Probability for SBI PO — Cards, Dice, Coins and Combinatorics Approach

Concept

Probability is the measure of how likely an event is to occur, expressed as a number between 0 (impossible) and 1 (certain). The formal definition is clean:

$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$

But here's where most aspirants go wrong — they rush to apply this formula without first mapping out what the "total" actually is. The total is your sample space: the complete list of all things that can happen.

Think of it like this. Imagine you're dealing cards to a table of 7 people. Before you can say "what's the chance Ravi gets an ace," you need to know exactly how many cards are in the deck, how many people are drawing, and whether cards are going back or not. The same logic governs every probability question on SBI PO.

The best analogy for SBI PO probability is a voting urn. You have a jar full of balls (or cards, or people). Someone reaches in blindly. Your job is to count the ways the draw can go right, divide by all possible draws, and get a fraction. Everything else — dice, coins, groups of people — is just a variation of the urn.

Three things determine your setup:

1. With or without replacement? Drawing with replacement means the total stays constant for every draw. Without replacement means the denominator shrinks.
2. Order matters or not? If you're selecting a committee, order doesn't matter (use combinations, $C$). If you're filling seats in a row, order matters (use permutations, $P$).
3. Independent or dependent events? Two events are independent when the outcome of one doesn't affect the other (two coins tossed together). They're dependent when one outcome changes the landscape for the next (drawing two cards without replacement).

At the SBI PO level, you'll see questions layered with these considerations at once. A single question might ask about a group selection (combinations) where at least one person from a subgroup must be included (complementary approach). Recognise the type first; calculate second.

Deep Dive

Core Probability Rules

Rule 1 — Addition (OR): When you want the probability that event A or event B occurs: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$

For mutually exclusive events (can't happen together), $P(A \cap B) = 0$, so: $P(A \cup B) = P(A) + P(B)$

Rule 2 — Multiplication (AND): When you want the probability that both A and B occur:

• If independent: $P(A \cap B) = P(A) \times P(B)$
• If dependent: $P(A \cap B) = P(A) \times P(B|A)$, where $P(B|A)$ is the probability of B given A has already occurred.

Rule 3 — Complementary Events: $P(\text{not } A) = 1 - P(A)$

This is your single most powerful tool for SBI PO. Any question with "at least one" should be attacked with complement.

Standard Problem Types

Type 1: Dice Problems

A standard die has faces 1–6. Two dice together give $6 \times 6 = 36$ total outcomes.

Key sums to memorise (two dice):

| Sum | Count | Combinations | |-----|-------|--------------| | 2 | 1 | (1,1) | | 7 | 6 | (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) | | 11 | 2 | (5,6),(6,5) | | 12 | 1 | (6,6) |

Sum of 7 is the most frequent — 6 ways out of 36. This comes up in almost every exam.

Type 2: Card Problems

A standard deck has 52 cards: 4 suits (Hearts, Diamonds, Clubs, Spades) × 13 values (A, 2–10, J, Q, K).

• 4 Kings, 4 Queens, 4 Jacks (face cards = 12)
• 13 Hearts, 13 Spades (black), 13 Diamonds, 13 Clubs (red)
• The King of Hearts is both a King AND a Heart — when using the addition rule, subtract it once.

For "king or heart": $4 + 13 - 1 = 16$ favorable outcomes, probability $= 16/52 = 4/13$.

Type 3: Balls/Urns with Combinations

When selecting multiple items at once, always use $\binom{n}{r}$.

• Total ways to pick $r$ balls from $n$ = $\binom{n}{r}$
• Favorable ways = product of $\binom{}{}$ for each subgroup's contribution

For "one each of three colors from bags of 5 red, 6 blue, 4 green — pick 3": Favorable $= \binom{5}{1} \times \binom{6}{1} \times \binom{4}{1} = 5 \times 6 \times 4 = 120$ Total $= \binom{15}{3} = 455$ Probability $= 120/455 = 24/91$

Type 4: "At Least One" Problems

Never enumerate "at least one man in a group of 7." Instead:

$P(\text{at least one man}) = 1 - P(\text{no man}) = 1 - \frac{\binom{3}{2}}{\binom{7}{2}} = 1 - \frac{3}{21} = \frac{18}{21} = \frac{6}{7}$

Type 5: Independent Events — Problem Solved by Neither/Both

When A and B attempt independently: $P(\text{problem solved}) = 1 - P(A \text{ fails}) \times P(B \text{ fails})$

This avoids listing three separate cases (only A solves, only B solves, both solve).

Type 6: Odd/Even Sum from Cards/Numbers

For sum to be odd: one number must be odd, one even. For sum to be even: both must be odd or both must be even.

With numbers 1 to 27: 14 odd, 13 even.

$P(\text{odd sum}) = \frac{14}{27} \times \frac{13}{26} + \frac{13}{27} \times \frac{14}{26} = 2 \times \frac{182}{702} = \frac{364}{702} = \frac{14}{27}$

Notice: $\frac{364}{702}$ simplifies by dividing both by 26 → $\frac{14}{27}$.

Type 7: Set-Based Probability (Venn Diagram Logic)

$P(\text{neither A nor B}) = 1 - P(A \cup B) = 1 - \frac{|A| + |B| - |A \cap B|}{n}$

For the class problem: $1 - \frac{18+15-10}{30} = 1 - \frac{23}{30} = \frac{7}{30}$.

