Difference between revisions of "Aufgaben:Exercise 1.5: Drawing Cards"
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| − | [[File:P_ID77__Sto_A_1_5.gif|right|frame|Wish result "Three aces"]] | + | [[File:P_ID77__Sto_A_1_5.gif|right|frame|Wish result "Three aces"]] |
| − | From a deck of $32$ cards, including four aces, three cards are drawn in succession. | + | 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. | + | *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)''' | + | *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 $i$ | + | 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 $i$. |
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*The topic of this chapter is illustrated with examples in the (German language) learning video | *The topic of this chapter is illustrated with examples in the (German language) learning video | ||
| − | :[[Statistische_Abhängigkeit_und_Unabhängigkeit_(Lernvideo)|Statistische Abhängigkeit und Unabhängigkeit]] $\Rightarrow$ "Statistical dependence and independence". | + | ::[[Statistische_Abhängigkeit_und_Unabhängigkeit_(Lernvideo)|Statistische Abhängigkeit und Unabhängigkeit]] $\Rightarrow$ "Statistical dependence and independence". |
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<quiz display=simple> | <quiz display=simple> | ||
| − | {First, consider the case of "drawing with putting back". What is the probability $p_1$, that three aces are drawn? | + | {First, consider the case of "drawing with putting back". What is the probability $p_1$, that three aces are drawn? |
|type="{}"} | |type="{}"} | ||
$p_1 \ = \ $ { 0.002 3% } | $p_1 \ = \ $ { 0.002 3% } | ||
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$p_2 \ = \ $ { 0.0008 3% } | $p_2 \ = \ $ { 0.0008 3% } | ||
| − | { | + | {Consider further the case "drawing without putting back". What is the probability $p_3$ that not a single ace is drawn? |
|type="{}"} | |type="{}"} | ||
$p_3 \ = \ $ { 0.6605 3% } | $p_3 \ = \ $ { 0.6605 3% } | ||
| − | {What is the probability $p_4$ that exactly one ace is drawn | + | {What is the probability $p_4$ that exactly one ace is drawn |
|type="{}"} | |type="{}"} | ||
$p_4 \ = \ $ { 0.3048 3% } | $p_4 \ = \ $ { 0.3048 3% } | ||
| − | {What is the probability $p_5$ that two of the drawn cards are aces? <br> | + | {What is the probability $p_5$ that two of the drawn cards are aces? <br>Hint: Consider that the four events [exactly $i$ aces are drawn] with $i \in \{ 0, 1, 2, 3 \}$ describe a complete system. |
|type="{}"} | |type="{}"} | ||
$p_5 \ = \ $ { 0.0339 3% } | $p_5 \ = \ $ { 0.0339 3% } | ||
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===Solution=== | ===Solution=== | ||
{{ML-Kopf}} | {{ML-Kopf}} | ||
| − | '''(1)''' For each card, the probability of an ace is exactly equal $ | + | '''(1)''' For each card, the probability of an ace is exactly equal $1/8$: |
:$$p_{\rm 1} = {\rm Pr} (3 \hspace{0.1cm} {\rm Asse}) = {\rm Pr} (A_{\rm 1})\cdot {\rm Pr} (A_{\rm 2})\cdot {\rm Pr}(A_{\rm 3}) = \rm ({1}/{8})^3 \hspace{0.15cm}\underline {\approx 0.002}.$$ | :$$p_{\rm 1} = {\rm Pr} (3 \hspace{0.1cm} {\rm Asse}) = {\rm Pr} (A_{\rm 1})\cdot {\rm Pr} (A_{\rm 2})\cdot {\rm Pr}(A_{\rm 3}) = \rm ({1}/{8})^3 \hspace{0.15cm}\underline {\approx 0.002}.$$ | ||
| − | + | '''(2)''' Now, using the general multiplication theorem, we obtain: | |
| − | '''(2)''' Now, using the general multiplication theorem, we obtain: | ||
:$$p_{\rm 2} = {\rm Pr} (A_{\rm 1}\cap A_{\rm 2} \cap A_{\rm 3} ) = {\rm Pr} (A_{\rm 1}) \cdot {\rm Pr} (A_{\rm 2}\hspace{0.05cm}|\hspace{0.05cm}A_{\rm 1} ) \cdot {\rm Pr} \big[A_{\rm 3} \hspace{0.05cm}|\hspace{0.05cm}( A_{\rm 1}\cap A_{\rm 2} )\big].$$ | :$$p_{\rm 2} = {\rm Pr} (A_{\rm 1}\cap A_{\rm 2} \cap A_{\rm 3} ) = {\rm Pr} (A_{\rm 1}) \cdot {\rm Pr} (A_{\rm 2}\hspace{0.05cm}|\hspace{0.05cm}A_{\rm 1} ) \cdot {\rm Pr} \big[A_{\rm 3} \hspace{0.05cm}|\hspace{0.05cm}( A_{\rm 1}\cap A_{\rm 2} )\big].$$ | ||
