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Numerical Aptitude Syllabus / Module 13 · Probability / Week 13

Week 13: Probability

Probability is Week 12's counting, one layer up: favorable outcomes over total outcomes, both counted using last week's permutation and combination formulas. This week adds the one genuinely new idea — how probabilities combine across multiple events — plus the "at least one" pattern that trips up more students than any single formula in the chapter.

Module 13 of 20 Week 13 of 20 ~2–3 Hours Practice Problems Included

By the end of this week, you'll be able to

  • Build a correct sample space and count favorable outcomes using Week 12's formulas
  • Tell independent and dependent events apart, and combine their probabilities correctly
  • Solve "at least one" problems using the complement shortcut
1 Classical Probability & Sample Space 2 Independent vs. Dependent Events 3 Conditional Probability 4 "At Least One" Problems 5 Practice Problems 6 Knowledge Check

1. Classical Probability & Sample Space

Probability, in its simplest form, is a ratio of counts — which is exactly why Week 12's counting tools are the actual prerequisite for this chapter, not new machinery.

the core formula
P(event) = (number of FAVORABLE outcomes) / (number of TOTAL
            outcomes in the sample space)

Worked: a bag has 5 red and 3 blue balls. Find the probability
of drawing 2 red balls (without replacement).

Total ways to draw 2 balls from 8: 8C2 = 28
Favorable ways to draw 2 red from 5: 5C2 = 10

P(2 red) = 10/28 = 5/14

Both the numerator and denominator here are Week 12 combinations — building the sample space (denominator) and counting favorable outcomes (numerator) correctly, using the SAME counting method for both, is the entire skill. A common error is counting the numerator with combinations but the denominator with something inconsistent (or vice versa) — always use the same counting approach for both.

2. Independent vs. Dependent Events

Two events are independent if one's outcome doesn't affect the other's probability; dependent if it does. Rolling two separate dice — independent. Drawing two cards without replacement — dependent, since the first draw changes what's left for the second.

combining probabilities
Independent events: P(A and B) = P(A) x P(B)

Worked: probability of getting heads on a coin AND rolling a
6 on a die (two unrelated events)
  P(heads) = 1/2, P(six) = 1/6
  P(both) = 1/2 x 1/6 = 1/12

Dependent events: P(A and B) = P(A) x P(B | A)
  (P(B|A) = probability of B GIVEN A already happened --
   Section 3 covers this notation properly)

Worked: probability of drawing 2 aces in a row from a deck,
WITHOUT replacement
  P(first ace) = 4/52
  P(second ace | first was an ace) = 3/51  (one ace and one
                                             card gone from
                                             the deck)
  P(both aces) = 4/52 x 3/51 = 12/2652 = 1/221

The "without replacement" phrase is the reliable signal for dependent events — each draw changes the deck/bag for the next draw. "With replacement," or drawing from genuinely separate populations (two different dice, two different bags), signals independence.

3. Conditional Probability

P(B|A) — "the probability of B, given that A has already happened" — is worth understanding precisely, since it's both the general formula behind Section 2's dependent-event calculation and a common question format on its own.

the conditional probability formula
P(B|A) = P(A and B) / P(A)

Worked: in a class, 60% of students play cricket, 40% play
football, and 25% play both. If a student plays cricket,
what's the probability they also play football?

  P(football | cricket) = P(both) / P(cricket)
                         = 0.25 / 0.60
                         = 0.4167 (41.67%)

This is worth contrasting with the unconditional P(football) = 40% directly: knowing a student plays cricket actually raises the estimate of them also playing football (from 40% to 41.67%) in this example — conditioning on new information genuinely changes the probability, which is the entire point of the concept.

4. "At Least One" Problems

"Find the probability of getting at least one head in 3 coin flips" is far easier to solve by its complement — the probability of the opposite event — than by directly counting every way "at least one" can happen.

the complement shortcut
P(at least one) = 1 - P(none)

Worked: 3 coin flips, find P(at least one head)

Directly: would need to count "exactly 1 head," "exactly 2,"
"exactly 3" separately and add them -- three separate cases

Via complement: the ONLY way to NOT get at least one head is
to get ZERO heads (all tails)
  P(no heads) = P(TTT) = 1/2 x 1/2 x 1/2 = 1/8
  P(at least one head) = 1 - 1/8 = 7/8

"At least one" almost always has exactly one clean complement ("none"), while "at least one" itself can happen in many different ways (exactly 1, exactly 2, ... up to all of them) — which is precisely why the complement route is faster nearly every time this phrase appears.

"At least one" is a strong trigger phrase — train yourself to spot it

Whenever a probability question contains the words "at least one," reach for the complement (1 minus "none") as the default first move, before attempting to count the "at least one" cases directly. This single habit resolves a large fraction of the chapter's harder-looking questions almost instantly.

5. Practice Problems

Practice

30 problems, no calculator — Easy, Medium & Tough

Watch for "at least one" and reach for the complement immediately. Easy problems should take under 30 seconds each; Medium under 60 seconds; Tough under 90 seconds.

