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AFCAT Numerical Ability · Formula sheet

Probability formulas

24 formulas, 1 reference table and 45 common traps for AFCAT Numerical Ability Probability, grouped by subtopic.

Full notes with worked examples

Probability Basics: Equally Likely Outcomes and Three Rules

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Favourable over total for one pick

Classical probability

P(E)=n(E)n(S)P(E) = \dfrac{n(E)}{n(S)}

Three rules: not, or for exclusive events, and for independent events

The three rules

P(not A)=1−P(A);P(A or B)=P(A)+P(B) (exclusive);P(A and B)=P(A) P(B) (independent)P(\text{not } A) = 1 - P(A); \quad P(A \text{ or } B) = P(A) + P(B) \ (\text{exclusive}); \quad P(A \text{ and } B) = P(A)\,P(B) \ (\text{independent})

Common traps

1 is a perfect square

From 1 to 30 the perfect squares are 1, 4, 9, 16 and 25: five numbers, so 530=16\tfrac{5}{30} = \tfrac{1}{6}. Starting the list at 4 gives 430\tfrac{4}{30}.

Read which ends are included

'From 1 to 40' has 40 numbers. 'Between 1 and 40' can leave out both ends, which leaves 38. The total changes, so the answer does too.

Adding when you should multiply

For independent events with probabilities 0.4 and 0.9, 'both' is 0.4×0.9=0.360.4 \times 0.9 = 0.36. Adding gives 1.3, which is more than 1 and cannot be a probability.

Exclusive is not the same as independent

Mutually exclusive events cannot both happen, so P(A and B)=0P(A \text{ and } B) = 0. Independent events can happen together. Multiplying the probabilities of exclusive events gives a number where the answer is 0.

Coin Tosses: Counting Heads, At Least One, Order and Given That

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Listing the outcomes for up to three coins

Outcomes for n coins

n(S)=2n,P(E)=n(E)2nn(S) = 2^n, \qquad P(E) = \dfrac{n(E)}{2^n}

Exactly k heads from n coins: choose the positions

Exactly k heads in n tosses

P(k heads)=nCk2nP(k \text{ heads}) = \dfrac{^nC_k}{2^n}

At least one: one minus none

At least one

P(at least one)=1−P(none)P(\text{at least one}) = 1 - P(\text{none})

Order matters: given tosses, a first head, patterns and a biased coin

One sequence of n tosses

P(a given sequence)=(12)nP(\text{a given sequence}) = \left(\tfrac12\right)^n

Given that: count inside the condition

Conditional probability

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

Common traps

HT and TH are two outcomes

Two coins have four outcomes, not three. 'One head and one tail' happens in 2 of the 4, so its probability is 12\tfrac12, not 13\tfrac13.

Write the list in order

Listing three coins as HHH, HHT, HTT, TTT misses half of them. Branch on the first coin, then the second, and you get all 8.

Exactly is not at least

'At least 3 heads from 4 coins' means 3 or 4 heads: 4+116=516\tfrac{4 + 1}{16} = \tfrac{5}{16}. Stopping at exactly 3 gives 416\tfrac{4}{16}.

The total is 2 to the power n

Six coins have 26=642^6 = 64 outcomes, not 2×6=122 \times 6 = 12.

None is one outcome

With 4 coins, 'no head' is the single outcome TTTT, probability 116\tfrac{1}{16}. Writing 1−121 - \tfrac12 for 'at least one' ignores how many coins there are.

On every go means multiply the goes

'At least one head on each of 3 goes of two coins' is (34)3=2764\left(\tfrac34\right)^3 = \tfrac{27}{64}. The number 1−(14)31 - \left(\tfrac14\right)^3 answers a different question: a head on at least one of the goes.

Positions are not counts

'Heads on the first two of 5 tosses' is 14\tfrac14. 'Exactly two heads in 5 tosses' is 1032\tfrac{10}{32}, because the two heads can sit in any two places.

The pattern must not finish early

Tossing until two heads come in a row, exactly 3 tosses needs THH, so 18\tfrac18. HHH does not count: it stopped after 2 tosses.

Divide by the condition, not by everything

Given at least one head with two coins, divide by 3, not by 4. Dividing by the whole sample space answers 'both heads', which is 14\tfrac14.

