Probability Calculator

Introduction to the Probability Calculator

The Probability Calculator computes P(A) = favourable outcomes / total outcomes for any event A. It also handles combined events: P(A and B) = P(A) x P(B) for independent events, P(A or B) = P(A) + P(B) - P(A and B) for union, and conditional probability P(A|B) = P(A and B)/P(B).

Indian students from Class 9 to JEE Main use this tool for chapter problems on coins, dice, cards, balls in bags, and Bayes' theorem applications. CBSE Class 12 probability is a 10-mark chapter that includes random variables, binomial distribution, and conditional probability based on real-world scenarios.

You enter favourable and total outcomes for event A (and optionally B). The calculator returns P(A), P(B), P(A and B), P(A or B), P(A|B) and P(not A). It also computes binomial probability P(X = r) = nCr x p^r x (1-p)^(n-r) for repeated trials and odds form k:m.

Who Should Use This Probability Calculator

Class 10 students in Coimbatore solving NCERT chapter 15 probability problems on coin tosses, dice rolls and bag-of-balls questions use it to verify their homework against book answers.

JEE Main aspirants in Kota practicing the 4-mark single correct questions on conditional probability and Bayes' theorem need step-by-step verification before competitive exam day.

Class 12 board candidates in Ahmedabad working through CBSE chapter on random variables and binomial distribution check P(X = r) values for n = 5, 10 trial scenarios.

CAT and bank PO aspirants in Mumbai handling Quantitative Aptitude sections often face 2-3 probability questions per section, where speed matters as much as accuracy.

Statistics undergraduates in Bengaluru studying business analytics, ML and data science apply probability fundamentals to real datasets and need quick sanity checks during coursework.

Tips for Probability Problems

Smart Probability Tips

Always check whether events are independent or mutually exclusive. Two coins are independent (P(both H) = 0.5 x 0.5 = 0.25), but drawing without replacement is conditional, not independent.

For "at least one" problems, use complement: P(at least one) = 1 - P(none). If you roll 3 dice, P(at least one six) = 1 - (5/6)^3 = 1 - 125/216 = 91/216, faster than direct counting.

Sample space matters. For two coins, sample space is {HH, HT, TH, TT}, size 4. For three dice, sample space is 6^3 = 216. Always compute the denominator first to avoid silly mistakes in exams.

For card problems, remember the standard 52-card deck: 13 spades, 13 hearts, 13 diamonds, 13 clubs. 4 aces, 4 kings, 12 face cards. P(king or heart) = 4/52 + 13/52 - 1/52 = 16/52.

For Indian competitive exams like SSC and bank PO, learn binomial distribution shortcuts: P(X = r) for symmetric p = 0.5 simplifies to nCr / 2^n. Memorise nCr triangle up to n = 10.

Formula Explanation

Core Probability Formulas

P(A) = favourable / total

P(A and B) = P(A) x P(B|A)

P(A or B) = P(A) + P(B) - P(A and B)

P(A|B) = P(A and B) / P(B)

Where:

• P(A) = probability of event A
• P(A and B) = both events occur
• P(A or B) = at least one occurs
• P(A|B) = A given B has occurred

Example: A bag has 5 red and 7 blue balls. P(red) = 5/12 = 0.417. After drawing one red without replacement, P(red on second draw|red first) = 4/11 = 0.364.

Probability Quick Reference Table

Event	Favourable	Total	P (decimal)	P (%)
Head on one coin	1	2	0.5	50%
Six on one die	1	6	0.167	16.7%
Ace from deck	4	52	0.077	7.7%
Sum 7 on two dice	6	36	0.167	16.7%
Two heads in two tosses	1	4	0.25	25%
At least one head in 3 tosses	7	8	0.875	87.5%

Real-World Example

Example: Ishita's Card Game Probability

Meet Ishita, 17, a Class 12 student from Pune preparing for CBSE board exams and JEE Main. Her teacher gives her a problem: from a standard 52-card deck, what is the probability of drawing a face card or a heart?

Step 1: Ishita identifies the events. Event A = face card (Jack, Queen, King). Total face cards = 12. Event B = heart. Total hearts = 13.

Step 2: She finds the overlap: face cards that are also hearts = 3 (Jack, Queen, King of hearts). P(A) = 12/52, P(B) = 13/52, P(A and B) = 3/52.

Step 3: Apply union formula: P(A or B) = P(A) + P(B) - P(A and B) = 12/52 + 13/52 - 3/52 = 22/52 = 11/26.

Result: Ishita's answer is 11/26, or about 0.423 (42.3%). She verifies with the calculator and writes the step-by-step working in her board exam practice paper.

Frequently Asked Questions About Probability

Indian students often ask about the difference between independent and mutually exclusive events, when to use Bayes' theorem versus simple conditional probability, how the binomial distribution simplifies for fair coins, and how probability problems on cards and dice differ in CBSE board versus JEE Main difficulty. The FAQ below addresses each with NCERT and JEE-specific examples.