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--- tanszek:oktatas:techcomm:statistical_properties [2024/09/30 09:54] – knehez
+++ tanszek:oktatas:techcomm:statistical_properties [2024/10/04 19:17] (current) – knehez
@@ Line 17: / Line 17: @@
 In the \( E_i \) event space, an event happened \(k_i\) times then the frequency of that given event may be calculated with the following formula:
-\(g_i=\frac{k_i}{k} \)
+\(freq_i=\frac{k_i}{k} \)
 This means that we divide the number of all events by the number (frequency) of that given event. In case of a large number of experiments this number will show us the probability of that event.
-$$ \lim_{k \to \infty} g_i = \frac{k_i}{k} =  P(E_i) $$
+$$ \lim_{k \to \infty} freq_i = \frac{k_i}{k} =  P(E_i) $$
   * if \(k_i = k \), then the occurrence will be **certain** \(P(E) = 1 \)
@@ Line 33: / Line 33: @@
 {{:tanszek:oktatas:techcomm:pasted:20240826-164556.png}}
-The probability of every event will be 1/6 which can be calculated with the following formula. The rolls (of the dice) will form a so-called //full event system//. In the case of a //full event system// the sum of the probabilities (of each event) is 1 by definition.
+The probability of every event will be 1/6 which can be calculated with the following formula. The rolls (of the dice) will form a so-called //full event system//. In the case of a //full event system// the sum of the probabilities (of each event) is 1 (by definition).
-If one the events happen then the others can not happen:
+If one of the events happens then the others can not happen:
 $$ \sum_{i=1}^{n} p(E_i) = 1 \text{ where } E_i \cap E_j = 0 \text{ and } i \not = j $$
-This formula applies to a set of //mutually exclusive// events. Since these events are disjoint (no overlap between any two different events) and together they cover all possible outcomes (the sum of their probabilities is 1).
+This formula applies to a set of //mutually exclusive// events. Since these events are disjoint (no overlap between any two different events) and they together cover all possible outcomes (the sum of their probabilities is 1).
 === Sum of events ===
@@ Line 53: / Line 53: @@
 We can only form a general formula, if we extract the probability of the intersection of both sets from the total sum:
-$$ p(A \cap B) = p(A) + p(B) - p(A \cup B) $$
+$$ p(A \cup B) = p(A) + p(B) - p(A \cap B) $$
 So we can say that (according to this formula) the total probability for both events is 4/6.
@@ Line 59: / Line 59: @@
 **Example 2**: What is the probability that we roll the dice twice and the results will be 6 in both cases?
-The probability for throwing a 6 is 1/6, but we cannot multiply it by 2 and take the result (2/6). We can not do that because both events are independent. There is no connection between the 2 trials, so we have to look at them as independent events. So the result will be the multiplication of both events:
+The probability of throwing a 6 is 1/6, but we cannot multiply it by 2 and take the result (2/6). We can not do that because both events are independent. There is no connection between the 2 trials, so we have to look at them as independent events. So the result will be the multiplication of both events:
-$$ p(A) p(B) = \frac{1}{6} \frac{1}{6} = \frac{1}{36} $$
+$$ p(A) \times p(B) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36} $$
+**Example 3**: What is the probability of getting at least one 5 when rolling two dice?
+Let //Event A// represent rolling a 5 on the first die and //Event B// represent rolling a 5 on the second die. These are independent events.
+  * The probability of rolling a 5 on the first die is: \( p(A) = \frac{1}{6} \)
+  * The probability of rolling a 5 on the second die is \( p(B) = \frac{1}{6} \)
+  * The probability of both dice showing a 5 (i.e, \( A \cap B \) is \( p(A \cap B) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36} \)
+$$ p(A \cup B) = p(A) + p(B) - p(A \cap B) $$
+$$ p(A \cup B) = \frac{1}{6} + \frac{1}{6} - \frac{1}{36} =  \frac{2}{6} - \frac{1}{36} = \frac{12}{36} - \frac{1}{36} = \frac{11}{36} $$
+**Example 4:**:  Woman is expecting twins, what is the probability that one of the children will be a girl?
+Create sets containing all the possibilities.
+Try the following c code here: https://www.onlinegdb.com/online_c_compiler and examine the results.
+<sxh c>
+#include <stdio.h>
+#include <stdlib.h>
+#include <time.h>
+#define SIMULATIONS 1000
+// Function to simulate the birth of twins
+int simulate_twins() {
+    // Randomly determine the gender of the twins (0 = boy, 1 = girl)
+    int first_child = rand() % 2;
+    int second_child = rand() % 2;
+    // Return 1 if there is at least one girl, otherwise 0
+    return (first_child == 1 || second_child == 1);
+}
+int main() {
+    int at_least_one_girl = 0;
+    // Initialize the random number generator
+    srand(time(0));
+    // Simulate 1000 pairs of twins
+    for (int i = 0; i < SIMULATIONS; i++) {
+        if (simulate_twins()) {
+            at_least_one_girl++;
+        }
+    }
+    // Calculate and print the result
+    double probability = (double)at_least_one_girl / SIMULATIONS * 100;
+    printf("In %.2f%% of cases, there is at least one girl in the twin pair.\n", probability);
+    return 0;
+}
+</sxh>
+**Example 5:**: https://en.wikipedia.org/wiki/Monty_Hall_problem