Friday, November 23, 2018

Probability Formula

In probability theory, experiments that cannot be ascertained or controlled by values ​​of certain variables where the first experiment will differ from the subsequent experiment are called random experiments even though most of the conditions are the same. For example, if we throw a dice, then the results of this experiment will appear one of the sets {1, 2, 3, 4, 5, 6}. We do not know exactly what numbers will appear and cannot control the numbers that have to appear, but we only know what possible numbers will appear. All possible results from a random experiment are called sample rooms. After we throw a dice then the number 1 appears then this is called an event which is the result of an experiment conducted. In other words, the event is a subset of the sample space.

The probability value represents how much we believe in the event that will occur. If the opportunity value is close to 1, we have a great confidence in the occurrence of the event. Conversely, if the opportunity value is close to 0, our small belief will occur. An event that is certain to occur has an probability value of 1. For example, the chance of the appearance of images or numbers in one coin toss Because if we do one coin toss then what will definitely appear is a picture or number. While events that might not occur have an probability value of 0. For example, the probability for the emergence of images and numbers in one throw of a coin because it is not possible the results of the occurrence of images and numbers appear simultaneously. This event means mutually exclusive events, ie if the number that appears means an image that does not appear, or vice versa. Calculating Opportunities with the Nisbi Frequency Approach

Suppose that a random experiment is carried out n times. If event A appears as many times ($ 0 \le k \le n $), then the relative frequency of occurrence of event A is determined by the formula:
$ F (A) = \frac{k}{n} $
If the value of n approaches infinity ($ n \rightarrow \infty $), then the value $ \frac{k}{n} $ tends to be close to a certain value. This particular value is the chance value of the occurrence of event A.
$ P (A) = \lim_{n \rightarrow \infty} F (A) = \lim_{n \rightarrow \infty} \frac{k}{n} $
Calculating Opportunities by Approaching Classical Opportunity Definitions
Suppose that in an experiment there are possible results with each of them having equally likely opportunities. If event A can appear as many times as k, then the chance of occurrence A is determined by the formula:
$ P (A) = \frac{k}{n}$
Calculating Opportunities Using Sample Space
To be able to work on probability questions from random experiments, we must understand what the sample space means (which is usually symbolized by S) and what is the event (usually symbolized by A) in a random experiment. The opportunity formula to be used is:
$ P (A) = \frac{n (A)}{n (S)} $
where P (A) states the chance of occurrence A; n (A) number of events A; and n (S) the number of sample rooms.

Problems example:
1. A student conducts an experiment by throwing a coin several times. The results of these experiments are shown in the following table.

On throws as much as 100 times, namely by summing the throws 10 times, 20 times, 30 times and 40 times, the frequency of the appearance of the image is equal to (6 + 9 + 16 + 21) = 52 so the frequency is = $ \frac{52}{100} = 0,52$. If there are more numbers of throws in the experiment, then the relative frequency of the image will be closer to the value of 0,5. Number $ 0,5 = \frac{1}{2} $ This is called the chance of the occurrence of an image in an attempt to throw a coin.

2. A natural number is taken randomly from natural numbers 1, 2, 3, ..., 7, 8, and 9. If E is the occurrence of prime numbers, calculate the probability of occurrence A!
Answer:
Because taking numbers is random, the numbers have the same chance to be taken so that n = 9. Event A is the number of occurrences of prime numbers, namely 2, 3, 5, and 7, so k = 4.
$ P (A) = \frac{k}{n} = \frac{4}{9} $
So, the chance value of event A is $\frac{4}{9} $.

3. Two dice are thrown together one time. Calculate the chance value of event A, which is the occurrence of 9 dice
Answer:
The sample room in the experiment was S={(1,1), (1,2), (1,3), ..., (6,4), (6.5), (6,6)} with many members $ S = 6 \times 6 = 36 $ so that $ n (S) = 36 $.
Event A = {(3,6), (4,5), (5,4), (6.3)} so that $ n (A) = 4 $.
So, the chance of the occurrence of the second number of dice equals 9 is
$ \begin{align} P (A) & = \frac{n (A)}{n (S)} \\ & = \frac{4}{36} \\ & = \frac{1}{9} \end{align} $.
REFERENCE:
Wirodikromo, Sartono. 2001. Mathematics for High School Class XI. Jakarta: Erlangga.

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