Aufgaben:Exercise 3.1Z: Drawing Cards

• Return to book

The desired result
»Three aces are drawn«

From a deck of  $32$  cards, including four aces, three cards are drawn in succession.  For question  (1),  it is assumed that after a card has been drawn

• it is put back into the deck,
• the deck is reshuffled and
• then the next card is drawn.

In contrast, for the other sub-questions from  (2)  onwards, you should assume that the three cards are drawn all at once 
("card draw without putting back").

• In the following, we use  $A_i$  to denote the event that the card drawn at time  $i$  is an ace.  
  Here we have to set  $i = 1,\ 2,\ 3$ .
• The complementary event  $\overline{\it A_i}$  then states that at time  $i$  no ace is drawn, but any other card.

Hints:

• The exercise belongs to the chapter  Some preliminary remarks on 2D random variables.
• In particular, the subject matter of the chapter  Statistical Dependence and Independence in the book "Stochastic Signal Theory" is repeated here.
• A summary of the theoretical basics with examples can be found in the (German language) learning video 
        Statistische Abhängigkeit und Unabhängigkeit   ⇒   "Statistical Dependence and Independence".

Questions

1 First, consider the case of  "card draw with putting back".  What is the probability  $p_1$, that three aces will be drawn?

$p_1 \ = \ $

2 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 \ = \ $

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

$p_3 \ = \ $

4 What is the probability  $p_4$ that exactly one ace is drawn in the case  "card draw without putting back"?

$p_4 \ = \ $

5 What is the probability that two of the three drawn cards are aces?
Note:   The events „exactly  $i$  aces are drawn” with  $i = 0,\ 1,\ 2,\ 3$  describe a so-called  "complete system".

$p_5 \ = \ $

Solution

Solution

(1)  If the cards are put back after being drawn, the probability of an ace is the same at every time  $(1/8)$:

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

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

$$ p_{\rm 2} = \rm Pr (\it A_{\rm 1}\cap \it A_{\rm 2} \cap \it A_{\rm 3} \rm ) = \rm Pr (\it A_{\rm 1}\rm ) \cdot \rm Pr (\it A_{\rm 2} |\it A_{\rm 1}\rm ) \cdot \rm Pr (\it A_{\rm 3} |( \it A_{\rm 1}\cap \it A_{\rm 2} \rm )).$$

• The conditional probabilities can be calculated according to the classical definition.
• One thus obtains the result  $k/m$  $($with  $m$  cards there are still  $k$  aces$)$:

$$p_{\rm 2} =\rm \frac{4}{32}\cdot \frac{3}{31}\cdot\frac{2}{30}\hspace{0.15cm}\underline{ \approx 0.0008}.$$

• $p_2$  is smaller than  $p_1$, because now the second and third aces are less likely than before.

(3)  Analogous to sub-task  (2),  we obtain here:

$$p_{\rm 3} = \rm Pr (\overline{\it A_{\rm 1}})\cdot \rm Pr (\overline{\it A_{\rm 2}} \hspace{0.05cm}|\hspace{0.05cm}\overline{\it A_{\rm 1}})\cdot \rm Pr (\overline{\it A_{\rm3}}\hspace{0.05cm}|\hspace{0.05cm}(\overline{\it A_{\rm 1}} \cap \overline{\it A_{\rm 2}} )) =\rm \frac{28}{32}\cdot\frac{27}{31}\cdot\frac{26}{30}\hspace{0.15cm}\underline{\approx 0.6605}.$$

(4)  This probability can be expressed as the sum of three probabilities   ⇒   $p_{\rm 4} = \rm Pr (\it D_{\rm 1} \cup \it D_{\rm 2} \cup \it D_{\rm 3}) $.

• The corresponding events  $D_1$,  $D_2$  and  $D_3$  are disjoint:

$${\rm Pr} (D_1) = {\rm Pr} (A_1 \cap \overline{ \it A_{\rm 2}} \cap \overline{\it A_{\rm 3}}) = \rm \frac{4}{32}\cdot \frac{28}{31}\cdot \frac{27}{30}=\rm 0.1016,$$

$${\rm Pr} (D_2) = {\rm Pr} ( \overline{A_1} \cap A_2 \cap \overline{A_3}) = \rm \frac{28}{32}\cdot \frac{4}{31}\cdot\frac{27}{30}=\rm 0.1016,$$

$${\rm Pr} (D_3) = {\rm Pr} ( \overline{A_1} \cap \overline{A_2} \cap A_3) = \rm \frac{28}{32}\cdot \frac{27}{31}\cdot \frac{4}{30}=\rm 0.1016.$$

• 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 this as the first, second or third card.
• This gives us for the sum:

$$p_{\rm 4}= {\rm Pr} (D_1 \cup D_2 \cup D_3) \rm \hspace{0.15cm}\underline{= 0.3084}.$$

(5)  If one defines the events  $E_i =$  »Exactly  $i$  aces are drawn«  with the indices  $i = 0,\ 1,\ 2,\ 3$,

• then  $E_0$,  $E_1$,  $E_2$  and $E_3$  describe a  "complete system".
• Therefore:

$$p_{\rm 5} = {\rm Pr} (E_2) = 1 - p_{\rm 2} - p_{\rm 3} - p_{\rm 4} \hspace{0.15cm}\underline{= \rm 0.0339}.$$