Kasus dan hukum probabilitas total

The Law of Total Probability

Before diving into this topic, consider the following scenario.

There is a nation called ExpiiLand this country is made up of four provinces.

• ExLand --> Exland is $38$ miles wide and $20$ miles long.
• Xorland --> XORland is $22$ miles wide and $29$ miles long.
• PoLandia --> PoLandia is $36$ miles by $18$ miles.
• Infinite Isle --> Infinite Isle is $39$ miles by $42$ miles.

Yes! When finding the area of all of ExpiiLand we must first find the area of each province then add them together.

e.g. $$\begin{align} (38 \cdot 20) + (22 \cdot 29) + (36 \cdot 18) +39 \cdot 42) &= \\ 760+638+648+1,638 &= \\ 3,684 \ \text{miles}^2. \end{align}$$

This same logic can be applied to the law of total probability!

We can find total probability the same way we just found the area of ExpiiLand. We simply find the probability of each individual event (the area of each individual province), then add those probabilities together to get the total probability.

Let's try an example problem!

Consider the following: There are four bags on the table, each bag contains a mixture of grapes. The contents of each bag are listed below:

• Bag 1: $\dfrac{1}{3}$ red grapes, $\dfrac{1}{3}$ green grapes, and $\dfrac{1}{3}$ purple grapes
• Bag 2: $25\%$ red grapes, $46\%$ green grapes, and $29\%$ purple grapes
• Bag 3: $36\%$ red grapes, $22\%$ green grapes, and $42\%$ purple grapes
• Bag 4: $29\%$ red grapes, $19\%$ green grapes, and $52\%$ purple grapes

Assume that we randomly pick one of the bags so each bag is equally likely to be selected, and then randomly pick a grape out of that bag.

So basically, the law of total probability states that the total probability is equal to the sum of the probabilities of each way something can occur or $P(A)=P(A_1)+P(A_2)+P(A_3)+...+P(A_n)$.

It can also be written with summation notation:

$$P(A) = \sum^{n}_{i = 1}P(A_i).$$

Casework

Casework is the method of solving probability problems by considering the different cases of a particular problem and adding them together. Casework is not the most glamorous of methods in solving problems, but it is a brute force way if you do not know any other ways to solve the problem. Let's do an example where the casework method is used, then you will see how much time it may consume.

Example of Casework

Question: How many three-digit numbers satisfy the property that the middle digit is the average of the first and last digits?

This may be a lot to process. But let's break it down. Let's say that we have three-digit number $ABC$. The question is asking how many three-digit numbers satisfy $B = (A+C)/2$? Let's do some casework.

We notice that if the middle digit is the average of the first and last digits, then that means that: $2B = A + C$, or twice the middle digit is equal to the summation of the first and last digit. Let's also denote a variable $n$ that is equal to the summation of all possible numbers that satisfy the given conditions.

If the middle digit is $1$, possible numbers that satisfy this case are $111$ and $210$. Therefore $n_1 = 2$

If the middle digit is $2$, possible numbers that satisfy this case are $123$, $222$, $321$, and $420$. Therefore $n_2 = 4$

If the middle digit is $3$, possible numbers that satisfy this case are $135$, $234$, $333$, $432$, $531$, and $630$. Therefore $n_3 = 6$

If the middle digit is $4$, possible numbers that satisfy this case are $147$, $246$, $345$, $444$, $543$, $642$, $741$, and $840$. Therefore $n_4 = 8$

If the middle digit is $5$, possible numbers that satisfy this case are $159$, $258$, $357$, $456$, $555$, $654$, $753$, $852$, and $951$. Therefore $n_5 = 9$

If the middle digit is $6$, possible numbers that satisfy this case are $369$, $468$, $567$, $666$, $765$, $864$, and $963$. Therefore $n_6 = 7$

If the middle digit is $7$, possible numbers that satisfy this case are $579$, $678$, $777$, $876$, and $975$. Therefore $n_7 = 5$

If the middle digit is $8$, possible numbers that satisfy this case are $789$, $888$, and $987$. Therefore $n_8 = 3$

If the middle digit is $9$, possible numbers that satisfy this case are $999$. Therefore $n_9 = 1$

So our total $n$ value will be $n = n_1 + n_2 + n_3...+n_8+n_9$ which is: $$n = 2 + 4 + 6 +8 + 9 + 7 + 5 + 3 + 1 = 45$$

Therefore, there are $45$ combinations that will satisfy our prompt. Now, that took quite a while to analyze. You may not have that much time on a test. Let's look at the law of total probability to give you another tool to use so you can save some time!

The Law of Total Probability

The Law of Total Probability is a rule that makes it easier to find the probability of an outcome which has multiple parts to it, also known as events.

If $B_1, B_2, B_3,\cdots$ is a partition, or part, of the sample space $S$, then for any event in the same sample space, $A$, we have $$P(A)=\sum_{i} P(A \cap B_i)=\sum_{i} P(A | B_i) P(B_i).$$

This can be read as "The probability of $A$ happening is equal to the probability of $A$, given that $B$ happens multiplied with the probability of $B$ happening."

We can use a form of casework with these problems, where we will find values for each event, then add them at the end to find the total probability. Here is an example.

