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Unconditional probability (also known as marginal probability) is simply the probability that the occurrence of an event does not, in any way, depend on any other preceding events. In other words, unconditional probabilities are **not **conditioned on the occurrence of any other events; they are ‘stand-alone’ events. Therefore, if we are interested in the probability of, say, event A, the standard annotation is P(A). Let’s look at a few examples:

- The probability that it will rain on a given day without considering the rainfall pattern of a given area or any other climatic factor.
- The probability that a given stock will earn a 10% annual return without considering the preceding annual returns.

In a future learning outcome statement, we will learn to calculate and interpret an unconditional probability using the total probability rule.

Conditional probability is the probability of one event occurring with some relationship to one or more other events. Our interest lies in the probability of event ‘A’ **given** that another event ‘B ‘**has already occurred**. Here’s what you should ask yourself:

“What is the probability of one event occurring **if **another event has already taken place?”

We pronounce P(A | B) as “the probability of A given B.,” and it is given by:

$$P(A│B)=\frac{P(A\cap B)}{P(B)}$$

The bar sandwiched between A and B simply indicates “given.”

In a group of 100 investors,

- 40 buy stocks.
- 30 purchase bonds.
- 20 purchase stocks and bonds.

If an investor chosen randomly bought bonds, what is the probability they also bought stocks?

**Solution**

$$

\begin{array}{c|c|c}

\textbf{Event} & \textbf{Notation} & \textbf{Probability} \\

\hline

\text{Buys stocks} & \text{A} & \text{0.4(=40/100)} \\

\text{Buys bonds} & \text{B} & \text{0.3 (=30/100)} \\

\text{Buys stocks and bonds} & \text{A and B} & \text{0.2 (=20/100)} \\

\end{array}

$$

We want the probability of an investor buying stocks, given that they have already bought bonds. This is P(A | B):

$$ \begin{align} P\left( A|B \right) & =\frac { P\left( A\cap B \right) }{ P\left( B \right) } \\ & =\frac { 0.2 }{ 0.3 } = 0.67 \end{align}$$

Note that we can also make the numerator the subject so that:

$$ P\left( A\cap B \right) =P\left( A|B \right) P\left( B \right) $$

For independent events, which we will see later, however,

$$ P\left( A|B \right) =\frac{P(A)P(B)}{P(B)}=P\left( A \right) $$

This is also true for \(P(B│A)=P(B)\).

QuestionIn a group of 100 investors,

- 40 buy stocks.
- 30 purchase bonds.
- 20 purchase stocks and bonds.
If an investor chosen randomly bought stocks, the probability they also bought bonds is

closest to:

- 0.50.
- 0.67.
- 0.70.

SolutionThe correct answer is

A.Consider the following table:

$$

\begin{array}{c|c|c}

\textbf{Event} & \textbf{Notation} & \textbf{Probability} \\

\hline

\text{Buys stocks} & \text{A} & \text{0.4(=40/100)} \\

\text{Buys bonds} & \text{B} & \text{0.3 (=30/100)} \\

\text{Buys stocks and bonds} & \text{A and B} & \text{0.2 (=20/100)} \\

\end{array}

$$The probability of an investor buying stocks given that they have already bought bonds. This is P(B|A):

$$ \begin{align} P\left( B|A \right) & =\frac { P\left( A\cap B \right) }{ P\left( A \right) } \\ & =\frac { 0.2 }{ 0.4 } = 0.50 \end{align}$$