Statistics and Probability — SSC CGL Quantitative Aptitude

These topics enter mainly in Tier 2, and they reward memory over cleverness. Statistics is three summaries of a list — mean, median, mode — plus one empirical bridge between them. Probability is a single ratio, favourable ÷ total, applied to coins, dice and cards. Know the deck (52 cards, 4 suits, 12 face cards) and the dice sums cold, and most questions are pure counting.

1. What SSC actually asks

Tier 1: rare · Tier 2: 2–4 Q. Statistics: find the mean/median/mode of a small list, use the empirical mode formula, or read range. Probability: single die/coin events, two coins or two dice, and card draws (single or two-card). Almost never anything requiring calculus or heavy formulas.

2. Measures of central tendency

• Mean $=\frac{\text{sum}}{\text{count}}$ — the balance point.
• Median = middle value of the sorted list. Odd $n$: the middle term; even $n$: the average of the two middle terms.
• Mode = the most frequently occurring value (a list can have more than one mode).
• Range = max − min.

Empirical relation (for a moderately skewed distribution):

$\text{Mode}=3\text{Median}-2\text{Mean}$

Given mean 30 and median 28, mode $=84-60=24$. The formula is a stock Tier-2 question — memorise it exactly.

3. Probability — the one ratio

$P(E)=\frac{\text{number of favourable outcomes}}{\text{total number of outcomes}},0\le P(E)\le 1$

• Complement: $P(\text{not }E)=1-P(E)$ — often faster ("at least one" problems).
• Mutually exclusive (OR): $P(A\cup B)=P(A)+P(B)$ if they can't co-occur; otherwise subtract the overlap: $P(A)+P(B)-P(A\cap B)$.
• Independent (AND): $P(A\cap B)=P(A)P(B)$ (e.g. two separate tosses).
• Without replacement: the total shrinks after the first draw — multiply the changing fractions.

4. The three standard experiments

Coins ($n$ coins → $2^n$ outcomes):

• One coin: $P(H)=\frac{1}{2}$. Two coins {HH, HT, TH, TT}: $P(\text{exactly one head})=\frac{2}{4}=\frac{1}{2}$.

Dice (one die → 6 outcomes; two dice → 36):

• $P(\text{even})=\frac{3}{6}=\frac{1}{2}$; $P(\text{prime: 2,3,5})=\frac{1}{2}$.
• Two dice sums: $P(7)=\frac{6}{36}=\frac{1}{6}$ (the most likely sum); $P(\text{doubles})=\frac{6}{36}=\frac{1}{6}$.

Cards (52-card deck — memorise):

• 26 red (♥♦), 26 black (♠♣); 4 suits of 13 each.
• 4 aces, 4 kings, 4 queens, 4 jacks; 12 face cards (J, Q, K); 2 red kings, 2 black kings, etc.
• $P(\text{king})=\frac{4}{52}=\frac{1}{13}$; $P(\text{red})=\frac{1}{2}$.

5. Solved PYQ-style examples

Q1. Median of 3, 7, 5, 1, 9? Solution. Sorted: 1, 3, 5, 7, 9 → middle value 5.

Q2. For a distribution, mean = 30 and median = 28. The mode is… Solution. $3(28)-2(30)=84-60=$ 24.

Q3. Two coins are tossed. Probability of exactly one head? Solution. {HH, HT, TH, TT}: two favourable → $\frac{2}{4}=$ $\frac{1}{2}$.

Q4. Two dice are thrown. Probability the sum is 7? Solution. 6 ways out of 36 → $\frac{1}{6}$.

Q5. A bag has 4 red and 6 blue balls. Two are drawn without replacement. Probability both are red? Solution. $\frac{4}{10}\times \frac{3}{9}=\frac{12}{90}=$ $\frac{2}{15}$.

Q6. One card is drawn from a pack. Probability it is a red card or a king? Solution. $\frac{26+4-2}{52}=\frac{28}{52}=$ $\frac{7}{13}$ (subtract the 2 red kings counted twice).

6. Exam protocol

1. Statistics: sort first for the median; count frequencies for the mode.
2. Memorise Mode = 3 Median − 2 Mean — it appears almost every Tier-2 GA/quant slot.
3. Probability = favourable ÷ total; write the total sample space size first (2ⁿ, 6, 36, 52).
4. "OR" with overlap → add and subtract the common part; "AND"/independent → multiply.
5. Without replacement: shrink the denominator each draw. Know the 52-card deck cold.