Conditional Probability and Multiplication Theorem on Probability

Concept Explanation: Probability measures the likelihood of an event occurring. For an event \(E\) in sample space \(S\), probability is \(P(E) = \frac{n(E)}{n(S)}\), where \(n(E)\) is the number of favorable outcomes and \(n(S)\) is the total number of outcomes.

Events can be:

• Mutually Exclusive: Events that cannot occur simultaneously, i.e., \(A \cap B = \emptyset\).
• Independent: Occurrence of one event does not affect the other.
• Exhaustive: Events whose union covers the entire sample space and are pairwise disjoint.

Conditional Probability is the probability of event \(A\) given event \(B\) has occurred, denoted \(P(A|B) = \frac{P(A \cap B)}{P(B)}\), provided \(P(B) \neq 0\).

Formula Derivation

By definition, the conditional probability of \(A\) given \(B\) is:

\[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]

Multiplying both sides by \(P(B)\), we get the multiplication theorem:

\[ P(A \cap B) = P(B) \times P(A|B) \]

Similarly, \(P(B \cap A) = P(A) \times P(B|A)\).

Worked Illustrations and Solved Examples

Example 1: A family has two children. Find the probability both are boys given at least one is a boy.

Solution:

Sample space \(S = \{(b,b), (b,g), (g,b), (g,g)\}\), \(n(S) = 4\).

Event \(E\): both children are boys \(= \{(b,b)\}\), \(n(E) = 1\).

Event \(F\): at least one boy \(= \{(b,b), (b,g), (g,b)\}\), \(n(F) = 3\).

\(E \cap F = \{(b,b)\}\), \(n(E \cap F) = 1\).

Calculate probabilities:

\[ P(F) = \frac{3}{4}, \quad P(E \cap F) = \frac{1}{4} \]

Conditional probability:

\[ P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3} \]

Example 2: Ten cards numbered 1 to 10 are mixed. If a card drawn is known to be greater than 3, find the probability it is even.

Solution:

Sample space \(S = \{1,2,3,4,5,6,7,8,9,10\}\), \(n(S) = 10\).

Event \(A\): card is even \(= \{2,4,6,8,10\}\), \(n(A) = 5\).

Event \(B\): card is greater than 3 \(= \{4,5,6,7,8,9,10\}\), \(n(B) = 7\).

\(A \cap B = \{4,6,8,10\}\), \(n(A \cap B) = 4\).

Calculate probabilities:

\[ P(A) = \frac{5}{10} = \frac{1}{2}, \quad P(B) = \frac{7}{10}, \quad P(A \cap B) = \frac{4}{10} = \frac{2}{5} \]

Conditional probability:

\[ P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{2}{5}}{\frac{7}{10}} = \frac{2}{5} \times \frac{10}{7} = \frac{4}{7} \]

Practice Set

• Level 1 – Easy: A die is rolled. Find the probability of getting a 4 given the number is even.
• Level 2 – Moderate: Two cards are drawn without replacement from a deck of 52. Find the probability both are kings.
• Level 3 – Challenging: In a box, there are 5 red and 7 blue balls. Two balls are drawn one after another without replacement. Find the probability both are red.

Answer Key

• Level 1: \(P(4|\text{even}) = \frac{P(4)}{P(2,4,6)} = \frac{\frac{1}{6}}{\frac{3}{6}} = \frac{1}{3}\)
• Level 2: \(P(\text{both kings}) = \frac{4}{52} \times \frac{3}{51} = \frac{12}{2652} = \frac{1}{221}\)
• Level 3: \(P(\text{both red}) = \frac{5}{12} \times \frac{4}{11} = \frac{20}{132} = \frac{5}{33}\)

Quick Reference

Formula	Description
\(P(A|B) = \frac{P(A \cap B)}{P(B)}\)	Conditional probability of \(A\) given \(B\)
\(P(A \cap B) = P(B) \times P(A|B)\)	Multiplication theorem
\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)	Probability of union of two events

Glossary

• Sample Space (S): Set of all possible outcomes.
• Event (E): Subset of sample space.
• Mutually Exclusive: Events that cannot occur together.
• Independent Events: Events where occurrence of one does not affect the other.
• Conditional Probability: Probability of an event given another event has occurred.

Bayes Theorem

Concept Explanation: Bayes theorem provides a way to update probabilities based on new information. If \(E_1, E_2, ..., E_n\) form a partition of sample space \(S\) and \(A\) is an event with \(P(A) > 0\), then the probability of \(E_i\) given \(A\) is:

Formula Derivation

By definition of conditional probability:

\[ P(E_i|A) = \frac{P(E_i \cap A)}{P(A)} \]

Since \(E_i\) and \(A\) occur together, and \(E_i\) are disjoint partitions,

\[ P(E_i \cap A) = P(E_i) \times P(A|E_i) \]

Also, total probability of \(A\) is:

\[ P(A) = \sum_{j=1}^n P(E_j) P(A|E_j) \]

Therefore, Bayes theorem is:

\[ P(E_i|A) = \frac{P(E_i) P(A|E_i)}{\sum_{j=1}^n P(E_j) P(A|E_j)} \]

Worked Illustrations and Solved Examples

Example 3: A construction job has a 0.65 probability of strike. Probability of completion on time is 0.80 if no strike, 0.32 if strike. Find probability job completes on time.

