Exercise 3.1Z: Drawing Cards

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
("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 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

$p_1 \ = \ $

$p_2 \ = \ $

$p_3 \ = \ $

$p_4 \ = \ $

$p_5 \ = \ $

Solution

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

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

(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 ${\rm Pr}(D_1)$, ${\rm Pr}(D_2)$ and ${\rm Pr}(D_3)$ are disjoint:

$$\rm Pr (\it D_{\rm 1}) = \rm Pr (\it A_{\rm 1} \cap \overline{ \it A_{\rm 2}} \cap \overline{\it A_{\rm 3}}) = \rm \frac{4}{32}\cdot \frac{28}{31}\cdot \frac{27}{30}=\rm 0.1016,$$

$$\rm Pr (\it D_{\rm 2}) = \rm Pr ( \overline{\it A_{\rm 1}} \cap \it A_{\rm 2} \cap \overline{\it A_{\rm 3}}) = \rm \frac{28}{32}\cdot \frac{4}{31}\cdot\frac{27}{30}=\rm 0.1016,$$

$$\rm Pr (\it D_{\rm 3}) = \rm Pr ( \overline{\it A_{\rm 1}} \cap \overline{\it A_{\rm 2}} \cap \it A_{\rm 3}) = \rm \frac{28}{32}\cdot \frac{27}{31}\cdot \frac{4}{30}=\rm 0.1016.$$

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

$$p_{\rm 4}= \rm Pr (\it D_{\rm 1} \cup \it D_{\rm 2} \cup \it D_{\rm 3}) \rm \hspace{0.15cm}\underline{= 0.3084}.$$

(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:

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