Conditional Probability

Have you ever wonder how weather forecasters can predict about the weather condition of a particular area? They use the previously available data. Is it always necessary that the previous data affect the occurrence of any event? If a die is rolled and the first throw is 4 does the next throw is also 4? How can this event be affected by the last one? How much they affect its happening? This gives rise to a new concept in probability. So what is a conditional probability?

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Conditional Probability

Suppose you have to calculate the chance of a person meeting with a road accident. You also have information regarding his driving skills. This information can be used to show a relationship between the two events. We can say if a person drives well there is less chance of him to meet with an accident.

This information of good driving skills put a condition on the chances of meeting with an accident. This is an example of conditional probability. This probability shows the chances of occurrence of some event for which some other events of the random experiment has already occurred. The occurrence of an event is related to one or more events.

This is similar to the fact the occurrence of some events based on the previously available information. This information can be used to calculate the probability of some new event. Mathematically, the calculation of the probability of an event A for which event B has already occurred. It is denoted by P(A | B).

Theorems on Conditional probability

Theorem 1

Let A and B be events of a sample space S of a random experiment. Then, P(S | B) = P(B | B) = 1.

Proof: P(S | B) = P(S ∩ B) ⁄ P(B) = P(B) ⁄ P(B) = 1.
[S ∩ B shows the outcomes common in S and B equals the outcomes in B].

Theorem 2

Let A and B be events of a sample S and E is an event of S such that P(E) ≠ 0, then

P((A ∪ B)| E) = P(A | E) + P(B | E) – P((A ∩ B) | E).

Proof: We know P(A ∪ B) = P(A) + P(B) – P(A ∩ B).

P((A ∪ B) |E) = P((A ∪ B) ∩ E) ⁄ P(E) = P((A ∩ E) ∪ (B ∩ E)) ⁄ P(E).

This is the distributive law of union of sets over the intersection.

or, P((A ∪ B) | E)  = [P(A ∩ E) + P(B ∩ E) – P(A ∩ B ∩ E)] ⁄ P(E) = [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).

If A and B are disjoint events, P(A ∩ B) = 0.

And, P((A ∪ B) |E) = P(A | E) + P(B | E).

Theorem 3

For any two events A and B of a sample space s, P(A ∩ B) = P(A). P(B | A), P(A) >0 or, P(A ∩ B) = P(B).P(A | B), P(B) > 0. This is the multiplication theorem of probability.

Proof: We know that P(B | A) = P(B ∩ A) / P(A), P(A) ≠ 0.

Or, P(B|A) = P(A ∩ B) ⁄ P(A) (as A ∩ B = B ∩ A).

P(A ∩ B) = P(A). P(B | A).

Similarly, P(A ∩ B) = P(B). P(A | B).

The multiplication theorem of probability can be extended to more than two events.

P(A ∩ B ∩ C) = P(A) . P(B | A). P(C | (A ∩ B))

Independent Events

As the name suggests the two events will be independent or free of each other. This means that the occurrence of one event does not affect the occurrence of the other. An event A is said to be independent of another event B, if the conditional probability of A given B, i.e., P(A | B) is equal to the unconditional probability of A, i.e., if P(A | B) = P(A).

Suppose from a pack of 52 well-shuffled cards we draw a card which turns out to be of heart. If the card is replaced or put back into the pack and another card is then drawn. It turns out to be of an ace.

Probability of getting a card of heart, P(H) = 13⁄52 = ¼  and probability of getting an ace, P(A) = 4⁄52 = 1⁄13. There is a possibility of getting a card of an ace of heart and the probability is P(A ∩ H) =  1⁄52. Here P(A | H) = P(A ∩ H)/P(H) = (1⁄52) ÷ (13⁄52) = 1⁄13 = P(A). This shows that the occurrence of an ace card is not affected by the occurrence of heart.

Theorem 4

For two events A and B such that P(A) ≠ 0, P(B) ≠ 0. If A is independent of B, then B is independent of A.

Proof: If A is independent of B, we have

P(A |B) = P(A) or, P(A ∩ B) ⁄ P(B) = P(A) or, P(A ∩ B) = P(A). P(B) … (I)

P(B |A) = P(A ∩ B)  ⁄ P(A) = P(A).P(B) ⁄ P(A) = P(B) [from I]

So, B is also independent of A.

Solved Examples for You

Question 1: A die is thrown. If A is the event of getting a multiple of 2, B is the event of getting an odd number and C is the event of getting an even number. Check the independence of A and B and that of A and C.

Answer: The sample space, S = {1, 2, 3, 4, 5, 6}. A = {2, 4, 6}, B = {1, 3, 5}, C= {2, 4, 6}.

P(A) = 3⁄6 = ½ , P(B) =3⁄6 = ½, P(C) = 3⁄6 = ½, P(A ∩ B) = 0, P(A ∩ C) = 3⁄6 = ½.

P(A ∩ B) = 0 ≠ P(A).P(B). Events A and B are not independent.

P(A ∩ C) = ½ ≠ P(A). P(C). Events A and C are not independent.

Question 2: The probability of a person speaking truth is 5⁄7 and of another person is 4⁄9. What is the probability they will contradict each other?

Answer : Let the events A: person 1 speaks the truth, B: person 2 speaks the truth.

P(A) = 5⁄7 and P(B) = 4⁄9.

P(A’) = 1 – (5⁄7) = 2⁄7 and P(B’) = 1 – (4⁄9) = 5⁄9.

The events A and B are independent of each other. The two persons will contradict each other if either of the two speaks a lie when the other speaks the truth. There are two possible cases.

• Case 1: Person 1 speaks truth and person 2 tells a lie. The event is A ∩ B’.
• Case 2: Person 1 tells a lie and person 2 speaks the truth. The event is A’ ∩ B.

The required probability = P(A ∩ B’) + P(A’ ∩ B) = P(A). P(B’) + P(A’). P(B) = (5⁄7 × 5⁄9) + (2⁄7 × 4⁄9) = 25⁄63 + 8⁄63 = 33⁄63 = 11⁄21.

Question 3: Why wee need conditional probability?

Answer: First of all, conditional probability is of fundamental importance. In addition, in the example of classification, the evidence is the values of the measurements or the features on which the classification is based. Also, the possible results are the possible classes.

Question 4: Explain the joint, marginal, and conditional probability?

Answer: Joint probability refers to the two events that occur simultaneously. Marginal probability is the probability of an event irrespective of the outcome of another variable. Lastly, conditional probability is the probability of one event occurring in the presence of a second event.

Question 5: State the difference between marginal and conditional probability?

Answer: Both conditional and marginal probabilities are ways to look at specific conditions of bivariate data. Furthermore, the marginal probability is the probability of occurrence of a single event, whereas, conditional probability is the probability of an event that will occur at another specific event that has already occurred.

Question 6: What is the reverse of conditional probability?

Answer: The reverse of conditional probability is called the positive predictive value that can be denoted by PPV: PPV = Positive Predictive Value = is the probability that someone has the disease if they test positive.