Probability

What is probability?

P(A) = \dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}

For example, 4 of the 52 cards in a standard deck are Kings, so $P(\text{King}) = \dfrac{4}{52} = \dfrac{1}{13}$ .

The addition rule

P(A \cup B) = P(A) + P(B) - P(A \cap B)

Worked example: the probability of drawing a heart or a King is $\dfrac{13}{52} + \dfrac{4}{52} - \dfrac{1}{52} = \dfrac{16}{52} = \dfrac{4}{13}$ , subtracting the King of hearts so it isn't counted twice.

Independent vs. dependent events

Independent events don't affect each other: tossing a coin and rolling a die, $P(\text{Head and 3}) = \dfrac{1}{2}\times\dfrac{1}{6} = \dfrac{1}{12}$ .

Dependent events do: drawing a King and then a Queen without replacement, $P = \dfrac{4}{52}\times\dfrac{4}{51} = \dfrac{4}{663}$ .

Conditional probability

P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}

Worked example: of 50 students surveyed, 35 play basketball, and 25 of those are boys. The probability that a basketball player is a boy is $\dfrac{25/50}{35/50} = \dfrac{5}{7} \approx 0.71$ .

Worked example: Bayes' theorem

Machine A makes 60% of a factory's parts and 2% of its parts are defective; machine B makes the other 40%, with 5% defective. A part is picked at random and found to be defective. What is the probability it came from machine A?

P(\text{defective}) = 0.6 \times 0.02 + 0.4 \times 0.05 = 0.012 + 0.020 = 0.032

P(A \mid \text{defective}) = \dfrac{0.012}{0.032} = \dfrac{3}{8} = 0.375

So although A makes most of the parts, a defective part is more likely (62.5%) to come from B.

Worked example: the complement rule

What is the probability of getting at least one six in four rolls of a fair die?

Solution: "At least one" has many cases; its complement, "no six at all", has one. Each roll misses a six with probability $\tfrac{5}{6}$ , independently, so:

P(\text{at least one six}) = 1 - \left(\dfrac{5}{6}\right)^4 = 1 - \dfrac{625}{1296} = \dfrac{671}{1296} \approx 0.518

Common mistakes

Confusing independent with mutually exclusive. Mutually exclusive events can't both happen; independent ones can.
Swapping the condition. $P(A \mid B)$ and $P(B \mid A)$ are different numbers, as the Bayes example shows.
Ignoring replacement. Without replacement, the second draw's total drops by one.
Double counting in a union. Subtract the overlap $P(A \cap B)$ .

Where this comes up in exams

JEE Main 2026 lists the addition and multiplication theorems of probability and Bayes' theorem, and JEE Advanced 2026 lists conditional probability, independence, total probability and Bayes' theorem. The 2025-26 CBSE Class XII syllabus lists conditional probability, the multiplication theorem, independent events, total probability and Bayes' theorem, and the GATE 2027 CS syllabus lists conditional probability and Bayes theorem.

Checked against JEE (Main) 2026 syllabus (NTA), JEE (Advanced) 2026 syllabus, CBSE Mathematics curriculum, Classes XI-XII (2025-26) and GATE 2027 CS syllabus. Syllabi can change from year to year, so confirm with the latest official notification for your exam.

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