# Law of Total Probability Examples



The law of total probability is explained and used to solve examples including detailed explanations.

## Explanation of the Law of Total Probability

Let the sample space $$S$$ be partioned into $$n$$ mutually exclusive events $$E_1$$, $$E_2$$, $$E_3$$ ... $$E_n$$ and collectively exhaustive (covering the entire sample space $$S$$ ).

Note that
$$A = (A \cap E_1) \cup (A \cap E_2) \cup (A \cap E_3) ... \cup (A \cap E_n)$$
and since the events $$E_i$$ are all mutually exclusive, then $$(A \cap E_i)$$ are also all mutually exclusive and therefore we can use the addition rule of probabilities to write
$P(A) = P(A \cap E_1) + P(A \cap E_2) + P(A \cap E_3) ... P(A \cap E_n)$
Using the definition of conditional probabilities we write
$$P(A \cap E_n) = P(A | E_n) P(E_n)$$,
and rewrite the law of total probability as
$P(A) = P(A | E_1) P(E_1) + P(A | E_2) P(E_2) + P(A | E_3) P(E_3) ... P(A | E_n) P(E_n)$ $= \sum_{i=1}^{n} P(A | E_i) P(E_i)$ Note that because $$E_1$$, $$E_2$$, $$E_3$$ ... $$E_n$$ are collectively exhaustive, we have
$$\sum_{i=1}^{n} P(E_i) = 1$$

## The Law of Total Probability Examples with Detailed Solutions

We start with a simple example that may be solved in two different ways and one of them is using the the Law of Total Probability.

Example 1
We have three similar bags B1, B2 and B3 containing 4 balls each. B1 contains 2 red and 2 blue balls, B2 contains 3 red and 1 blue balls and B3 contains 1 red and 3 blue balls. One bag is selected at random and a ball is selected at random from that bag.
What is the probability that the ball is red?

Solution to Example 1
We start by a diagram that explains what is given.

We present T two methods to solve the above question:
Method 1: Use classical method of calculating probabilities
In all three bags there is a total of 12 balls 6 of which are red balls. Let event $$A$$ be that of selecting a red ball. Hence the probability that the ball selected is red is given by
$$P(A) = 6/12 = 1/2$$
Method 2: Use law of total probability
Let $$E_1$$, $$E_2$$ and $$E_3$$ be the event of selectiong bag $$B_1$$, $$B_2$$ and $$B_3$$ respectively and $$A$$ be the event of selecting a red ball.
$$P(A) = P(A | E_1) P(E_1) + P(A | E_2) P(E_2) + P(A | E_3) P(E_3)$$
The bags are similar and are therefore equally likely to be selected. The probabilities of selecting bags $$B_1$$, $$B_2$$ or $$B_3$$ at random are given by
$$P(E_1) = 1/3$$ (one bag B1 out of 3 bags)
$$P(E_2) = 1/3$$ (one bag B2 out of 3 bags)
$$P(E_3) = 1/3$$ (one bag B3 out of 3 bags)
The conditional probabilities of selecting a red ball given that it is in bag $$B_1$$, $$B_2$$ and $$B_3$$ are given by
$$P(A | E_1) = 2/4 = 1/2$$ (2 red balls out of a total of 4 balls in B1)
$$P(A | E_2) = 3/4$$ (3 red balls out of a total of 4 balls in B2)
$$P(A | E_3) = 1/4$$ (1 red ball out of a total of 4 balls in B3)
We now use the law of total probability
$$P(A) = (1/2)(1/3) + (3/4)(1/3) + (1/4)(1/3) = 1/2$$

Note that the law of total probability will be used in situations where classical methods of calculating probabilities cannot be used as we will see in the examples below.

Example 2
Students in a math class where 40% are males and 60% are females took a test. 50% of the males and 70% of the females passed the test. What percent of students passed the test?

Solution to Example 2
We start by drawing a diagram with the whole class including the males and females groups.
Let events $$E_1$$ "be a male" and $$E_2$$ "be a female", and event $$A$$ "passed the test".

The law of total probability gives
$$P(A) = P(A | E_1) P(E_1) + P(A | E_2) P(E_2)$$
$$= 50\% \times 40\% + 70\% \times 60\% = 62\%$$

Example 3
Three factories produce the same tool and supply it to the market. Factory A produces 30% of the tools for the market and the remaining 70% of the tools are produced in factories B and C. 98% of the tools produced in factory A, 95% of the tools produced in factory B and 97% of the tools produced in factory C are not defective.
What percent of tools should be produced by factories B and C so that a tool picked at random in the market will have a probability of being non defective equal to 96%?

Solution to Example 3
We start by drawing a diagram of all three factories.
Let $$A$$, $$B$$ and $$C$$ be the events of tools produced by factories A, B and C and $$ND$$ be the event "non defective".
$$x$$ and $$y$$ are the percentages of tools to be produced by companies B and C respectively.

Using the above diagram, we can write
$$P(A) = 30\%$$ , $$P(B) = x$$ , $$P(C) = y$$
The three factories supply the whole market; hence
$$30\% + x + y = 100\%$$ which gives $$y = 70\% - x$$
Use the law of total probability to write the equation
$$P(ND) = P(ND|A) P(A) + P(ND|B) P(B) + P(ND|C) P(C)$$
$$= 98\% \times 30\% + 95\% \times x + 97\% \times (70\% - x) = 96\%$$
Solve for $$x$$
Company B produces: $$x = 0.65 = 65\%$$
Company C produces: $$y =70\% - 65\% = 5\%$$

Example 4
5% of a population have a flu and the remaining 95% do not have this flu.
A test is used to detect the flu and this test is positive in 95% of people with a flu and is also (falsely) positive in 1% of the people with no flu.
If a person from this population is randomly selected and tested, what is the probability that the test is positive?

Solution to Example 4
Use the total probability theorem to find the percentage as follows:
$$5\% \times 95\% + 95\% \times 1\% = 5.7\%$$