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Multiplication Theorem of Probability

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Estimated time: 10 minutes
CBSE: Class 12

Introduction

The Multiplication Theorem of Probability is used to find the probability that two or more events occur together. It is closely connected with conditional probability, because the chance of one event may depend on whether another event has already occurred.

This theorem is especially useful in questions involving successive events, such as drawing cards, selecting balls without replacement, or performing actions step by step.

CBSE: Class 12
Maharashtra State Board: Class 12

Theorem: Multiplication Theorem

For two events:

  • \[P(E \cap F) = P(F) \cdot P(E | F)\]

  • \[P(E \cap F) = P(E) \cdot P(F | E)\]

For three events:

  • \[P(E \cap F \cap G) = P(E) \cdot P(F | E) \cdot P(G | E \cap F)\]

When to Use This Theorem

CBSE: Class 12

Example 1

An urn contains 10 black balls and 5 white balls. Two balls are drawn one after another without replacement. Find the probability that both balls are black.

Step 1: Define events

  • Let E = {first ball is black.}

  • Let F = {second ball is black.}

Step 2: Find probabilities

  • \[P(E) = \frac{10}{15}\]

  • After one black ball is removed, 9 black balls remain out of 14 balls. Therefore, \[P(F | E) = \frac{9}{14}\].

Step 3: Apply multiplication theorem

\[P(E \cap F) = P(E) \cdot P(F | E)\]
\[P(E \cap F) = \frac{10}{15} \cdot \frac{9}{14} = \frac{3}{7}\]

Answer: Probability that both balls are black = \[\frac{3}{7}\].

CBSE: Class 12

Real Life Examples

  • If a student attends class regularly and then prepares properly, the probability of scoring well in a test can be thought of as a sequence of related events.

  • In a cricket match, the chance that a team wins after a strong opening partnership can be viewed through conditional probability.

CBSE: Class 12

Key Points: Multiplication Theorem on Probability

  • Multiplication theorem is used to find the probability of simultaneous occurrence of events.

  • For two events: \[P(E \cap F) = P(E) \cdot P(F | E)\]

  • Another equivalent form is \[P(E \cap F) = P(F) \cdot P(E | F)\].

  • For three events: \[P(E \cap F \cap G) = P(E) \cdot P(F | E) \cdot P(G | E \cap F)\].

  • Most “without replacement” questions are solved using this theorem.

  • Always define events before solving a probability problem.

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