tanszek:oktatas:techcomm:statistical_properties

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tanszek:oktatas:techcomm:statistical_properties [2024/09/30 10:44] kneheztanszek:oktatas:techcomm:statistical_properties [2024/10/04 19:17] (current) knehez
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 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:  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. 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** \(P(E) = 1 \)   * if \(k_i = k \), then the occurrence will be **certain** \(P(E) = 1 \)
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 So we can say that (according to this formula) the total probability for both events is 4/6. So we can say that (according to this formula) the total probability for both events is 4/6.
  
-**Example 3**: What is the probability that we roll the dice twice and the results will be 6 in both cases?+**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: 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:
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 $$ p(A) \times p(B) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36} $$ $$ p(A) \times p(B) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36} $$
  
-**Example 2**: What is the probability of getting at least one 5 when rolling two dice?+**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. 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.
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 $$ p(A \cup B) = \frac{1}{6} + \frac{1}{6} - \frac{1}{36} =  \frac{2}{6} - \frac{1}{36} = \frac{12}{36} - \frac{1}{36} = \frac{11}{36} $$ $$ p(A \cup B) = \frac{1}{6} + \frac{1}{6} - \frac{1}{36} =  \frac{2}{6} - \frac{1}{36} = \frac{12}{36} - \frac{1}{36} = \frac{11}{36} $$
  
 +**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
  
tanszek/oktatas/techcomm/statistical_properties.1727693098.txt.gz · Last modified: 2024/09/30 10:44 by knehez