01 · Explore
Understanding Conditional Probability
Conditional probability measures the likelihood of an event occurring given that another event has already taken place. This concept shifts our focus from the original sample space to a reduced subset defined by the known information.
In many real-world scenarios, the probability of an event is influenced by additional information. For instance, if we toss three fair coins, the original sample space S contains 8 outcomes. If we are told the first coin resulted in a tail (Event F), we no longer consider outcomes starting with a head. Our 'new' sample space becomes F = {THH, THT, TTH, TTT}.
The conditional probability of Event E given Event F is denoted as P(E|F). It is calculated by finding the ratio of the probability of the intersection of E and F to the probability of F. Formally, P(E|F) = P(E ∩ F) / P(F), provided that P(F) is not zero. This formula effectively scales the favorable outcomes within the new restricted sample space.
Calculating Conditional Probability in a Family
P(E|F) = P(E ∩ F) / P(F) = (1/4) / (3/4) = 1/3
Consider a family with two children. Let E be the event that both are boys and F be the event that at least one is a boy. The sample space S = {(b,b), (b,g), (g,b), (g,g)}. Here F = {(b,b), (b,g), (g,b)} so P(F) = 3/4. The intersection E ∩ F = {(b,b)} so P(E ∩ F) = 1/4. The conditional probability is 1/3.
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Question
If P(A) = 0.5, P(B) = 0.6, and P(A ∩ B) = 0.2, what is P(B|A)?
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02 · Explore
Properties of Conditional Probability
Conditional probability follows specific mathematical properties that mirror the axioms of standard probability but within a restricted sample space.
The first property states that the conditional probability of the sample space S given an event F is 1, i.e., P(S|F) = 1. Similarly, P(F|F) = 1. This confirms that within the context of F occurring, the occurrence of F itself is a certainty.
The second property is the addition rule for conditional probability: P((A ∪ B)|F) = P(A|F) + P(B|F) - P((A ∩ B)|F). If A and B are disjoint events, then P((A ∪ B)|F) = P(A|F) + P(B|F).
The third property relates an event to its complement: P(E'|F) = 1 - P(E|F). This is useful for calculating the probability of an event not happening given that another event has occurred.
| Property | Formula | Condition |
|---|---|---|
| Certainty | P(S|F) = 1 | P(F) > 0 |
| Addition Rule | P((A ∪ B)|F) = P(A|F) + P(B|F) - P((A ∩ B)|F) | General events |
| Complement | P(E'|F) = 1 - P(E|F) | Standard complement |
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Question
If P(E|F) = 0.35, what is the value of P(E'|F)?
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03 · Explore
Multiplication Theorem on Probability
The multiplication theorem provides a method to calculate the probability of the simultaneous occurrence of two or more events.
Derived from the definition of conditional probability, the multiplication rule states that P(E ∩ F) = P(E) · P(F|E). This means the probability of both E and F happening is the probability of E happening multiplied by the probability of F happening given that E has already occurred.
This rule can be extended to three or more events. For three events E, F, and G, the joint probability is P(E ∩ F ∩ G) = P(E) · P(F|E) · P(G|E ∩ F). This is particularly useful in experiments involving sampling without replacement, where the outcome of the first draw changes the probabilities for the second draw. The theorem can be further extended to four or more events in a similar sequential manner.
Drawing Balls Without Replacement
P(E ∩ F) = (10/15) × (9/14) = 3/7
An urn contains 10 black and 5 white balls. Two balls are drawn without replacement. Let E be 'first ball is black' and F be 'second ball is black'. P(E) = 10/15. After drawing one black ball, 9 black and 5 white balls remain, so P(F|E) = 9/14. The probability both are black is 3/7.
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Question
Two cards are drawn from a deck of 52 without replacement. What is the probability both are Aces?
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04 · Explore
Independent Events
Two events are independent if the occurrence of one does not change the probability of the occurrence of the other.
Formally, events E and F are independent if P(E|F) = P(E) and P(F|E) = P(F). Using the multiplication rule, this leads to the most common test for independence: P(E ∩ F) = P(E) · P(F). If this equality does not hold, the events are said to be dependent.
It is vital to distinguish between independent events and mutually exclusive events. Mutually exclusive events cannot happen at the same time (their intersection is empty), whereas independent events can happen together; their occurrence simply doesn't influence each other's likelihood. In fact, two mutually exclusive events with non-zero probabilities can never be independent because if one happens, the probability of the other happening becomes zero.
