Class 12 — Mathematics (NCERT)
Probability
Class 12
- ✓Define and calculate conditional probability for given events.
- ✓Apply the multiplication theorem on probability to find the probability of the intersection of events.
- ✓Understand and apply the Theorem of Total Probability and Bayes' Theorem to solve problems involving inverse probability.
- ✓Define a random variable and distinguish between discrete and continuous random variables.
- ✓Determine the probability distribution of a discrete random variable and calculate its mean and variance.
Key concepts
The probability of an event E occurring, given that another event F has already occurred, is called the conditional probability of E given F. It is denoted by P(E|F).
This theorem provides a way to find the probability of the simultaneous occurrence of two events. From the definition of conditional probability, we have P(E ∩ F) = P(F)P(E|F) or P(E ∩ F) = P(E)P(F|E). This can be extended to more than two events.
Two events E and F are said to be independent if the occurrence of one does not affect the probability of the occurrence of the other. In this case, P(E|F) = P(E) and P(F|E) = P(F).
Let E₁, E₂, ..., Eₙ be a partition of the sample space S, and let A be any event associated with S. Then the probability of event A can be expressed as the sum of probabilities of A occurring with each event Eₒ.
Bayes' theorem describes the probability of an event, based on prior knowledge of conditions that might be related to the event. It is used to find the "inverse probability". If E₁, E₂, ..., Eₙ are a partition of the sample space S, and A is any event with P(A) ≠ 0, then the posterior probability P(Eₒ|A) is given by:
A random variable is a real-valued function whose domain is the sample space of a random experiment. It assigns a real number to each outcome of the random experiment. Random variables can be discrete (taking a finite or countably infinite number of values) or continuous (taking any value within an interval).
The probability distribution of a discrete random variable X is a list of the possible values of X along with their corresponding probabilities. It satisfies two conditions: (i) P(X=xₒ) ≥ 0 for all i, and (ii) ∑P(X=xₒ) = 1.
The mean or expectation of a discrete random variable X, denoted by E(X) or μ, is the weighted average of its possible values, where the weights are the corresponding probabilities.
The variance of a discrete random variable X, denoted by Var(X) or σ², measures the spread or dispersion of the distribution around its mean.
Key facts to remember
- 1Conditional probability P(E|F) = P(E ∩ F) / P(F), where P(F) > 0.
- 2For independent events E and F, P(E ∩ F) = P(E)P(F).
- 3The Multiplication Theorem states P(E ∩ F) = P(E)P(F|E) = P(F)P(E|F).
- 4The Theorem of Total Probability: P(A) = ∑ P(Eₒ)P(A|Eₒ) for a partition E₁, ..., Eₙ of the sample space.
- 5Bayes' Theorem: P(Eₒ|A) = [P(Eₒ)P(A|Eₒ)] / ∑[P(E)P(A|E)].
- 6A random variable assigns a real number to each outcome of a random experiment.
- 7For a valid probability distribution of a discrete random variable X, P(X=xₒ) ≥ 0 for all i and ∑P(X=xₒ) = 1.
- 8The mean (expectation) of a discrete random variable X is E(X) = ∑xₒ * P(X=xₒ).
- 9The variance of a discrete random variable X is Var(X) = E(X²) - [E(X)]², where E(X²) = ∑xₒ² * P(X=xₒ).
Worked examples
Example 1
A die is thrown twice. What is the probability that the sum of the numbers appearing is 8, given that the first number appearing is 3?
Answer
The probability that the sum of the numbers appearing is 8, given that the first number appearing is 3, is 1/6.
Alternatively, since F has occurred, the reduced sample space is F. Out of these 6 outcomes, only (3,5) results in a sum of 8. So, the probability is 1/6.
Example 2
A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the bags is chosen at random and a ball is drawn from it which is found to be red. Find the probability that the ball was drawn from the first bag.
Answer
The probability that the red ball was drawn from the first bag is 2/3.
Always clearly define your events and use the correct notation for conditional probabilities before applying Bayes' Theorem.
Example 3
Two cards are drawn successively without replacement from a well-shuffled pack of 52 cards. Find the probability distribution of the number of aces. Also, find the mean and variance of this distribution.
Answer
Probability Distribution:\n| X | 0 | 1 | 2 |\n|------|-----------|-----------|-----------|\n| P(X) | 188/221 | 32/221 | 1/221 |\nMean (E(X)) = 2/13.\nVariance (Var(X)) = 400/2873.
Remember to calculate combinations correctly for 'without replacement' problems. Always verify that the sum of probabilities in your distribution is 1.
Common mistakes
- ✗Confusing P(E|F) with P(F|E) or P(E ∩ F).
- ✗Assuming independence of events without verification, leading to incorrect use of P(E ∩ F) = P(E)P(F).
- ✗Not ensuring that the sum of probabilities in a probability distribution equals 1.
- ✗Errors in calculating combinations or permutations when determining probabilities for random variables.
- ✗Incorrectly applying the formula for variance, especially forgetting to square E(X) or calculating E(X) instead of E(X²).
Exam tips
- ★Always clearly define the events involved using proper notation (e.g., E, F, A, E₁, E₂).
- ★For problems involving Bayes' Theorem or Total Probability, drawing a tree diagram can help visualise the problem and organise the probabilities.
- ★Carefully read whether events are independent or dependent, and whether sampling is with or without replacement.
- ★When dealing with random variables, list all possible values of the variable and their corresponding probabilities systematically.
- ★Show all intermediate steps in calculations for mean and variance to avoid errors and gain partial marks.
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