Easy (1–10)

  1. A die is rolled once. Find the probability of getting a 4.
  2. A coin is tossed once. Find the probability of getting heads.
  3. A bag has 3 red and 2 blue balls. Find the probability of drawing a red ball.
  4. Find the probability of getting an even number on a single die roll.
  5. A card is drawn from a deck of 52. Find the probability it's a king.
  6. Find the probability of getting a number less than 5 on a die.
  7. A bag has 5 white and 5 black balls. Find the probability of drawing a white ball.
  8. Two coins are tossed. Find the total number of possible outcomes.
  9. Find the probability of NOT getting a 6 on a single die roll.
  10. A card is drawn from a deck of 52. Find the probability it's a heart.
Easy — Answers

1) 1/6. 2) 1/2. 3) 3/5. 4) 3/6=1/2. 5) 4/52=1/13. 6) 4/6=2/3. 7) 5/10=1/2. 8) 2²=4. 9) 5/6. 10) 13/52=1/4.

Medium (11–20)

  1. Two dice are rolled. Find the probability of getting a sum of 7.
  2. A bag has 4 red and 6 green balls. Find the probability of drawing a red ball, then (without replacement) another red ball.
  3. Find the probability of getting at least one head in 2 coin tosses.
  4. A card is drawn from a deck. Find the probability it's a red card or a king.
  5. Two dice are rolled. Find the probability that both show the same number.
  6. A bag has 3 red, 4 blue and 5 green balls. Find the probability of drawing a blue or green ball.
  7. Find the probability of getting exactly 2 heads in 3 coin tosses.
  8. A number is chosen at random from 1 to 20. Find the probability it's a multiple of 3.
  9. Two cards are drawn from a deck without replacement. Find the probability both are aces.
  10. A die is rolled twice. Find the probability that the first roll shows a number greater than the second.
Medium — Answers

11) 6/36=1/6. 12) 4/10×3/9=2/15. 13) 1-1/4=3/4. 14) (26+4-2)/52=28/52=7/13. 15) 6/36=1/6. 16) (4+5)/12=3/4. 17) 3C2×(1/2)³=3/8. 18) Multiples of 3 in 1-20: 6 → 6/20=3/10. 19) 4/52×3/51=1/221. 20) 15/36=5/12 (15 of 36 ordered pairs have first > second).

Tough (21–30)

  1. Three dice are rolled. Find the probability of getting a sum of 18 (all sixes).
  2. A bag has 5 red and 7 black balls. Two balls are drawn without replacement. Find the probability that both are black.
  3. In a class of 40 students, 25 study Math, 20 study Physics and 10 study both. Find the probability that a randomly chosen student studies neither subject.
  4. Find the probability of getting at least one 6 in 4 rolls of a die.
  5. A bag contains 4 red, 5 white and 6 blue balls. Three balls are drawn at random. Find the probability all three are white.
  6. Two dice are rolled. Find the probability that the sum is greater than 9.
  7. A card is drawn from a deck. Given that it's a face card, find the probability it's a king.
  8. Three coins are tossed. Find the probability of getting at least 2 tails.
  9. A bag has 6 red and 4 blue balls. One ball is drawn and not replaced, then another is drawn. Find the probability that the first is red and the second is blue.
  10. In a lottery, there are 10 prizes and 25 blanks. Find the probability of getting a prize on one draw.
Tough — Answers

21) 1/6³=1/216. 22) 7/12×6/11=7/22. 23) At least one=25+20-10=35; neither=40-35=5 → 5/40=1/8. 24) P(no 6 in 4)=(5/6)⁴=625/1296; at least one=671/1296. 25) 15 balls total; C(15,3)=455; favorable=C(5,3)=10 → 10/455=2/91. 26) Sums >9 are 10,11,12: 3+2+1=6 ways → 6/36=1/6. 27) 12 face cards, 4 kings → 4/12=1/3. 28) Exactly 2 tails (3 ways) + exactly 3 tails (1 way) = 4/8=1/2. 29) 6/10×4/9=4/15. 30) 10/35=2/7.

6. Knowledge Check

Four quick questions. Expand each to check your answer.

Q1

Why must the numerator and denominator of a probability calculation use the same counting method (both combinations, or both permutations)?

A probability is a ratio of favorable outcomes to total outcomes drawn from the SAME sample space — if the numerator counts outcomes as unordered groups (combinations) while the denominator counts them as ordered sequences (permutations), the two counts aren't measuring the same underlying units, and their ratio doesn't correctly represent the true probability.

Q2

Why does drawing two cards "without replacement" make the two draws dependent events, while "with replacement" makes them independent?

Without replacement, the first card drawn is permanently removed from the deck, changing the composition (and therefore the probabilities) available for the second draw. With replacement, the first card is returned before the second draw, so the deck's composition — and the second draw's probabilities — are completely unaffected by what happened on the first draw.

Q3

Why is the complement approach ("1 minus P(none)") usually faster for "at least one" problems than counting the "at least one" cases directly?

"At least one" typically covers many separate sub-cases (exactly 1, exactly 2, exactly 3, etc.) that would each need to be counted and summed individually. Its complement, "none," is almost always exactly one single, simple case to compute directly, making the complement route dramatically less work for the same final answer.

Q4

Why can conditioning on new information (like "this student plays cricket") change a probability estimate, rather than leaving it the same as the unconditional probability?

Conditioning restricts attention to only the subset of the population where the given condition is true, and if the two events aren't independent, that subset can have a different rate of the second event than the population as a whole does. When knowing A changes the effective proportion of B within that narrower group, P(B|A) genuinely differs from the unconditional P(B).

← Back to Week 12: Permutations & Combinations Up next Week 14: Logarithms, Surds & Indices →