Solution:

Let \(F\): strike, \(E\): job completes on time.

\[ P(F) = 0.65, \quad P(\overline{F}) = 0.35 \] \[ P(E|F) = 0.32, \quad P(E|\overline{F}) = 0.80 \]

By total probability theorem:

\[ P(E) = P(F) P(E|F) + P(\overline{F}) P(E|\overline{F}) = 0.65 \times 0.32 + 0.35 \times 0.80 = 0.208 + 0.28 = 0.488 \]

Example 4: Three boxes I, II, III contain coins: I has 2 gold, II has 2 silver, III has 1 gold and 1 silver. A box is chosen at random and a coin drawn. If the coin is gold, find probability the other coin in the box is gold.

Solution:

Let \(A, B, C\) be choosing boxes I, II, III respectively.

\[ P(A) = P(B) = P(C) = \frac{1}{3} \]

Event \(E\): coin drawn is gold.

\[ P(E|A) = 1, \quad P(E|B) = 0, \quad P(E|C) = \frac{1}{2} \]

By Bayes theorem:

\[ P(A|E) = \frac{P(A) P(E|A)}{P(A) P(E|A) + P(B) P(E|B) + P(C) P(E|C)} = \frac{\frac{1}{3} \times 1}{\frac{1}{3} \times 1 + \frac{1}{3} \times 0 + \frac{1}{3} \times \frac{1}{2}} = \frac{\frac{1}{3}}{\frac{1}{3} + 0 + \frac{1}{6}} = \frac{\frac{1}{3}}{\frac{1}{2}} = \frac{2}{3} \]

Practice Set

• Level 1 – Easy: A box contains 3 red and 2 blue balls. One ball is drawn. If it is red, find the probability it came from the first box if two boxes have different compositions.
• Level 2 – Moderate: A factory has two machines producing items. Machine A produces 60% items with 2% defect rate, Machine B produces 40% items with 3% defect rate. If an item is defective, find probability it was produced by Machine A.
• Level 3 – Challenging: Three boxes contain different numbers of defective and non-defective items. Given probabilities of choosing each box and drawing a defective item, find the probability the defective item came from a specific box.

Answer Key

• Level 1: Use Bayes theorem with given box probabilities and red ball probabilities.
• Level 2: \(P(A|D) = \frac{P(A) P(D|A)}{P(A) P(D|A) + P(B) P(D|B)} = \frac{0.6 \times 0.02}{0.6 \times 0.02 + 0.4 \times 0.03} = \frac{0.012}{0.012 + 0.012} = \frac{1}{2}\)
• Level 3: Apply Bayes theorem with given data.

Quick Reference

Formula	Description
\(P(E_i|A) = \frac{P(E_i) P(A|E_i)}{\sum_{j=1}^n P(E_j) P(A|E_j)}\)	Bayes theorem for event \(E_i\) given \(A\)
\(P(A) = \sum_{j=1}^n P(E_j) P(A|E_j)\)	Total probability of \(A\)

Glossary

• Prior Probability: Probability of an event before new evidence.
• Posterior Probability: Updated probability after considering new evidence.
• Partition: A set of mutually exclusive and exhaustive events.

Random Variable and its Probability Distributions

Concept Explanation: A random variable \(X\) is a real-valued function defined on the sample space of an experiment. It assigns a numerical value to each outcome.

Probability Distribution: The set of values \(X\) can take along with their probabilities is called the probability distribution.

Types:

• Discrete Random Variable: Takes finite or countably infinite values.
• Continuous Random Variable: Takes any value in an interval.

Formula Derivation

Mean or Expectation \(\mu\) of \(X\):

\[ \mu = E(X) = \sum_i x_i P(x_i) \]

where \(x_i\) are values of \(X\) and \(P(x_i)\) their probabilities.

Worked Illustrations and Solved Examples

Example 5: Two cards drawn with replacement from 52 cards. Find probability distribution of number of aces.

Solution:

Random variable \(X\): number of aces drawn, possible values \(0,1,2\).

Probability of ace \(= \frac{4}{52} = \frac{1}{13}\), non-ace \(= \frac{48}{52} = \frac{12}{13}\).

Calculate probabilities:

\[ P(X=0) = P(\text{no ace in both draws}) = \left(\frac{12}{13}\right)^2 = \frac{144}{169} \] \[ P(X=1) = P(\text{one ace}) = 2 \times \frac{1}{13} \times \frac{12}{13} = \frac{24}{169} \] \[ P(X=2) = P(\text{two aces}) = \left(\frac{1}{13}\right)^2 = \frac{1}{169} \]

Probability distribution table:

\(X\)	0	1	2
\(P(X)\)	\(\frac{144}{169}\)	\(\frac{24}{169}\)	\(\frac{1}{169}\)

Practice Set

• Level 1 – Easy: Toss a coin 3 times. Find probability distribution of number of heads.
• Level 2 – Moderate: A die is rolled twice. Find probability distribution of sum of numbers.
• Level 3 – Challenging: Given a probability distribution with unknown constant \(k\), find \(k\) and calculate probabilities for given events.

Answer Key

• Level 1: \(P(X = k) = \binom{3}{k} \left(\frac{1}{2}\right)^k \left(\frac{1}{2}\right)^{3-k}\) for \(k=0,1,2,3\).
• Level 2: Calculate sums and their probabilities by counting outcomes.
• Level 3: Use \(\sum P(X=x_i) = 1\) to find \(k\), then compute required probabilities.

Quick Reference

Formula	Description
\(E(X) = \sum x_i P(x_i)\)	Mean or expectation of random variable
\(\sum P(x_i) = 1\)	Sum of probabilities equals 1

Glossary

• Random Variable: Function assigning numerical values to outcomes.
• Probability Distribution: Mapping of values to their probabilities.
• Expectation: Weighted average of values.
• Discrete Variable: Takes countable values.
• Continuous Variable: Takes values in an interval.