Memory Tricks & Shortcuts

Fast-Solving Framework

When a probability question appears on SBI PO, run this decision sequence before writing a single number:

Step 1 — Identify the setup: Is this dice / cards / balls / people selection / consecutive numbers? This tells you the sample space formula instantly.

Step 2 — Check for "at least one" or "neither": If yes — use complement. Write $1 - P(\text{opposite})$ and solve the simpler version.

Step 3 — Check if order matters: Committee / group / mixed color selection → combinations ($C$). Arrangement / sequence → permutations ($P$). Most SBI PO probability questions are combination-based.

Step 4 — Check if events are independent: Two people solving separately / two coins / two dice = independent. Multiply the failure probabilities for complement. Drawing without replacement = dependent. Adjust the denominator after each draw.

Step 5 — Apply the right formula, simplify: Compute numerator and denominator, find GCD, simplify. Cross-check: is your answer between 0 and 1? If not, you've inverted something.

Look — if you can classify the problem type in 10 seconds, the calculation follows in under 30. Most errors on these questions are classification errors, not arithmetic errors.

Solved PYQs

Why this question: Tests whether you recognise the odd/even parity structure rather than writing out conditional probability from scratch.

Solving path: Identify: 1 to 27 → 14 odd, 13 even. For odd sum: need one odd + one even. P = (14/27)(13/26) + (13/27)(14/26). Both terms are identical → multiply one by 2: $2 \times (14 \times 13)/(27 \times 26) = 364/702$. Divide numerator and denominator by 26: 14/27.

Why this question: Classic "at least one" trap. Many students enumerate; the complement is twice as fast.

Solving path: Total selections = $\binom{7}{2} = 21$. P(no man) = only women selected = $\binom{3}{2}/21 = 3/21 = 1/7$. P(at least one man) = $1 - 1/7 = 6/7$.

Why this question: Tests memorisation of dice sums and the addition rule for mutually exclusive events.

Solving path: Total outcomes = 36. Sum 7: 6 ways (anchor this). Sum 11: 2 ways — (5,6) and (6,5) only. Total favorable = 8. P = 8/36 = 2/9.

Why this question: Independent events with complement — tests whether you know the 2-step shortcut versus 3-case enumeration.

Solving path: P(A fails) = 1/3, P(B fails) = 1/4. P(neither solves) = 1/3 × 1/4 = 1/12. P(at least one solves) = 1 - 1/12 = 11/12.

Why this question: Multi-group combination selection — tests whether you multiply individual group selections and divide by total combinations.

Solving path: Total = $\binom{15}{3} = 455$. Favorable (one of each color) = $\binom{5}{1} \times \binom{6}{1} \times \binom{4}{1} = 5 \times 6 \times 4 = 120$. P = 120/455. GCD of 120 and 455 is 5 → 24/91.

Why this question: Card OR problem with overlap — tests the inclusion-exclusion principle in probability.

Solving path: P(King) = 4/52. P(Heart) = 13/52. P(King of Hearts) = 1/52. P(King or Heart) = 4/52 + 13/52 - 1/52 = 16/52 = 4/13.

Why this question: Set-theory probability using inclusion-exclusion — a Venn diagram problem disguised as a class question.

Solving path: Students studying at least one = 18 + 15 - 10 = 23. Neither = 30 - 23 = 7. P(neither) = 7/30.

Common Mistakes

• Forgetting the overlap in OR problems. When two events share an outcome (King of Hearts is both a King and a Heart), students add the counts without subtracting the overlap. Always check whether the two events can occur together; if yes, subtract the intersection once.

• Using permutations when combinations are needed. Selecting a committee of 3 from 10 does not depend on order. If you use $P(10,3) = 720$ instead of $\binom{10}{3} = 120$ as your denominator, every answer will be wrong. Ask yourself: does rearranging the selected group give a different outcome? If no, use $C$.

• Not adjusting the denominator for without-replacement draws. Drawing card 1 and card 2 without replacing card 1 changes the denominator from 52 to 51 for the second draw. Students who multiply (favorable/52) × (favorable/52) are treating it as with-replacement. Read the question — "one by one" without replacement is the default in most SBI PO problems.

• Counting odd numbers incorrectly in 1 to n ranges. In 1 to 27: there are 14 odd numbers (1, 3, 5, ... 27 — that's (27+1)/2 = 14). Students often write 13. Use the formula: odd count = $\lceil n/2 \rceil$ for 1 to n.

• Skipping simplification before multiplying. In 14/27 × 13/26, notice that 26 = 2 × 13 and 14 = 2 × 7. Cross-cancel before multiplying: (14 × 13)/(27 × 26) = (14 × 13)/(27 × 2 × 13) = 14/(27 × 2) = 7/27. Then double it for both orderings: 14/27. Multiplying first (182/702) then simplifying wastes ~20 seconds.

• Treating "at least one" as requiring enumeration. Listing "exactly one man" and "exactly two men" separately doubles your work. The moment you see "at least one" in a question involving groups, switch to complement. No exceptions.

Related Topics

• permutations
• combinations
• set theory
• data interpretation
• number system
• simplification

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