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Define independence and calculate probability of independent events

Define independence and calculate probability of independent events

Define independence and calculate probabilities of independent events

1.1 Defining Independence

In probability and statistics, two events, \(A\) and \(B\), are said to be independent if the occurrence of event A does not affect the probability that the other event B will occur.

Example 1.1

Consider the experiment rolling a fair die twice. The outcome of the first roll does not affect the outcome of the second roll. In both rolls, there is a 1/6 probability of getting any particular number between 1 and 6 regardless of the outcome of the other roll. Therefore, these events are considered independent.

Example 1.2

If we toss a fair coin twice, the outcome of the first toss has no bearing on the second toss. If we get a head in the first toss, we could get a tail in the second, but we might also end up with another head. The probability of both heads and tails remains the same (0.5) regardless of the number of times the coin is tossed.

How about Dependence?

Two events are considered dependent if the outcome of one event impacts the probability of the second event occurring.

Example 1.3

Consider an experiment where we draw two cards without replacement from a standard deck 52 of cards. The probability of drawing a specific 2nd card depends on the outcome of the 1st draw.

Before any card is drawn, the probability of drawing a heart is 13/52. But if the king of hearts is the first card drawn, then the probability of drawing a heart on the second draw becomes 12/51.

If a card is drawn and then replaced for the second random draw, the probability for the second draw is clearly independent of the first draw.

The Multiplication Rule for Independent Events

The general multiplication rule for any two events is as follows:

$$ P(A \cap B) = P(A|B) \cdot P(B) $$

Switching the role of \(A\) and \(B\), we can also write the rule as:

$$ P(A \cap B) = P(B|A) \cdot P(A) $$

If \(A\) and \(B\) are independent, then:

$$ P(A \cap B) = P(A) \cdot P(B) $$

For three independent events, \(A\), \(B\), and \(C\):

$$ P(A \cap B \cap C) = P(A) \cdot P(B) \cdot P(B) $$

Example 1.4

Suppose we wish to calculate the probability of obtaining “heads” during successive coin flips. We’d multiply the probability of getting heads during the first flip (0.5) by the probability of getting heads during the second flip (0.5). The joint probability of two consecutive heads is 0.25 (i.e., 0.5 \(\times\) 0.5)

Example 1.5

Suppose there are 1,200 policyholders in a cohort. Of these, 800 have only whole life insurance and the remaining 400 have term life insurance. Suppose two policyholders are selected from the cohort with replacement. What would be the probability that both have term life insurance?

Since there’s replacement, the outcome of the first selection is independent of the second outcome. If we let \(W\) represent whole life insurance and \(T\) represent term life,

$$ P(T \cap T) = P(T) \cdot P(T) = \frac {400}{1200} \times \frac {400}{1200} = 0.11 \text{ or } 11\% $$

Important Formulas

If \(A\) and \(B\) are independent events, then the following are also independent events:

\(A \text{ and } B^C\)

\(A^C \text{ and } A\)

\(A^C \text{ and } B^C\)

It follows that:

$$ \begin{align*} P(A \cap B^C) & = P(A) \cdot P(B^C) \\ P(A^C \cap B) & = P(A^C) \cdot P(B) \\ P(A^C \cap B^C) & = P(A^C) \cdot P(B^C) \end{align*} $$

If \(A\), \(B\), and \(C\) are mutually independent events:

  1. \(A\), \(B\) and \(C\) are pairwise independent, meaning \(P(A \cap B)=P(A) \cdot P(B), P(A \cap C)=P(A) \cdot P(C),\) and \(P(B \cap C) = P(B) \cdot P(C)\)
  2. \(P(A \cap B \cap C) = P(A) \cdot P(B) \cdot P(C)\)

If \(A\), \(B\) and \(C\) are mutually independent, then the following are also independent events:

\(A \text{ and } (B \cap C)\)

\(A \text{ and } (B \cup C)\)

\(A^C \text{ and } (B \cap C)\)

\(A^C, B^C \text{ and } C^C\)

It also follows that:

$$ P(A \cap B \cap C)^C = 1 − P(A \cap B \cap C) $$

The final equation above implies that the probability that a set of mutually independent events “fails” is 1 minus the probability that all three mutually independent events do occur.

Example 1.6

What would be the probability of flipping tails on 3 successive flips of a coin?

Since these are all independent events (each flip is independent of any other flip), then

$$ P(A \cap B \cap C) = P(A) \cdot P(B) \cdot P(C) =\frac {1}{2} \cdot \frac {1}{2} \cdot \frac {1}{2} = 18 $$

Question 1.1

After studying the insurance preferences of pet owners, an actuary draws the following conclusions:

  1. A pet owner is twice as likely to purchase accidental death coverage as theft coverage.
  2. The event that a pet owner purchases accidental death coverage is independent of the event that he or she purchases theft coverage.
  3. The probability that a pet owner purchases both accidental death coverage and theft coverage is 0.20.

Calculate the probability that a prospective policyholder purchases neither theft coverage nor theft coverage.


D = event a pet owner purchases accidental death insurance

T = event a pet owner purchases theft coverage


\(P(D) = 2P(T)\) and \(P(D \cap T) = 0.20\)

We want to find \(P(D^C \cap T^C) = P(D \cap T)^C\)

Since \(D\) and \(T\) are independent, \(P(D \cap T) = P(D) \times P(T) = 0.20\)

But \(P(D) = 2P(T)\)

This means that \(2P(T) \times P(T) = 0.20\)

$$ P(T) = \frac {0.11}{2} = 0.3162 $$

and \(P(D) = 2(0.3162) = 0.6324\)

Again because \(D\) and \(T\) are independent, \(P(D^C \cap T^C) = P(D^C) \times P(T^C)\)

We also know that \(P(D^C) = 1 – P(D)\) and \(P(T^C) = 1 – P(T)\)

Therefore, \(P(D^C \cap T^C) = (1 – 0.6324) \times (1 – 0.3162) = 25.14\%\)

Question 1.2

A computer firewall is designed so that, if component C1 fails, component C2 is used instead. If component C2 fails, component C3 is used instead. Consider a scenario where the probability that C1 fails is 0.02, C2 fails is 0.05, and C3 fails is 0.1. Further, assume that the failures are mutually independent events. What is the probability that the firewall fails?

Since the failures of the three events are independent,

$$ \begin{align*} P(C1 \cap C2 \cap C3) & = P(C1) \cdot P(C2) \cdot P(C3) \\ & = 0.02 \times 0.05 \times 0.1 \\ & = 0.0001 \end{align*} $$

Learning Outcome

1.c Topic: General Probability – Define independence and calculate probabilities of independent events.

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