Dice: One Die, the 36 Pairs and Three Dice

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One die: six equally likely faces

One die

P=faces that fit6P = \dfrac{\text{faces that fit}}{6}

Sums on two dice

Ways to make a sum

n(sum s)=6−∣s−7∣,2≤s≤12n(\text{sum } s) = 6 - |s - 7|, \quad 2 \le s \le 12

Products, differences and doublets

Pairs with difference d

n(∣a−b∣=d)=2(6−d),1≤d≤5n(|a - b| = d) = 2(6 - d), \quad 1 \le d \le 5

Die by die: multiply, and one minus none

At least one six from n dice

P(at least one six)=1−(56)nP(\text{at least one six}) = 1 - \left(\tfrac56\right)^n

Three dice: 216 ordered triples

Exactly k sixes on three dice

P=3Ck(16)k(56)3−kP = {}^3C_k \left(\tfrac16\right)^k \left(\tfrac56\right)^{3 - k}

Common traps

Greater than or at least

'Greater than 3' on a die is 4, 5, 6, so 12\tfrac12. 'At least 3' is 3, 4, 5, 6, so 23\tfrac23.

2 is prime

The prime faces are 2, 3 and 5. Leaving out 2 gives 13\tfrac13 instead of 12\tfrac12.

Count both orders

A sum of 3 comes from (1, 2) and (2, 1): 2 ways, so 118\tfrac{1}{18}. Counting it once gives 136\tfrac{1}{36}.

The 11 sums are not equally likely

There are 11 possible sums, 2 to 12, but 7 is six times as likely as 2. Never write 111\tfrac{1}{11} for one sum.

A square product counts once

Product 9 is only (3, 3), one pair. Product 6 is (1, 6), (6, 1), (2, 3) and (3, 2): four pairs.

Differ by means either order

'The numbers differ by 4' includes (5, 1) as well as (1, 5). Counting one order halves the answer.

Do not add for at least one

At least one six on two dice is not 16+16=13\tfrac16 + \tfrac16 = \tfrac13: that counts (6, 6) twice. It is 1−2536=11361 - \tfrac{25}{36} = \tfrac{11}{36}.

Exactly one has two orders

'Exactly one die shows a 4' on two dice is 16×56\tfrac16 \times \tfrac56 for each die, so twice that: 1036\tfrac{10}{36}.

Choose which dice

Exactly two sixes on three dice is not (16)2×56\left(\tfrac16\right)^2 \times \tfrac56 alone. The die that is not a six can be any of the three, so multiply by 3: 15216\tfrac{15}{216}.

A sum of 4 has three triples

(1, 1, 2), (1, 2, 1) and (2, 1, 1) are three different outcomes, so a sum of 4 has probability 3216\tfrac{3}{216}, not 1216\tfrac{1}{216}.

Playing Cards: the Deck, Or, Given That, and Drawing Two

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Ranges of values, and two conditions on one card

Cards in some ranks

P=4×ranks that fit52P = \dfrac{4 \times \text{ranks that fit}}{52}

Or: add, then take off the overlap

Or with an overlap

P(A or B)=n(A)+n(B)−n(A and B)52P(A \text{ or } B) = \dfrac{n(A) + n(B) - n(A \text{ and } B)}{52}

Given that, or a deck with cards removed

Given that

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

Two draws with replacement

With replacement

P(A then B)=P(A)×P(B)P(A \text{ then } B) = P(A) \times P(B)

Two or three cards without replacement

Without replacement

P(A then B)=P(A)×P(B∣A)P(A \text{ then } B) = P(A) \times P(B \mid A)

The 52-card deck

GroupCardsProbability of one card
One colour (red, or black)2612\tfrac{1}{2}
One suit1314\tfrac{1}{4}
One rank (for example the kings)4113\tfrac{1}{13}
Face cards: J, Q, K12313\tfrac{3}{13}
The Ace is not a face card here.
Number cards: 2 to 1036913\tfrac{9}{13}
One named card (for example the 7 of clubs)1152\tfrac{1}{52}

Common traps

Twelve face cards, not sixteen

With the Jack, Queen and King as face cards there are 12. Adding the four Aces gives 16, which these questions do not use.

Number cards are 2 to 10

Each suit has 9 number cards, 2 to 10. The Ace is not one of them unless the question says it counts as 1.

Ace high or Ace low

With the Ace as 1, 'value less than 4' is Ace, 2, 3: 12 cards. With the Ace as 14 it is 2 and 3 only: 8 cards.

Inclusive ranges

'Between 3 and 7 inclusive' is 3, 4, 5, 6, 7: five ranks, 20 cards. Leaving out an end gives 16.

