Probability Solved Examples (Class 12 Maths)
Probability measures the likelihood of an event occurring. These examples cover basic probability rules, conditional probability, binomial distribution, an
TL;DR: Probability measures the likelihood of an event occurring. These examples cover basic probability rules, conditional probability, binomial distributio…
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Probability measures the likelihood of an event occurring. These examples cover basic probability rules, conditional probability, binomial distribution, an
Probability — Solved Numerical Examples (Step by Step)
Example 1: A coin is tossed 3 times. Find the probability of getting at least 2 heads.
Solution: Total possible outcomes = 2³ = 8: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT
At least 2 heads means 2 heads or 3 heads:
- Exactly 2 heads: HHT, HTH, THH (3 outcomes)
- Exactly 3 heads: HHH (1 outcome)
Total favorable outcomes = 3 + 1 = 4
Probability = 4/8 = 1/2
Example 2: A die is rolled. What is the probability of getting a number greater than 4?
Solution: Total possible outcomes = 6 (numbers 1, 2, 3, 4, 5, 6)
Favorable outcomes (greater than 4) = 5, 6 (2 outcomes)
Probability = Number of favorable outcomes / Total outcomes
P = 2/6 = 1/3
Example 3: In a bag, there are 5 red balls and 3 blue balls. If two balls are drawn without replacement, find the probability that both are red.
Solution: Total balls = 5 + 3 = 8
Probability of first ball being red = 5/8
After removing one red ball, remaining red balls = 4, total balls = 7
Probability of second ball being red = 4/7
Since the events are dependent:
P(both red) = P(1st red) × P(2nd red | 1st red)
= (5/8) × (4/7)
= 20/56 = 5/14
Example 4: Given P(A) = 0.4, P(B) = 0.5, and P(A ∩ B) = 0.2, find P(A ∪ B).
Solution: Using the formula for union of two events:
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
P(A ∪ B) = 0.4 + 0.5 - 0.2
P(A ∪ B) = 0.7
Example 5: Find the probability of rolling a 6 on a die given that an even number appears.
Solution: This is a conditional probability: P(6 | even)
Even numbers on a die: 2, 4, 6 (3 outcomes)
Number of favorable outcomes (rolling 6) = 1
P(6 | even) = Number of outcomes that are 6 among even numbers / Total even numbers
= 1/3
Example 6: In a binomial distribution with n = 5 and p = 0.4, find P(X = 2) where X is the number of successes.
Solution: Binomial probability: P(X = k) = C(n,k) × p^k × (1-p)^(n-k)
For n = 5, k = 2, p = 0.4:
P(X = 2) = C(5,2) × (0.4)² × (0.6)³
C(5,2) = 5! / (2! × 3!) = 10
(0.4)² = 0.16
(0.6)³ = 0.216
P(X = 2) = 10 × 0.16 × 0.216
= 10 × 0.03456
= 0.3456
Tips
- Always identify whether events are independent (P(A ∩ B) = P(A) × P(B)) or dependent before calculating combined probabilities.
- Conditional probability is defined as P(A | B) = P(A ∩ B) / P(B), provided P(B) > 0.
- When drawing without replacement, adjust the denominator in subsequent draws to reflect the reduced total.
- Complement rule states P(A') = 1 - P(A), which is useful when calculating probability of at least one or opposite events.
Frequently Asked Questions
What is the difference between mutually exclusive and independent events?
Mutually exclusive events cannot occur together: P(A ∩ B) = 0. Independent events have no influence on each other: P(A ∩ B) = P(A) × P(B). Events can be independent but not mutually exclusive, or mutually exclusive but not independent (in fact, mutually exclusive events are usually dependent).
Why do we use n! in permutation and combination formulas?
Factorial (n!) counts the number of ways to arrange n distinct objects. In combinations C(n,k) = n! / (k!(n-k)!), we divide by k! to account for the fact that order doesn't matter. In permutations P(n,k) = n! / (n-k)!, order does matter, so we use fewer factorials in the denominator.
More Maths Solved Examples
- Quadratic Equations
- Trigonometry
- Arithmetic Progressions
- Surface Areas and Volumes
- Coordinate Geometry
- Probability
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