tanszek:oktatas:techcomm:statistical_properties
Differences
This shows you the differences between two versions of the page.
Both sides previous revisionPrevious revisionNext revision | Previous revision | ||
tanszek:oktatas:techcomm:statistical_properties [2024/09/30 10:44] – 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: | 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. | ||
- | $$ \lim_{k \to \infty} | + | $$ \lim_{k \to \infty} |
* if \(k_i = k \), then the occurrence will be **certain** \(P(E) = 1 \) | * if \(k_i = k \), then the occurrence will be **certain** \(P(E) = 1 \) | ||
Line 57: | Line 57: | ||
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 | + | **Example |
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: | ||
Line 63: | Line 63: | ||
$$ p(A) \times p(B) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36} $$ | $$ p(A) \times p(B) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36} $$ | ||
- | **Example | + | **Example |
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. | ||
Line 75: | Line 75: | ||
$$ p(A \cup B) = \frac{1}{6} + \frac{1}{6} - \frac{1}{36} = \frac{2}{6} - \frac{1}{36} = \frac{12}{36} - \frac{1}{36} = \frac{11}{36} $$ | $$ p(A \cup B) = \frac{1}{6} + \frac{1}{6} - \frac{1}{36} = \frac{2}{6} - \frac{1}{36} = \frac{12}{36} - \frac{1}{36} = \frac{11}{36} $$ | ||
+ | **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:// | ||
+ | |||
+ | <sxh c> | ||
+ | #include < | ||
+ | #include < | ||
+ | #include < | ||
+ | |||
+ | #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; | ||
+ | if (simulate_twins()) { | ||
+ | at_least_one_girl++; | ||
+ | } | ||
+ | } | ||
+ | |||
+ | // Calculate and print the result | ||
+ | double probability = (double)at_least_one_girl / SIMULATIONS * 100; | ||
+ | printf(" | ||
+ | |||
+ | return 0; | ||
+ | } | ||
+ | </ | ||
+ | |||
+ | |||
+ | **Example 5:**: https:// | ||
tanszek/oktatas/techcomm/statistical_properties.1727693098.txt.gz · Last modified: 2024/09/30 10:44 by knehez