# Drawing from a deck without replacement (Checking my answer)

by BigBear   Last Updated June 30, 2020 02:20 AM

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

Tags :