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Probability - Independent events, total probability, Bayes’ theorem

Grade 12CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Independent Events: Two events AA and BB are considered independent if the occurrence or non-occurrence of one does not change the probability of the occurrence of the other. Mathematically, this is satisfied if P(A∩B)=P(A)⋅P(B)P(A \cap B) = P(A) \cdot P(B). Visually, imagine a sample space SS as a square; if AA and BB are independent, the ratio of the area of their intersection to the area of BB is the same as the ratio of the area of AA to the whole square.

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Multiplication Theorem: For any two events AA and BB, the probability of both occurring is P(A∩B)=P(A)⋅P(B∣A)P(A \cap B) = P(A) \cdot P(B|A). If the events are independent, the conditional probability P(B∣A)P(B|A) is simply P(B)P(B). This can be extended to multiple events using a chain-like structure in a tree diagram.

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Partition of a Sample Space: A collection of events E1,E2,…,EnE_1, E_2, \dots, E_n is said to form a partition of the sample space SS if they are pairwise disjoint (Ei∩Ej=∅E_i \cap E_j = \emptyset for i≠ji \neq j) and their union is SS (∪Ei=S\cup E_i = S). Visualizing a partition is like looking at a jigsaw puzzle where each piece is an event; no pieces overlap, and together they fill the entire frame of the puzzle.

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Theorem of Total Probability: This theorem is used to calculate the probability of an event AA that can happen via several mutually exclusive paths E1,E2,…,EnE_1, E_2, \dots, E_n. P(A)P(A) is the sum of probabilities of AA occurring under each partition. Visually, this is represented by a tree diagram where P(A)P(A) is the sum of the 'path probabilities' (products of probabilities along branches) for every branch that ends in event AA.

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Bayes' Theorem: Bayes' Theorem finds the 'inverse' or 'posterior' probability—the probability of a specific cause (partition EiE_i) given that an outcome AA has already occurred. It effectively reverses the direction of the conditional probability. Visualizing this involves taking the specific branch in a tree diagram that leads to outcome AA via cause EiE_i and dividing its value by the total probability of all branches leading to AA.

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Conditional Probability: The probability of an event AA occurring given that BB has already occurred is P(A∣B)=P(A∩B)P(B)P(A|B) = \frac{P(A \cap B)}{P(B)}. This concept restricts our focus to the subset of the sample space where BB is true. Visually, if BB is a circle in a Venn diagram, we treat the area of BB as the new 'total' universe and calculate how much of AA is contained within it.

📐Formulae

P(A∣B)=P(A∩B)P(B),P(B)≠0P(A|B) = \frac{P(A \cap B)}{P(B)}, P(B) \neq 0

P(A∩B)=P(A)⋅P(B∣A)P(A \cap B) = P(A) \cdot P(B|A)

P(A∩B)=P(A)⋅P(B)P(A \cap B) = P(A) \cdot P(B) (if A,BA, B are independent)

P(A)=∑i=1nP(Ei)P(A∣Ei)P(A) = \sum_{i=1}^{n} P(E_i)P(A|E_i)

P(Ei∣A)=P(Ei)P(A∣Ei)∑j=1nP(Ej)P(A∣Ej)P(E_i|A) = \frac{P(E_i)P(A|E_i)}{\sum_{j=1}^{n} P(E_j)P(A|E_j)}

💡Examples

Problem 1:

Bag I contains 3 red and 4 black balls while Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it was drawn from Bag II.

Solution:

Let E1E_1 be the event of choosing Bag I, E2E_2 be the event of choosing Bag II, and AA be the event of drawing a red ball. \ Since bags are chosen at random, P(E1)=P(E2)=12P(E_1) = P(E_2) = \frac{1}{2}. \ P(A∣E1)=P(Red from Bag I)=37P(A|E_1) = P(\text{Red from Bag I}) = \frac{3}{7} \ P(A∣E2)=P(Red from Bag II)=511P(A|E_2) = P(\text{Red from Bag II}) = \frac{5}{11} \ Using Bayes' Theorem: \ P(E2∣A)=P(E2)P(A∣E2)P(E1)P(A∣E1)+P(E2)P(A∣E2)P(E_2|A) = \frac{P(E_2)P(A|E_2)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)} \ P(E2∣A)=12⋅51112⋅37+12⋅511=522314+522P(E_2|A) = \frac{\frac{1}{2} \cdot \frac{5}{11}}{\frac{1}{2} \cdot \frac{3}{7} + \frac{1}{2} \cdot \frac{5}{11}} = \frac{\frac{5}{22}}{\frac{3}{14} + \frac{5}{22}} \ P(E2∣A)=52233+35154=522⋅15468=5⋅768=3568P(E_2|A) = \frac{\frac{5}{22}}{\frac{33 + 35}{154}} = \frac{5}{22} \cdot \frac{154}{68} = \frac{5 \cdot 7}{68} = \frac{35}{68}.

Explanation:

This problem uses Bayes' Theorem because we are given the result (the ball is red) and asked to find the probability of the cause (Bag II). We first establish the prior probabilities of choosing each bag, then the conditional probabilities of drawing a red ball from each, and finally apply the formula.

Problem 2:

A problem in mathematics is given to three students whose chances of solving it are 12,13, and 14\frac{1}{2}, \frac{1}{3}, \text{ and } \frac{1}{4} respectively. What is the probability that the problem is solved?

Solution:

Let A,B, and CA, B, \text{ and } C be the events that the first, second, and third students solve the problem respectively. These are independent events. \ P(A)=12,P(B)=13,P(C)=14P(A) = \frac{1}{2}, P(B) = \frac{1}{3}, P(C) = \frac{1}{4} \ The problem is solved if at least one student solves it. It is easier to find the probability that the problem is NOT solved and subtract from 1. \ P(A′)=1−12=12P(A') = 1 - \frac{1}{2} = \frac{1}{2} \ P(B′)=1−13=23P(B') = 1 - \frac{1}{3} = \frac{2}{3} \ P(C′)=1−14=34P(C') = 1 - \frac{1}{4} = \frac{3}{4} \ P(Problem not solved)=P(A′∩B′∩C′)=P(A′)P(B′)P(C′)P(\text{Problem not solved}) = P(A' \cap B' \cap C') = P(A')P(B')P(C') (due to independence) \ P(Not solved)=12⋅23⋅34=14P(\text{Not solved}) = \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} = \frac{1}{4} \ P(Problem solved)=1−P(Not solved)=1−14=34P(\text{Problem solved}) = 1 - P(\text{Not solved}) = 1 - \frac{1}{4} = \frac{3}{4}.

Explanation:

Since the students work independently, the probability of the intersection of their failures is the product of their individual failure probabilities. Subtracting the probability of complete failure from 1 gives the probability that at least one person succeeds.