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Probability - Conditional Probability and Multiplication Theorem

Grade 12ICSE

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

🔑Concepts

Conditional Probability Definition: The probability of an event AA occurring given that event BB has already occurred is denoted by P(AB)P(A|B). Visually, this is interpreted as a reduction of the sample space from the entire set SS to just the set BB. On a Venn diagram, we calculate the probability by finding the ratio of the area of the intersection AcapBA \\cap B to the total area of the set BB.

The Multiplication Theorem: This theorem provides a formula for the probability of the simultaneous occurrence of two events AA and BB. It states that P(AcapB)=P(A)cdotP(BA)P(A \\cap B) = P(A) \\cdot P(B|A), meaning the probability of both events occurring is the product of the probability of the first event and the conditional probability of the second event given that the first has occurred.

Independent Events: Two events are independent if the occurrence of one does not change the probability of the other. Visually, in a Venn diagram, independence implies that the proportion of circle AA that overlaps with BB is the same as the proportion of AA in the entire sample space SS. Mathematically, this is expressed as P(AB)=P(A)P(A|B) = P(A) and P(BA)=P(B)P(B|A) = P(B).

Condition for Independence: Two events AA and BB are independent if and only if P(AcapB)=P(A)cdotP(B)P(A \\cap B) = P(A) \\cdot P(B). This serves as the primary test for independence in problems. If this equality does not hold, the events are dependent, and the general multiplication theorem must be used.

Multiplication Theorem for Three Events: The theorem extends to multiple events. For three events A,B,A, B, and CC, the probability of their joint occurrence is P(AcapBcapC)=P(A)cdotP(BA)cdotP(CAcapB)P(A \\cap B \\cap C) = P(A) \\cdot P(B|A) \\cdot P(C|A \\cap B). This describes a chain of dependencies, which can be visualized as moving through three levels of a tree diagram.

Tree Diagrams: A tree diagram is a visual representation of multi-stage experiments. Each node splits into branches representing outcomes, and each branch is labeled with its conditional probability. To find the probability of a specific outcome sequence (path), you multiply the probabilities along the branches from the root to the leaf.

Complementary Conditional Probability: For any event AA and a given condition BB, the probability of the complement event AA' occurring given BB is P(AB)=1P(AB)P(A'|B) = 1 - P(A|B). This rule highlights that within the restricted sample space of BB, the sum of probabilities of all mutually exclusive and exhaustive events must equal 1.

📐Formulae

P(AB)=fracP(AcapB)P(B),P(B)neq0P(A|B) = \\frac{P(A \\cap B)}{P(B)}, P(B) \\neq 0

P(AcapB)=P(A)cdotP(BA)=P(B)cdotP(AB)P(A \\cap B) = P(A) \\cdot P(B|A) = P(B) \\cdot P(A|B)

P(AcapBcapC)=P(A)cdotP(BA)cdotP(CAcapB)P(A \\cap B \\cap C) = P(A) \\cdot P(B|A) \\cdot P(C|A \\cap B)

P(AcapB)=P(A)cdotP(B)text(ForIndependentEvents)P(A \\cap B) = P(A) \\cdot P(B) \\text{ (For Independent Events)}

P(AB)=1P(AB)P(A'|B) = 1 - P(A|B)

P(AcupBC)=P(AC)+P(BC)P(AcapBC)P(A \\cup B | C) = P(A|C) + P(B|C) - P(A \\cap B | C)

💡Examples

Problem 1:

A box contains 10 black and 5 white balls. Two balls are drawn from the box one after the other without replacement. What is the probability that both drawn balls are black?

Solution:

  1. Let B1B_1 be the event that the first ball is black and B2B_2 be the event that the second ball is black.
  2. The probability of the first ball being black is P(B1)=frac1015=frac23P(B_1) = \\frac{10}{15} = \\frac{2}{3}.
  3. Since the balls are drawn without replacement, after the first black ball is drawn, there are 9 black balls left out of a total of 14 balls.
  4. The conditional probability of the second ball being black given the first was black is P(B2B1)=frac914P(B_2|B_1) = \\frac{9}{14}.
  5. By the Multiplication Theorem: P(B1capB2)=P(B1)cdotP(B2B1)=frac23cdotfrac914=frac1842=frac37P(B_1 \\cap B_2) = P(B_1) \\cdot P(B_2|B_1) = \\frac{2}{3} \\cdot \\frac{9}{14} = \\frac{18}{42} = \\frac{3}{7}.

Explanation:

This problem illustrates the use of the multiplication theorem for dependent events. The probability of the second draw is affected by the outcome of the first draw because the total number of balls decreases.

Problem 2:

Given P(A)=0.4P(A) = 0.4, P(B)=0.7P(B) = 0.7, and the conditional probability P(BA)=0.6P(B|A) = 0.6, find P(AcupB)P(A \\cup B).

Solution:

  1. First, find P(AcapB)P(A \\cap B) using the multiplication theorem: P(AcapB)=P(A)cdotP(BA)=0.4cdot0.6=0.24P(A \\cap B) = P(A) \\cdot P(B|A) = 0.4 \\cdot 0.6 = 0.24.
  2. Use the general addition rule for probability: P(AcupB)=P(A)+P(B)P(AcapB)P(A \\cup B) = P(A) + P(B) - P(A \\cap B).
  3. Substitute the known values: P(AcupB)=0.4+0.70.24=1.10.24=0.86P(A \\cup B) = 0.4 + 0.7 - 0.24 = 1.1 - 0.24 = 0.86.

Explanation:

This example shows how to bridge conditional probability and the addition theorem. We first calculate the intersection using the conditional data and then find the union probability.