because when all the cards are drawn simultaneously, order doesn't matter. This probability is given by 4 C 1/ 52 C 1, hence the required probability is 4/52 = 1/13.`
But I don't think it is correct. Since chances of getting King in 4th trial will depend on the number of kings drawn in the first 3 draws? Am I correct?
So, solution would be
(4/52)(48/51)(47/50)(3/49) + (4/52)(3/51)(48/50)(2/49) + (4/52)(3/51)(2/50)*(1/49)
Is my solution correct? If not why not? If yes, then is there a faster way to get to the answer?
Best Answer
A priori the $k^{\rm th}$ card of the deck is a king with probability ${1\over13}$, whatever $k\in[52]$. Since we are not told the values of the first three cards drawn this probability is still the same, namely ${1\over13}$, at the moment we are asked the question about the fourth card.