5 questions from actual NDA papers · all shown free · click option to reveal solution
PROBABILITY — CORE CONCEPT Probability tells us how likely an event is to happen. Think of it as a number between 0 and 1. If something is impossible, probability = 0. If something is certain, probability = 1. Everything else falls in between.
Simple definition: Probability of an event = (Number of favourable outcomes) / (Total number of possible outcomes) P(E) = n(E) / n(S)
Where n(E) = number of ways the event can happen, n(S) = total outcomes in the sample space. ---
0 ≤ P(E) ≤ 1 always. Probability is never negative and never more than 1.
P(E) + P(E') = 1, where E' means the event does NOT happen. So P(E') = 1 - P(E).
If two events cannot happen at the same time, they are called Mutually Exclusive Events. P(A or B) = P(A) + P(B).
If two events do not affect each other, they are called Independent Events. P(A and B) = P(A) × P(B).
P(A or B) = P(A) + P(B) - P(A and B) — this is the Addition Rule.
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— WHAT GETS ASKED SSC CGL tests probability in 3 main ways: 1. Coin, Dice, and Card problems — most common. 2. Balls from a bag (red, blue, green balls drawn randomly). 3. Two-event problems using AND/OR logic. For cards: A standard deck has 52 cards — 4 suits (Hearts, Diamonds, Clubs, Spades), 13 cards each.
Each suit has Ace, 2 to 10, Jack, Queen, King. Face cards = Jack, Queen, King = 12 total. Red cards = 26 (Hearts + Diamonds). Black cards = 26 (Clubs + Spades). For dice: One die has 6 faces (1 to 6).
Two dice give 6 × 6 = 36 total outcomes. For coins: One coin gives 2 outcomes (H or T). Two coins give 4 outcomes. Three coins give 8 outcomes. ---
Instead of counting what you WANT, count what you DON'T want. P(at least one) = 1 - P(none). This saves massive time in exam
If odds in favour = m : n, then P(E) = m / (m + n). If odds against = m : n, then P(E) = n / (m + n). Trick 3 — Multiply for AND, Add for OR (only if mutually exclusive): 'AND' means multiply the probabilities. 'OR' means add the probabilities (when events cannot happen together). ---
Total outcomes = 6 × 6 = 36
Pairs that give sum 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) = 6 pairs
P(sum = 7) = 6/36 = 1/6 Answer: 1/6 --- WORKED EXAMPLE 2 — Balls in a Bag (At Least One) Question: A bag has 4 red and 6 blue balls. Two balls are drawn randomly. What is the probability that at least one is red?
Total balls = 10. Total ways to pick 2 = 10C2 = 45
Use complement. P(at least one red) = 1 - P(no red ball)
P(no red) = ways to pick 2 blue balls = 6C2 = 15
P(no red) = 15/45 = 1/3
P(at least one red) = 1 - 1/3 = 2/3 Answer: 2/3 ---
— THE NUMBER ONE TRAP Students confuse 'at least one' problems. They try to count all cases individually (one red + one blue, two red) and often miss combinations. ALWAYS use the complement method: P(at least one) = 1 - P(none).
It is faster and error-free. Also, students forget that drawing cards or balls without replacement changes the total for the second draw — always check if replacement is mentioned.
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