# Two cards are drawn one after another…

Two cards are drawn one after another (without replacement) from a well shuffled pack of $52$ playing cards. Show the probability of getting or not getting a face catd by drawing a tree diagram.

My Approach

For first drawn

$$n(S)=52$$ $$n(F)=12$$ $$n(N)=40$$

Then, $$P(F)=\frac {12}{52}$$ $$P(N)=\frac {40}{52}$$

Note, S=sample space F=face cards N= non face cards What should I do next? Help please.

Than

• Do you know what a tree diagram (mentioned in the question) is? How would you use it here? – shardulc says Reinstate Monica Aug 22 '16 at 23:41
• Seems a tree diagram here could represent whether the first draw was a face card or not, then give the probability of a second draw being a face card. Visually depicts conditional probability – manofbear Aug 23 '16 at 1:53

Find the probability of getting two non-face cards and subtract from 1. $$P(\text{Both non-face}) = P(\text{non-face on 1st})P(\text{non-face on 2nd | non-face on first}) = (40/52)(39/51)= 0.5882353.$$ Then $$P(\text{At least one face}) = 1 - 0.5882353 = 0.4117647.$$ Now, can you figure out the probability of getting exactly one face card?

Here is a simulation in R statistical software of the distribution of $X =$ number of face cards. It is based on a million performance of the 2 card experiment, each performance starting with a full shuffled deck. (Expect 2 or 3 place accuracy.)

m = 10^6; deck = 1:52; n = 2; x = numeric(m)
for(i in 1.:m) {
draw = sample(deck, n)
x[i] = sum(draw > 40) }  # face cards numbered 41 thru 52
table(x)/m
x
0        1        2
0.587444 0.362590 0.049966


The random variable $X$ has a 'hypergeometric distribution'. You can see if that is discussed in your text or look at the Wikipedia article. The histogram below shows the simulated distribution of $X$ and the dots atop bars show exact hypergeometric probabilities. (The resolution of the graph on the probability scale is about two decimal places, so the agreement looks perfect.)

R code for the graph:

hist(x, prob=T, br=(0:3)-.5, col="wheat", main="Face Cards in 2 Draws")
i = 0:2;  pdf = dhyper(i, 12, 40, 2)  # 12 face, 40 non-face, 2 drawn
points(i, pdf, pch=19, col="blue")


A Tree diagram looks roughly:

$$\begin{array}{cc} &&\mathsf P(F_1,F_2) \\ & _{\small\mathsf P(F_2\mid F_1)}\nearrow \\ & \bullet \\ & _{\small\mathsf P(N_2\mid F_1)}\searrow \\ ^{\small\mathsf P(F_1)}\nearrow&& \mathsf P(F_1, N_2) \\ \circ \\ ^{\small\mathsf P(N_1)}\searrow&& \mathsf P(N_1, F_2) \\ & _{\small\mathsf P(F_2\mid N_1)}\nearrow \\ & \bullet \\ & ^{\small\mathsf P(N_2\mid N_1)}\searrow \\ &&\mathsf P(N_1,N_2) \end{array}$$

Supply the numbers.