Independence vs Dependence
- 1
Check P(E ∩ F)
Calculate the probability of both events occurring simultaneously.
- 2
Compare with P(E)P(F)
Multiply the individual probabilities of the two events.
- 3
Conclusion
If P(E ∩ F) = P(E)P(F), they are independent. Otherwise, they are dependent.
The logical flow to determine if two events are mathematically independent.
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Question
A die is thrown. E = {3, 6} and F = {2, 4, 6}. Are E and F independent?
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05 · Explore
Theorem of Total Probability
The Theorem of Total Probability allows us to calculate the probability of an event by considering all the different ways (partitions) it can occur.
Suppose a sample space S is partitioned into n mutually exclusive and exhaustive events E1, E2, ..., En. A critical condition for this partition is that P(Ei) > 0 for all i = 1, 2, ..., n. If A is any event associated with S, then A can be viewed as the union of its intersections with each partition: A = (A ∩ E1) ∪ (A ∩ E2) ∪ ... ∪ (A ∩ En).
The total probability P(A) is the sum of the probabilities of these intersections: P(A) = P(E1)P(A|E1) + P(E2)P(A|E2) + ... + P(En)P(A|En). This is essentially a weighted average of the conditional probabilities of A across all possible scenarios represented by the partitions.
Construction Job Completion
P(A) = P(B)P(A|B) + P(B')P(A|B') = (0.65 × 0.32) + (0.35 × 0.80) = 0.488
Let A be 'job completed on time' and B be 'strike occurs'. P(B) = 0.65, so P(B') = 0.35. Given P(A|B) = 0.32 and P(A|B') = 0.80, the total probability of completing the job on time is 0.488.
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Question
If a bag is chosen from two (each with 0.5 probability) and Bag 1 has 3 red balls out of 10, while Bag 2 has 5 red balls out of 10, what is the total probability of drawing a red ball?
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06 · Explore
Bayes' Theorem
Bayes' Theorem provides a way to find the 'reverse probability'—the probability of a specific cause given that an effect has been observed.
While the theorem of total probability finds the probability of an outcome, Bayes' Theorem finds the probability that a specific partition (or 'cause') was responsible for that outcome. In this context, the events E1, E2, ..., En that constitute the partition are often called 'hypotheses'. If we know event A has occurred, the probability that it was caused by hypothesis Ei is given by P(Ei|A) = [P(Ei)P(A|Ei)] / Σ [P(Ej)P(A|Ej)].
In this formula, P(Ei) is called the 'priori' probability (the probability before the evidence is known), and P(Ei|A) is the 'posteriori' probability (the updated probability after observing event A). This theorem is widely used in medical testing, quality control, and machine learning.
Identifying the Source Bag
P(E2|A) = (1/2 × 5/11) / [(1/2 × 3/7) + (1/2 × 5/11)] = 35/68
Bag I has 3 red, 4 black balls. Bag II has 5 red, 6 black balls. A red ball is drawn. To find the probability it came from Bag II: P(E2) = 1/2, P(A|E2) = 5/11. P(E1) = 1/2, P(A|E1) = 3/7. Applying Bayes' formula gives 35/68.
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Question
A man speaks truth 3/4 of the time. He reports a 6 on a die. What is the probability it is actually a 6?
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07 · Explore
Introduction to Random Variables
A random variable is a mathematical bridge that links the qualitative outcomes of a sample space to quantitative real numbers.
A random variable X is formally defined as a real-valued function whose domain is the sample space S of a random experiment. This means for every outcome in the sample space, the random variable assigns a specific real number.
For example, if we toss a coin twice, the sample space is {HH, HT, TH, TT}. We can define a random variable X as the 'number of heads'. In this case, X(HH) = 2, X(HT) = 1, X(TH) = 1, and X(TT) = 0. Note that multiple random variables can be defined on the same sample space. Another variable Y could represent 'number of heads minus number of tails', resulting in values like 2, 0, 0, and -2.
| Outcome | X (No. of Heads) | Y (Heads - Tails) |
|---|---|---|
| HH | 2 | 2 |
| HT | 1 | 0 |
| TH | 1 | 0 |
| TT | 0 | -2 |
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If a die is rolled and X is the number shown, what is the domain of X?
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