Conditional Probability
The conditional probability, as its name suggests, is the probability of happening an event that is based upon a condition. For example, assume that the probability of a boy playing tennis in the evening is 95% (0.95) whereas the probability that he plays given that it is a rainy day is less which is 10% (0.1). Then the former case is just normal probability whereas the latter case is the conditional probability. In this example, we represent the two probabilities as P(Play tennis) = 0.95 and P(Play tennis | Rainy day) = 0.1.
Let us learn more about conditional probability along with its formula, examples, and practice questions.
1. | What Is Conditional Probability? |
2. | Conditional Probability Formula |
3. | Properties of Conditional Probability |
4. | Dependent and Independent Events |
5. | FAQs on Conditional Probability |
What Is Conditional Probability?
Conditional probability is one of the important concepts in probability and statistics. The "probability of A given B" (or) the "probability of A with respect to the condition B" is denoted by the conditional probability P(A | B) (or) P (A / B) (or) P\(_B\)(A). Thus, P(A | B) represents the probability of A which happens after event B has happened already. the probability of an event may alter if there is a condition given.
Definition of Conditional Probability
If A and B are two events associated with the same sample space of a random experiment, the conditional probability of event A given that B has occurred is given by P(A/B) = P( A ∩ B)/ P (B), provided P(B) ≠ 0.
Let us understand conditional probability with an example. Let us find the conditional probability of getting at least two tails given that it is a head on the first toss when 3 coins are tossed. The sample space, S (the list of all outcomes) when 3 coins are tossed is given as follows:
Let us assume the two events A and B as follows:
- A = the event of getting at least two tails
- B = the event of getting a head on the first toss
Then, A = {HTT, THT, TTH, TTT} and B = {HHH, HHT, HTH, HTT}.
Then P(A) = 4/8 = 1/2 and P(B) = 4/8 = 1/2.
We have to find the probability of getting at least two tails given that it is a head on the first toss. It means, out of all elements of B, we have to choose only the ones with two tails. We can see that among the elements of B, there is only one element (which is HTT) with two tails. Thus, the required probability is P(A | B) = 1/4 (only 1 outcome of B is favorable to A out of 4 outcomes of B).
Conditional Probability Formula
In the above example, we have got P(A | B) = 1/4, here 1 represents the element HTT which is present both in "A and B" and 4 represents the total number of elements in B. Using this, we can derive the formula of conditional probability as follows.
P(A | B) = P(A ∩ B) / P(B) (Note that P(B) ≠ 0 here)
Similarly, we can define P(B | A) as follows:
P(B | A) = P(A ∩ B) / P(A) (Note that P(A) ≠ 0 here)
These formulas are also known as the "Kolmogorov definition" of conditional probability.
Here:
- P(A | B) = The probability of A given B (or) the probability of A which happens after B
- P(B | A) = The probability of B given A (or) the probability of B which happens after A
- P(A ∩ B) = The probability of happening of both A and B
- P(A) = The probability of A
- P(B) = The probability of B
Derivation of Conditional Probability
Note that the elements of B which favor the event A are the common elements of A and B. i.e. the sample points of A ∩ B.
Thus P(A/B) = Number of events favorable to A ∩ B ÷ Number of events favorable to B.
P(A/B) = \(\dfrac{\dfrac{n(A ∩ B)}{n(S)}}{\dfrac{n(B)}{n(S)}}\)
Thus P(A | B) = P(A ∩ B) / P(B)
Properties of Conditional Probability
Here are some properties of conditional probability along with their proofs (derivations) which we may need to use while solving the problems. All these properties depend on the conditional probability formula (which is mentioned in the previous section).
Property 1
Let S be the sample space of an experiment and A be any event. Then P(S | A) = P(A | A) = 1.
Proof:
By the formula of conditional probability,
P(S | A) = P(S ∩ A) / P(A) = P(A) / P(A) = 1
P(A | A) = P(A ∩ A) / P(A) = P(A) / P(A) = 1
Hence property 1 is proved.
Property 2
Let S be the sample space of an experiment and A and B be any two events. Let E be any other event such that P(E) ≠ 0. Then P((A ⋃ B) | E) = P(A | E) + P(B | E) - P((A ∩ B) | E).
Proof:
By the formula of conditional probability,
P((A ⋃ B) | E) = [P((A ⋃ B) ∩ E)] / P(E)
= [ P(A ∩ E) ⋃ P(B ∩ E) ] / P(E) (using a property of sets)
= [P(A ∩ E) + P(B ∩ E) - P(A ∩ B ∩ E)] / P(E) (using addition theorem of probability)
= P(A ∩ E) / P(E) + P(B ∩ E) / P(E) - P(A ∩ B ∩ E) / P(E)
= P(A | E) + P(B | E) - P((A ∩ B) | E) (By conditional probability formula)
Hence property 2 is proved.
Property 3
P(A' | B) = 1 - P(A | B), where A' is the complement of the set A.
Proof:
By Property 1, we have P(S | B) = 1.
We know that S = A ⋃ A'. Thus by the above property,
P( A ⋃ A' | B) = 1
Since A and A' are disjoint events,
P(A | B) + P(A' | B) = 1
P(A' | B) = 1 - P(A | B)
Hence property 3 is proved.
Dependent and Independent Events
The definition of independent and dependent events is connected to conditional probability. Let us see the definitions of independent and dependent events along with their formulas.
Dependent Events
Dependent events, as the name suggests, are any two events in which the happening of one event depends on the happening of the other event.
- If A depends on B, then the probability of A is P(A | B).
- If B depends on A, then the probability of B is P(B | A).
