- CFA Exams
- 2023 Level I
- Topic 1. Quantitative Methods
- Learning Module 3. Probability Concepts
- Subject 4. Multiplication Rule for Independent Events

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##### Subject 4. Multiplication Rule for Independent Events PDF Download

Two events, A and B, are P(A and B) = P(A) x P(B) P(E

P (shares L and M both rise in price on the same day) = 0.5 x 0.8 = 0.4

**independent**if and only if P(A|B) = P(A), or equivalently, P(B|A) = P(B). That is, the occurrence of one event has no influence on the probability of the occurrence of the other event.In more detail, whether or not B occurs will have no effect on the probability of A and vice versa. Thus, there will be no difference between P(A|B) and P(A), and similarly there will be no difference between P(B|A) and P(B).

For example, suppose you flip a coin twice. The event of getting heads on the first flip does not affect the probability of getting heads on the second flip. Therefore, the event of getting heads on the second flip is independent of the event of getting heads on the first flip.

When two events are not independent, they must be

**dependent**: the occurrence of one is related to the probability of the occurrence of the other.If we are trying to forecast one event, information about a dependent event will be useful but information about an independent event will not be useful.

*Example 1*If C = {the price of insurance share C goes up} and D = {the price of insurance share D goes up}, then clearly C and D are dependent events, because the market as a whole might be bullish or the insurance sector alone might be having a good day. Although the increase in share price C might not affect share price D at all, there is clearly a good chance that the two shares will move in the same direction.

*Example 2*If A = {a person in Europe drives a red car} and B = {a person in Australia drives a white car}, then events A and B are clearly independent, as the one almost certainly has no bearing upon the other.

Remember that if A and B are independent, Ac and Bc will also be independent.

A and B are two events. If A and B are independent, the probability that events A and B both occur is:

In other words, the probability of A and B both occurring is the product of the probability of A and the probability of B. This relationship is known as the

**multiplication rule for independent events**.What is the probability that a fair coin will come up with heads twice in a row? Two events must occur: heads on the first toss and heads on the second toss. Since the probability of each event is 1/2, the probability of both events is: 1/2 x 1/2 = 1/4.

Now consider a similar problem: Someone draws a card at random out of a deck, replaces it, and then draws another card at random. What is the probability that the first card is the ace of clubs and the second card is a club (any club)? Since there is only one ace of clubs in the deck, the probability of the first event is 1/52. Since 13/52 = 1/4 of the deck is composed of clubs, the probability of the second event is 1/4. Therefore, the probability of both events is: 1/52 x 1/4 = 1/208.

Similarly, for any number of independent events E

_{1}, E_{2}.....E_{n}, the probability that all of them occur is:

_{1}and E

_{2}..... and E

_{n}) = P(E

_{1}) x P(E

_{2}) x ..... x P(E

_{n})

*Example*In a bullish market, three shares, chosen from different sectors of the market, have probabilities of 0.6, 0.5 and 0.8 that their share prices will rise on any particular day. Let's call the shares K, L and M respectively.

If we make the assumption that prices of shares on successive days are independent, and also that the price movement of one of the shares above is independent of the others, we can then carry out the following calculations.

Note that although this may seem slightly unrealistic, the fact that the shares come from different sectors of the market lends some credence to the assumption, and also simplifies our calculations considerably.

P (share K has a price rise on two consecutive days) = 0.6 x 0.6 = 0.36

P (shares L and M both rise in price on the same day) = 0.5 x 0.8 = 0.4

Thus, the independent assumptions make our work much easier.

To calculate the probability that share M rises in price on four consecutive days, we can use the results above to calculate this probability as: 0.8 x 0.8 x 0.8 x 0.8 = 0.4096 (i.e., 0.8 multiplied four times, once for each day).

Although the chance that the share will rise in price on any one day is 80%, the chance that this will happen for four days in a row is just over half of this total, or 40.96%.

If we wish to calculate the probability that all three shares will rise in price on the same day, we can use the results above to get: 0.6 x 0.5 x 0.8 = 0.24 (i.e., the individual probabilities multiplied together)

Warning: It is important to note that multiplying individual probabilities together can only be done if the events that make up those probabilities are independent. If the events are dependent, this process is not valid.

To calculate the probability that, of the three shares above, none will have a price rise on a particular day, we can multiply the probabilities of the complementary events together to get: 0.4 x 0.5 x 0.2 = 0.04.

Thus, there is only a 4% chance that no shares will have a price rise on a particular day.

You might be wondering, if the chance that all three shares rise in price is 24% and the chance that no shares rise in price is 4%, about what has happened to the remaining 100% - 24% - 4% = 72%. The answer is that this percentage covers situations where some shares rise in price while others don't. You'll see this type of example in more detail in later lessons.

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**User Contributed Comments**
2

User |
Comment |
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dethepp1 |
Why is the example of independent events correct? In the exercise stands, "and then draws another card". in my understanding that means the probability of getting the ace club with the first card is P(A) 1/52 and therefore taking another card and also getting a Club must be P(B l A) is 12/51 and not 1/4. The P(B) is 1/4 but with the conditions given the example is not correct to present an example for independent probabilities or am I wrong? |

DWright87 |
"Someone draws a card at random out of a deck, replaces it, and then draws another card at random. What is the probability that the first card is the ace of clubs and the second card is a club (any club)?" The replacing of the card makes them independent events. |

I was very pleased with your notes and question bank. I especially like the mock exams because it helped to pull everything together.