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

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

\( \sum_{i=1}^{n} P(E_i) = 1 \)

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

\( P(A) = (1/2)(1/3) + (3/4)(1/3) + (1/4)(1/3) = 1/2 \)

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:

\( 5\% \times 95\% + 95\% \times 1\% = 5.7\% \)

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