Drawing from a deck without replacement (Checking my answer)

Question

From Carol Ash's The Probability Tutoring Book,

Draw from a deck without replacement. Find the probability that

a) the $10^{th}$ draw is a king and the $11^{th}$ draw is a non-king

b) the first king occurs on the $10^{th}$ draw

c) it takes $10$ draws to get $3$ kings

My attempts:

a) $$P(10^{th}\text{ is king and }11^{th}\text { is non king}) = P(1^{st}\text{ is king and }2^{nd}\text { is non king}) \\ = \frac{4}{52} \frac{48}{51}$$

b) To get the probability that the first king is on the $10^{th}$ we want $9$ non-kings first. Here is where my confusion begins. I first try a combinations approach: Of $48$ non-kings, pick 9 of them and then of $4$ kings pick $1$ of them.

This results in: $$\frac{{48 \choose 9}{4 \choose 1}} {52 \choose 10}$$

Using a different approach: Draw 9 non-kings one at a time, and then draw a king $$\frac{(48)(47)(46)...(40)}{(52)(51)(51)...(44)} \frac{4}{43}$$

I feel as though these two should produce the same numerical answer, yet the first method produces an answer(0.424) that's a factor of 10 away from the latter method(.0424)

c)$$P(10 \text{ draws to get } 3 \text{ Kings}) = P(10^{th} = \text{King}|2 \text{Kings before})P(2 \text{ Kings before}) = \frac{2}{43} \frac{{4 \choose 2}{48 \choose 7}}{52 \choose 10}$$

Answer 1

a) It is intuitively clear that you answer is correct as there are altogether $52\cdot51$ ways to choose the 10th and 11th cards and $4\cdot 48$ ways to draw a king and a non-king cards. This can be confirmed by the direct count: $$ \frac{\binom{50}{9} 9!\binom{4}{1}\binom{48}{1}}{\binom{52}{11}11!}=\frac{4\cdot 48}{52\cdot51}. $$

b) The probability is: $$ \frac{\binom{48}{9} 9!\binom{4}{1}}{\binom{52}{10}10!}=\frac{\binom{48}{9}}{\binom{52}{9}}\frac4{43}. $$ Of course the r.h.s. can be obtained directly as $\frac4{43}$ is the probability to draw a king card from 43 remaining cards. The missing permutation factors explain the deviation by factor $10$ in your results.

c) It is not quite clear what is here meant. If it is (as you assumed) that the 10th card is a king and there were two king cards before the probability is: $$ \frac{\binom{48}{7}\binom{4}2}{\binom{52}{9}}\frac2{43}, $$ which is except for one important point similar to your answer. The important point is the number $\binom{52}{\color{red}{9}}$ in the denominator. Here you made the same mistake as in the first approach to part b).

Answer 2

I may give my opinion on the part b first. The approach you’ve tried seems right, it is a hypergeometric case. We could consider it as the probability of 9 cards randomly drawn out from 52 cards, but 0 king. According to the hypergeometric distribution:

$P(0)=\left(\frac {{4\choose 0}{48\choose 9}}{{52\choose 9}}\right)$

After that, you can use $P(0)•\left(\frac{4} {48}\right)$ to get the probability that first king be drawn at the 10th.

In terms of part c, the concept of hypergeometric distribution can be applied again. It is nothing more than: $P(3)=\left(\frac {{4\choose 3}{48\choose 7}}{{52\choose 10}}\right)$