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--- tanszek:oktatas:techcomm:statistical_properties [2024/09/30 10:29] – 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 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.
-**Example 2**: What is the probability of getting at least one 5 when rolling two dice?
+**Example 2**: What is the probability that we roll the dice twice and the results will be 6 in both cases?
+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:
+$$ p(A) \times p(B) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36} $$
+**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.
@@ Line 65: / Line 71: @@
   * The probability of both dice showing a 5 (i.e, \( A \cap B \) is \( p(A \cap B) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36} \)
-**Example 3**: What is the probability that we roll the dice twice and the results will be 6 in both cases?
-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:
+$$ p(A \cup B) = p(A) + p(B) - p(A \cap B) $$
+$$ p(A \cup B) = \frac{1}{6} + \frac{1}{6} - \frac{1}{36} =  \frac{2}{6} - \frac{1}{36} = \frac{12}{36} - \frac{1}{36} = \frac{11}{36} $$
+**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);
-$$ p(A) p(B) = \frac{1}{6} \frac{1}{6} = \frac{1}{36} $$
+    return 0;
+}
+</sxh>
+**Example 5:**: https://en.wikipedia.org/wiki/Monty_Hall_problem