| − | *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): | + | *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): |
:$$p_{\rm 2} ={4}/{32}\cdot {3}/{31}\cdot{2}/{30} \hspace{0.15cm}\underline { \approx 0.0008}.$$ | :$$p_{\rm 2} ={4}/{32}\cdot {3}/{31}\cdot{2}/{30} \hspace{0.15cm}\underline { \approx 0.0008}.$$ | ||
| − | *We can see: $p_2$ is smaller than $p_1$, | + | *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)''' | + | '''(3)''' Analogous to subtask '''(2)''', we obtain here: |
:$$p_{\rm 3} = {\rm Pr}(\overline{A_{\rm 1}})\cdot {\rm Pr} (\overline{A_{\rm 2}} \hspace{0.05cm}|\hspace{0.05cm}\overline{A_{\rm 1}})\cdot {\rm Pr} (\overline{A_{\rm 3}}\hspace{0.05cm}|\hspace{0.05cm}(\overline{A_{\rm 1}} \cap \overline{A_{\rm 2}} )) = {28}/{32}\cdot{27}/{31}\cdot {26}/{30}\hspace{0.15cm}\underline {\approx 0.6605}.$$ | :$$p_{\rm 3} = {\rm Pr}(\overline{A_{\rm 1}})\cdot {\rm Pr} (\overline{A_{\rm 2}} \hspace{0.05cm}|\hspace{0.05cm}\overline{A_{\rm 1}})\cdot {\rm Pr} (\overline{A_{\rm 3}}\hspace{0.05cm}|\hspace{0.05cm}(\overline{A_{\rm 1}} \cap \overline{A_{\rm 2}} )) = {28}/{32}\cdot{27}/{31}\cdot {26}/{30}\hspace{0.15cm}\underline {\approx 0.6605}.$$ | ||
| − | '''(4)''' This probability can be expressed as the sum of three probabilities, since the associated events are disjoint: | + | '''(4)''' This probability can be expressed as the sum of three probabilities, since the associated events are disjoint: |
| − | :$$p_{\rm 4} = {\rm Pr} (D_{\rm 1} \cup D_{\rm 2} \cup D_{\rm 3}) \rm \hspace{0.1cm} | + | :$$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 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 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,$$ | ||
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*These probabilities are all the same – why should it be any different? | *These probabilities are all the same – why should it be any different? | ||
| − | *If you draw exactly one ace from three cards, it is just as likely whether you draw it first, second, or third. | + | *If you draw exactly one ace from three cards, it is just as likely whether you draw it first, second, or third. |
| − | *This gives $p_4 \; \underline{= 0.3048}$ for the sum. | + | *This gives $p_4 \; \underline{= 0.3048}$ for the sum. |
| − | '''(5)''' If we define the events $E_i :=$ "Exactly $i$ aces are drawn on three cards" with index $i \in \{ 0, 1, 2, 3 \}$, <br> then $E_0$, $E_1$, $E_2$ and $E_3$ describe a complete system. Therefore: | + | '''(5)''' If we define the events $E_i :=$ "Exactly $i$ aces are drawn on three cards" with index $i \in \{ 0, 1, 2, 3 \}$, <br> then $E_0$, $E_1$, $E_2$ and $E_3$ describe a complete system. Therefore: |
:$$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}.$$ | :$$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}.$$ | ||
{{ML-Fuß}} | {{ML-Fuß}} | ||
Latest revision as of 16:04, 30 November 2021
Wish result "Three aces"
From a deck of $32$ cards, including four aces, three cards are drawn in succession.
- For subtask (1) it is assumed that after drawing a card it is put back into the deck, then the deck is reshuffled and the next card is drawn.
- In contrast, for the other subtasks starting with (2), you are supposed to assume that the three cards are drawn at once ("draw without putting back").
In the following, we denote by $A_i$ the event that the card drawn at time $i$ is an ace. Here $i \in \{ 1, 2, 3 \}$. The complementary event then states that some card other than an ace is drawn at time $i$.
Hints:
- The exercise belongs to the chapter Statistical dependence and independence.
- The topic of this chapter is illustrated with examples in the (German language) learning video
- Statistische Abhängigkeit und Unabhängigkeit $\Rightarrow$ "Statistical dependence and independence".
Questions
Solution
(1) For each card, the probability of an ace is exactly equal $1/8$:
- $$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}.$$
