- CFA Exams
- 2023 Level I
- Topic 1. Quantitative Methods
- Learning Module 3. Probability Concepts
- Subject 3. Addition Rule for Probabilities: the Probability that at Least One of Two Events Will Occur
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Subject 3. Addition Rule for Probabilities: the Probability that at Least One of Two Events Will Occur PDF Download
If we have two events, A and B, that we are interested in, we often want to know the probability that either A or B occurs. Note the use of the word "or," the key to this rule. The "or" is what we call an "inclusive or." In other words, either one event can occur or both events can occur.P(A or B) = P(A) + P(B) - P(AB)
P(ace) = 4/52; P(spade) = 13/52
The only way in which an ace and a spade can both be drawn is to draw the ace of spades. There is only one ace of spades, so:
P(ace and spade) = 1/52.
The probability of an ace or a spade can be computed as:
P(ace) + P(spade) - P(ace and spade) = 4/52 + 13/52 - 1/52 = 16/52 = 4/13.
Event A: 6 on the first roll: p(A) = 1/6
Event B: 6 on the second roll: p(B) = 1/6
P(A and B) = 1/6 x 1/6
P(A or B) = 1/6 + 1/6 - 1/6 x 1/6 = 11/36
The probability of getting a number from 1 to 5 on the first roll is 5/6. Likewise, the probability of getting a number from 1 to 5 on the second roll is 5/6. Therefore, the probability of getting a number from 1 to 5 on both rolls is: 5/6 x 5/6 = 25/36. This means that the probability of not getting a 1 to 5 on both rolls (getting a 6 on at least one roll) is: 1-25/36 = 11/36.
Such probabilities are calculated using the addition rule for probabilities.
The logic behind this formula is that when P(A) and P(B) are added, the occasions on which A and B both occur are counted twice. To adjust for this, P(AB) is subtracted.
If events A and B are mutually exclusive, the joint probability of A and B is 0. Consequently, the probability that either A or B occurs is simply the sum of the unconditional probabilities of A and B: P (A or B) = P(A) + P(B).
What is the probability that a card selected from a deck will be either an ace or a spade? The relevant probabilities are:
P(ace) = 4/52; P(spade) = 13/52
The only way in which an ace and a spade can both be drawn is to draw the ace of spades. There is only one ace of spades, so:
P(ace and spade) = 1/52.
The probability of an ace or a spade can be computed as:
P(ace) + P(spade) - P(ace and spade) = 4/52 + 13/52 - 1/52 = 16/52 = 4/13.
Consider the probability of rolling dice twice and getting a 6 on at least one of the rolls. The events are defined in the following way:
Event A: 6 on the first roll: p(A) = 1/6
Event B: 6 on the second roll: p(B) = 1/6
P(A and B) = 1/6 x 1/6
P(A or B) = 1/6 + 1/6 - 1/6 x 1/6 = 11/36
The same answer can be computed using the following (admittedly convoluted) approach: Getting a 6 on either roll is the same thing as not getting a number from 1 to 5 on both rolls. This is equal to: 1 - P(1 to 5 on both rolls).
The probability of getting a number from 1 to 5 on the first roll is 5/6. Likewise, the probability of getting a number from 1 to 5 on the second roll is 5/6. Therefore, the probability of getting a number from 1 to 5 on both rolls is: 5/6 x 5/6 = 25/36. This means that the probability of not getting a 1 to 5 on both rolls (getting a 6 on at least one roll) is: 1-25/36 = 11/36.
Despite the convoluted nature of this method, it has the advantage of being easy to generalize to three or more events. For example, the probability of rolling dice three times and getting a six on at least one of the three rolls is: 1 - 5/6 x 5/6 x 5/6 = 0.421
User Contributed Comments 6
| User | Comment |
|---|---|
| cleopatraliao | I actually prefer the admittedly convluted approach as to me it is more intuitive...I'm a bit confused about the last bit "Consider the probablility of rolling a dice twice&getting a 6 on at least one of the rolls. P(A or B)=1/6+1/6-1/6*1/6=11/36(even though I know it's the correct answer) but why does it have to minus 1/6*1/6...isn't p(A&B) is also part of the "prob of getting a 6 on at least one of the rolls?) |
| breakman | The key word is atleast once, so you should not count it roll a 6 and another 6 |
| rsanfo | Think of it like a Venn diagram with overlapping circles. The left circle is 1/6 or 6/36 (first roll). The right circle is 6/36 (second roll). The union (1/36 for both rolls being a 6) is subtracted. So 6/36 + 6/36 - 1/36 = 11/36. |
| darin3200 | But "at least" is meaning the probability of getting one, or more, 6s. So why would we exclude the probability of getting 2? |
| johntan1979 | I agree. Getting 6 on one of the rolls = P(A or B) Getting 6 on "at least" one of the rolls = P(A or B) + P(AB) = P(A) + P(B) = 1/3 |
| Shaan23 | Better for AT LEAST questions to always do the complement. Probability of getting NO 6's is (5/6)(5/6) = 25/36 1-25/36 = 11/36 <--at least one 6 |
I am using your study notes and I know of at least 5 other friends of mine who used it and passed the exam last Dec. Keep up your great work!
Barnes
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