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Probability - Independent Events

Grade 12ICSE

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

🔑Concepts

Definition of Independence: Two events AA and BB are said to be independent if the occurrence or non-occurrence of one does not affect the probability of the occurrence of the other. Visually, in a Venn diagram, independence is not shown by disjoint circles but by a specific proportional relationship where the area of the intersection P(AB)P(A \cap B) is exactly the product of the areas of AA and BB.

The Multiplication Theorem: For independent events, the probability of both events occurring simultaneously is the product of their individual probabilities, expressed as P(AB)=P(A)×P(B)P(A \cap B) = P(A) \times P(B). In a tree diagram, this is visualized by multiplying the probabilities along the specific branches that lead to the desired outcome.

Conditional Probability Property: If AA and BB are independent, the conditional probability of AA given BB is simply the probability of AA, i.e., P(AB)=P(A)P(A|B) = P(A). This implies that the 'reduced sample space' BB contains AA in the same proportion as the original sample space SS contains AA.

Independence of Complements: If events AA and BB are independent, then their complements are also independent. This means pairs like (A and B)(A' \text{ and } B), (A and B)(A \text{ and } B'), and (A and B)(A' \text{ and } B') also satisfy the multiplication rule. This is often visualized in a contingency table where row and column totals maintain consistent ratios.

Mutual Independence of Multiple Events: Three events A,B,A, B, and CC are mutually independent if they are pairwise independent (e.g., P(AB)=P(A)P(B)P(A \cap B) = P(A)P(B)) AND the probability of all three occurring is the product of all three probabilities: P(ABC)=P(A)P(B)P(C)P(A \cap B \cap C) = P(A)P(B)P(C). This ensures that no combination of events provides information about another.

Sampling with Replacement: A practical visual scenario for independent events is 'Sampling with Replacement'. When a ball is drawn from an urn and put back before the next draw, the composition of the urn remains unchanged. In a probability tree, this results in identical probability values on branches at every level of the experiment.

Distinction from Mutually Exclusive Events: It is vital to note that independent events are different from mutually exclusive events. Mutually exclusive events cannot happen at the same time (P(AB)=0P(A \cap B) = 0), whereas independent events usually have a non-zero intersection area. Visually, mutually exclusive events are separate circles, while independent events overlap.

📐Formulae

P(AB)=P(A)P(B)P(A \cap B) = P(A) \cdot P(B)

P(AB)=P(A)P(A|B) = P(A)

P(BA)=P(B)P(B|A) = P(B)

P(AB)=P(A)+P(B)P(A)P(B)P(A \cup B) = P(A) + P(B) - P(A) \cdot P(B)

P(ABC)=P(A)P(B)P(C)P(A \cap B \cap C) = P(A) \cdot P(B) \cdot P(C)

P(At least one of A,B)=1P(A)P(B)P(\text{At least one of } A, B) = 1 - P(A') \cdot P(B')

💡Examples

Problem 1:

A fair coin is tossed and a card is drawn at random from a well-shuffled pack of 52 cards. Find the probability of getting a Head and an Ace.

Solution:

Step 1: Define the events. Let HH be the event of getting a Head and AA be the event of drawing an Ace. \ Step 2: Calculate individual probabilities. P(H)=12P(H) = \frac{1}{2} and P(A)=452=113P(A) = \frac{4}{52} = \frac{1}{13}. \ Step 3: Check for independence. Tossing a coin and drawing a card are independent trials. \ Step 4: Apply the multiplication rule. P(HA)=P(H)×P(A)=12×113=126P(H \cap A) = P(H) \times P(A) = \frac{1}{2} \times \frac{1}{13} = \frac{1}{26}.

Explanation:

Since the outcome of the coin toss does not influence the card selection, we use the multiplication theorem for independent events.

Problem 2:

The probability of student A solving a specific math problem is 13\frac{1}{3} and student B solving it is 14\frac{1}{4}. If both try to solve the problem independently, find the probability that the problem is solved.

Solution:

Step 1: Identify given probabilities. P(A)=13P(A) = \frac{1}{3} and P(B)=14P(B) = \frac{1}{4}. \ Step 2: Calculate the probabilities of not solving. P(A)=113=23P(A') = 1 - \frac{1}{3} = \frac{2}{3} and P(B)=114=34P(B') = 1 - \frac{1}{4} = \frac{3}{4}. \ Step 3: Find the probability that neither solves the problem. P(AB)=P(A)×P(B)=23×34=24=12P(A' \cap B') = P(A') \times P(B') = \frac{2}{3} \times \frac{3}{4} = \frac{2}{4} = \frac{1}{2}. \ Step 4: Calculate the probability that at least one solves it. P(Solved)=1P(None solve)=112=12P(\text{Solved}) = 1 - P(\text{None solve}) = 1 - \frac{1}{2} = \frac{1}{2}.

Explanation:

The problem is considered 'solved' if at least one person gets the answer. It is mathematically simpler to subtract the probability of total failure from 1.