Exercise 1.5: Drawing Cards

From LNTwww

Wish result  "Three aces"

From a deck of  $32$  cards,  including four aces,  three cards are drawn in succession.

  • For subtask  (1)  it is assumed that after drawing a card it is put back into the deck,  then the deck is reshuffled and the next card is drawn.

  • In contrast,  for the other subtasks starting with  (2),  you are supposed to assume that the three cards are drawn at once  ("draw without putting back").

In the following,  we denote by  $A_i$  the event that the card drawn at time  $i$  is an ace.  Here  $i \in \{ 1, 2, 3 \}$.  The complementary event then states that some card other than an ace is drawn at time  $i$.


  • The topic of this chapter is illustrated with examples in the  (German language)  learning video
Statistische Abhängigkeit und Unabhängigkeit   $\Rightarrow$   "Statistical dependence and independence".



First,  consider the case of  "drawing with putting back".  What is the probability  $p_1$,  that three aces are drawn?

$p_1 \ = \ $


What is the probability  $p_2$  that three aces will be drawn if the cards are not put back?  Why is  $p_2$  smaller/equal/larger than  $p_1$?

$p_2 \ = \ $


Consider further the case  "drawing without putting back".  What is the probability  $p_3$  that not a single ace is drawn?

$p_3 \ = \ $


What is the probability  $p_4$  that exactly one ace is drawn

$p_4 \ = \ $


What is the probability  $p_5$  that two of the drawn cards are aces? 
Hint:  Consider that the four events  [exactly  $i$  aces are drawn]  with  $i \in \{ 0, 1, 2, 3 \}$  describe a complete system.

$p_5 \ = \ $


(1)  For each card,  the probability of an ace is exactly equal  $1/8$:

$$p_{\rm 1} = {\rm Pr} (3 \hspace{0.1cm} {\rm Asse}) = {\rm Pr} (A_{\rm 1})\cdot {\rm Pr} (A_{\rm 2})\cdot {\rm Pr}(A_{\rm 3}) = \rm ({1}/{8})^3 \hspace{0.15cm}\underline {\approx 0.002}.$$

(2)  Now,  using the general multiplication theorem,  we obtain:

$$p_{\rm 2} = {\rm Pr} (A_{\rm 1}\cap A_{\rm 2} \cap A_{\rm 3} ) = {\rm Pr} (A_{\rm 1}) \cdot {\rm Pr} (A_{\rm 2}\hspace{0.05cm}|\hspace{0.05cm}A_{\rm 1} ) \cdot {\rm Pr} \big[A_{\rm 3} \hspace{0.05cm}|\hspace{0.05cm}( A_{\rm 1}\cap A_{\rm 2} )\big].$$
  • The conditional probabilities are computable according to the classical definition.  For this,  one obtains  $k/m$  (with  $m$  cards, there are still  $k$  aces in the deck):
$$p_{\rm 2} ={4}/{32}\cdot {3}/{31}\cdot{2}/{30} \hspace{0.15cm}\underline { \approx 0.0008}.$$
  • We can see:   $p_2$  is smaller than  $p_1$,  since now the second and third aces are less probable than before.

(3)  Analogous to subtask  (2),  we obtain here:

$$p_{\rm 3} = {\rm Pr}(\overline{A_{\rm 1}})\cdot {\rm Pr} (\overline{A_{\rm 2}} \hspace{0.05cm}|\hspace{0.05cm}\overline{A_{\rm 1}})\cdot {\rm Pr} (\overline{A_{\rm 3}}\hspace{0.05cm}|\hspace{0.05cm}(\overline{A_{\rm 1}} \cap \overline{A_{\rm 2}} )) = {28}/{32}\cdot{27}/{31}\cdot {26}/{30}\hspace{0.15cm}\underline {\approx 0.6605}.$$

(4)  This probability can be expressed as the sum of three probabilities,  since the associated events are disjoint:

$$p_{\rm 4} = {\rm Pr} (D_{\rm 1} \cup D_{\rm 2} \cup D_{\rm 3}) \rm \hspace{0.1cm}with\hspace{-0.1cm}:$$
$$ {\rm Pr} (D_{\rm 1}) = {\rm Pr}( A_{\rm 1} \cap \overline{ A_{\rm 2}} \cap \overline{A_{\rm 3}}) = \rm \frac{4}{32}\cdot \frac{28}{31}\cdot \frac{27}{30}=\rm 0.1016,$$
$${\rm Pr} (D_{\rm 2}) = \rm Pr ( \overline{A_{\rm 1}} \cap A_{\rm 2} \cap \overline{A_{\rm 3}}) = \rm \frac{28}{32}\cdot \frac{4}{31}\cdot \frac{27}{30}=\rm 0.1016,$$
$${\rm Pr} (D_{\rm 3} \rm) = Pr ( \overline{\it A_{\rm 1}} \cap \overline{\it A_{\rm 2}} \cap A_{\rm 3}) = \rm \frac{28}{32}\cdot \frac{27}{31}\cdot \frac{4}{30}=\rm 0.1016.$$
  • These probabilities are all the same – why should it be any different?
  • If you draw exactly one ace from three cards,  it is just as likely whether you draw it first,  second, or third.
  • This gives   $p_4 \; \underline{= 0.3048}$  for the sum.

(5)  If we define the events  $E_i :=$  "Exactly  $i$  aces are drawn on three cards"  with index  $i \in \{ 0, 1, 2, 3 \}$,
        then  $E_0$,  $E_1$,  $E_2$  and  $E_3$  describe a complete system.  Therefore:

$$p_{\rm 5} = {\rm Pr}(E_2) = 1 - {\rm Pr}(E_0) -{\rm Pr}(E_1) - {\rm Pr}(E_3) = 1 - p_3 -p_4 - p_2 \hspace{0.15cm}\underline {= \rm 0.0339}.$$