The overlap is counted twice

A queen or a club is 4+13−1=164 + 13 - 1 = 16 cards. Writing 17 counts the queen of clubs twice.

Groups that cannot overlap

A diamond or a black card has no card in both groups, so it is 13+26=3913 + 26 = 39 cards with nothing taken off.

Removed cards change both counts

Taking out the 4 kings leaves 48 cards and 12 spades, so a spade is 1248\tfrac{12}{48}. Keeping 13 spades forgets that the king of spades went out with the kings.

Replacement keeps the totals

With replacement both draws are out of 52. Writing 51 for the second draw answers the without-replacement question.

Same suit as the first card

For two cards of the same suit the first card can be any card, then 12 of the 51 left match: 1251=417\tfrac{12}{51} = \tfrac{4}{17}. Multiplying by 1352\tfrac{13}{52} as well answers 'both are hearts' instead.

Shrink both counts

After one face card is drawn and kept, 11 face cards remain in 51 cards. Keeping 12 or 52 for the second draw gives the with-replacement answer.

Drawing from a Bag: One Ball, One After Another, With Replacement

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One ball: favourable over total

One ball

P(colour)=balls of that colourall ballsP(\text{colour}) = \dfrac{\text{balls of that colour}}{\text{all balls}}

One after another, without replacement

Without replacement

P(A then B)=an×bn−1P(A \text{ then } B) = \dfrac{a}{n} \times \dfrac{b}{n - 1}

With replacement: the bag resets

j of k draws, with replacement

P(j of k)=kCj pj(1−p)k−jP(j \text{ of } k) = {}^kC_j\, p^j (1 - p)^{k - j}

Common traps

The total is every ball

In a bag of 3 red, 5 blue and 2 white, white is 210\tfrac{2}{10}, not 28\tfrac{2}{8}: the white balls are part of the total.

Neither is not either

'Neither red nor blue' is the colours left over. 'Either red or blue' is those two colours. Reading one as the other gives the complement.

Change both numbers

After a red ball is kept out of 5 red and 3 blue, the bag has 4 red among 7. Writing 57\tfrac57 or 48\tfrac48 for the next red changes only one of the two.

Order is part of the event

'First red, then blue' is one order. 'One red and one blue' in two draws adds the other order too: 2×58×37=15282 \times \tfrac58 \times \tfrac37 = \tfrac{15}{28}.

With replacement nothing shrinks

Three draws with replacement from 2 red and 8 white give all red as (210)3=1125\left(\tfrac{2}{10}\right)^3 = \tfrac{1}{125}. Writing 210×19×08\tfrac{2}{10} \times \tfrac19 \times \tfrac08 treats the balls as kept out.

Choosing Several Balls at Once: Combinations

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A given mix of colours: choose from each colour and multiply

A mix of colours

P=aCx×bCynCrP = \dfrac{{}^aC_x \times {}^bC_y}{{}^nC_r}

Same colour: add over every colour

Same colour

P(same)=aCr+bCr+cCrnCrP(\text{same}) = \dfrac{{}^aC_r + {}^bC_r + {}^cC_r}{{}^nC_r}

Different colours: one minus same, or one of each

Different colours

P(different)=1−P(same)P(\text{different}) = 1 - P(\text{same})

At least: add the exact cases, or take away the rest

At least one of a colour

P(at least one)=1−n−aCrnCrP(\text{at least one}) = 1 - \dfrac{{}^{n - a}C_r}{{}^nC_r}

Common traps

Every ball is in the total

Choosing 2 balls from 3 red, 4 blue and 5 white, the total is 12C2=66^{12}C_2 = 66, even when the question asks only about red.

Exactly two red needs the rest non-red

Exactly 2 red among 3 chosen means the third ball is not red. Multiply by the ways to choose that ball; rC2^rC_2 over the total alone misses the factor.

Do not stop at one colour

Same colour from 2 red, 4 blue and 7 white is 1+6+21=281 + 6 + 21 = 28 pairs. Counting only the white pairs gives 2178\tfrac{21}{78}.

One of each is a product, not a sum

One red, one blue and one green from 2, 3 and 4 balls is 2×3×4=242 \times 3 \times 4 = 24 choices, not 2+3+42 + 3 + 4.

More colours than balls

Three balls from four colours: first choose which three colours, then one ball of each, and add over all four choices of colours.

At least includes the top case

At least 2 red among 3 drawn includes 3 red. Counting only exactly 2 red leaves that case out.

The opposite of at least one

The complement of 'at least one red' is 'no red', not 'exactly one red'.

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