By the conditional probability formulas,
P(A | B) = P(A ∩ B) / P(B) ⇒ P(A ∩ B) = P(A | B) · P(B)
P(B | A) = P(A ∩ B) / P(A) ⇒ P(A ∩ B) = P(B | A) · P(A)
Thus, two event A and B are said to be dependent events if one of the conditions is satisfied.
- P(A ∩ B) = P(A | B) · P(B) (or)
- P(A ∩ B) = P(B | A) · P(A)
Independent Events
Independent events, as the name suggests, are any two events in which the happening of one event does not depend on the happening of the other event. i.e., if A and B are independent then P(A | B) = P(A) and P(B | A) = P(B). Thus, to get the formula of independent events, we just need to replace P(A | B) with P(A) (or P(B | A) with P(B)) in one of the above (dependent events) formulas. Hence, two events are said to be independent if
P(A ∩ B) = P(A) · P(B)
This is also called as multiplication rule of probability.
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Important Notes:
- The probability of A given B is called the conditional probability and it is calculated using the formula P(A | B) = P(A ∩ B) / P(B).
- The events that are part of conditional probability are dependent events. For example, if we have P(A | B) anywhere in the problem, then it means that A and B are dependent.
- If two events A and B are independent, then P(A | B) = P(A) and P(B | A) = P(B).
- For any two events A and B, P(A ∩ B) = P(A) · P(B). This is called the multiplication theorem of probability.
Examples of Conditional Probability
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Example 1: The table below shows the occurrence of diabetes in 100 people. Let D and N be the events where a randomly selected person "has diabetes" and "not overweight". Then find P(D | N).
Diabetes (D) No Diabetes (D') Not overweight (N) 5 45 Overweight (N') 17 33 Solution:
From the given table, P(N) = (5+45) / 100 = 50/100.
P(D ∩ N) = 5/100.
By the conditional probability formula,
P(D | N) = P(D ∩ N) / P(N)
= (5/100) / (50/100)
= 5/50
= 1/10
Answer: P(D | N) = 1/10.
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Example 2: The probability that it will be sunny on Friday is 4/5. The probability that an ice cream shop will sell ice creams on a sunny Friday is 2/3 and the probability that the ice cream shop sells ice creams on a non-sunny Friday is 1/3. Then find the probability that it will be sunny and the ice cream shop sells the ice creams on Friday.
Solution:
Let us assume that the probabilities for a Friday to be sunny and for the ice cream shop to sell ice creams be S and I respectively. Then,
P(S) = 4/5.
P(I | S) = 2/3.
P(I | S') = 1/3.
We have to find P(S ∩ I).
We can see that S and I are dependent events. By using the dependent events' formula of conditional probability,
P(S ∩ I) = P(I | S) · P(S) = (2/3) · (4/5) = 8/15.
Answer: The required probability = 8/15.
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Example 3: If a fair die is rolled twice, observe the numbers that face up. Find the conditional probability that the sum of the numbers is 7, given that the first number is 2.
Solution:
Let us determine the sample space of rolling a die twice. S = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6)(4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) }
Considering events A and B as given: we have
A : the sum of the numbers is 7. Thus set A = {(1,6),(2,5), (3,4), (4,3), (5,2),(6,1) }
B: the first number is 2. Thus set B = {(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)}
A ∩ B: {(2,5)}
By the conditional probability, we know that
P(A ) = P(A ∩ B) / P(B)P(A ) = \(\dfrac{\dfrac{1}{36}}{\dfrac{6}{36}}\)
P(A ) = 1/6
Answer: The conditional probability that the sum of the numbers is 7, given that the first number is 2 is 1/6
FAQs on Conditional Probability
What Is Conditional Probability?
The conditional probability is the probability of happening of an event of A given that another event B has already occurred. It is denoted by P(A | B) and it is calculated by the formula P(A | B) = P(A ∩ B) / P(B).
What Is Conditional Probability Formula?
The conditional probability of A given B is given as P(A | B) = P(A ∩ B) / P(B) and the conditional probability of B given A is P(B | A) = P(A ∩ B) / P(A).
What Are the Properties of Conditional Probability?
Here are the important properties of conditional probability. In all the properties, assume that S is the sample space and A, B, and E are the events.
- P(S | A) = P(A | A) = 1.
- P((A ⋃ B) | E) = P(A | E) + P(B | E) - P((A ∩ B) | E)
- P(A' | B) = 1 - P(A | B)
Which Example Does Best describe Conditional Probability?
Assume that there are 100 blood donors available in a hospital. Among them, a non-diabetic person has to be chosen given that his blood group is O+. This situation best describes conditional probability. If N and O are the events of selecting a non-diabetic person and a person with the blood group O+ respectively, then the conditional probability representing the above situation is P(N | O) and is calculated using the formula, P(N | O) = P(N ∩ O) / P(O).
Which Theorem Best Explains Conditional Probability and Independence?
The multiplication theorem of probability (which is derived from conditional probability) best describes the independent events. According to this, two events A and B are said to be independent if P(A ∩ B) = P(A) · P(B).
How To Read Conditional Probability P(A | B)?
The conditional probability P(A | B) is read as "the probability of A given B" (or) "the probability of A after B has happened". P(A | B) can also be written as P(A/B) (or) P\(_B\)(A).
Why Is Conditional Probability Important?
The conditional probability is important when we have to find the probability of an event that depends on another event. If event A depends on another event B (i.e., event A happens after B has happened), then the probability of event A is denoted by the conditional probability P(A | B) and is calculated using the formula P(A/B) = P(A ∩ B)/P(B).
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