Example of the Law of Total Probability

Let's say that we have 3 boxes of crayons, each box contains a certain number of Frankenstein green, Pumpkin orange, or Witchy purple crayons (spooky colors for Halloween). The three boxes contain the following crayons:

Box 1: 3 Frankenstein green, 1 Pumpkin orange, 4 Witchy purple

Box 2: 1 Frankenstein green, 5 Pumpkin orange, 2 Witchy purple

Box 3: 4 Frankenstein green, 3 Pumpkin orange, 1 Witchy purple

Now, I select a box by drawing a stick out of a bag. Each stick has a number, $0$ through $4$. If I draw $0$, I pick Box $1$. If I draw an otherwise even number $(2, 4),$ I pick Box $2$; otherwise I pick Box $3$.

If I want to use a Pumpkin orange crayon, what is the probability that I get a Pumpkin orange crayon from one of the boxes? Let's elaborate.

The probability that I select a Pumpkin orange crayon from Box 1 is: $\frac{1}{5} \times \frac{1}{8}$. Since I can pick $1$ out of $5$ sticks that allow me to pick Box $1$, then from Box $1$, I can pick $1$ out of $8$ crayons that are Pumpkin orange. Therefore, my total probability of getting a Pumpkin orange crayon from Box 1 is: $\frac{1}{5} \times \frac{1}{8} = \frac{1}{40}$

The probability that I select a Pumpkin orange crayon from Box 2 is: $\frac{2}{5} \times \frac{5}{8} = \frac{10}{40} = \frac{1}{4}$

The probability that I select a Pumpkin orange crayon from Box 3 is: $\frac{2}{5} \times \frac{3}{8} = \frac{6}{40} = \frac{3}{20}$

Now, we need to add each probability together to get the total probability. This looks like: $$\frac{1}{40} + \frac{1}{4} + \frac{3}{20} = \frac{17}{40}$$

Therefore, out of the $40$ ways I can get a crayon, there are $17$ instances in which I can get a specifically Pumpkin orange crayon.

The law of total probability is a rule allowing us to compute the probability of an outcome that can be achieved through multiple different events.

For example, suppose we have a three bags containing the following marbles:

• Bag A: $10$ red marbles, $5$ blue marbles
• Bag B: $5$ red marbles, $7$ blue marbles
• Bag C: $7$ red marbles, $13$ blue marbles

Now, I select a bag by rolling a standard $6$-sided die. If the die shows $1$, I draw a random marble from Bag A; if the die shows $2$ or $3,$ I draw a random marble from Bag B; otherwise, I draw a random marble from Bag C.

What is the probability that I draw a red marble?

Here, there are three ways I could have drawn a red marble: I have drawn it from each of Bags A, B, and C. Let's do some casework!

The probability that I selected Bag A and drew a red marble is $\frac16 \times \frac{10}{15}$.

The probability that I selected Bag B and drew a red marble is $\frac26 \times \frac{5}{12}$.

The probability that I selected Bag C and drew a red marble is $\frac36 \times \frac{7}{20}$.

So hence, the total probability that I drew a red marble is $$\frac16 \times \frac{10}{15} + \frac26 \times \frac{5}{12} + \frac36 \times \frac{7}{20} = \boxed{\frac{17}{40}}.$$

(Video) Law of Total Probability Problem 1

by Tutorials Point

In this video, we walkthrough a problem involving the law of total probability.

$\large{\text{Summary}}$

Bag $1$: Has $4$ red, and $3$ blue balls. Bag $2$: Has $4$ blue and $2$ red balls. One bag is selected at random, find probability drawn ball is red.

Bag $1$ is Event $1$. Bag $2$ is Event $2$. Probability of event $1$ happens which leads to $p(A|E_1)$. Probability of event $2$ happens which leads to $p(A|E_2)$.

Law of probability states: $$p(A) = p(E_1)\cdot{p(A|E_1)} + p(E_2)\cdot{p(A|E_2)}$$

The probability of choosing the first bag: $p(E_1)$. This is $\frac{1}{2}$ since there are $2$ bags.

The probability the ball picked from the first bag is red: $p(A|E_1)$. This is $\frac{4}{7}$ since there are $7$ balls total and $4$ are red.

The probability of choosing the second bag: $p(E_2)$. This is $\frac{1}{2}$ since there are $2$ bags.

The probability the ball picked from the second bag is red: $p(A|E_2)$. This is $\frac{2}{6}$ since there are $6$ balls total and $2$ are red.

Now, $$\begin{align} p(A) &= p(E_1)\cdot{p(A|E_1)} + p(E_2)\cdot{p(A|E_2)}\\ &= \frac{1}{2}\cdot{\frac{4}{7}} + \frac{1}{2}\cdot{\frac{2}{6}}\\ &= \frac{1}{\cancel{2}}\cdot{\frac{\cancel{4}^2}{7}} + \frac{1}{\cancel{2}}\cdot{\frac{\cancel{2}}{6}}\\ &={\frac{2}{7}} + \frac{1}{6}\\ &={\frac{2}{7}}\cdot{\frac{6}{6}} + \frac{1}{6}\cdot{\frac{7}{7}}\\ &={\frac{12}{42}}+ \frac{7}{42}\\ &={\frac{19}{42}} \end{align}$$