1
$\begingroup$

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.

Thanks in advance.

Than

$\endgroup$
  • 1
    $\begingroup$ Do you know what a tree diagram (mentioned in the question) is? How would you use it here? $\endgroup$ – shardulc says Reinstate Monica Aug 22 '16 at 23:41
  • $\begingroup$ 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 $\endgroup$ – manofbear Aug 23 '16 at 1:53
0
$\begingroup$

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.)

enter image description here

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")
$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.