Probability: AFCAT Numerical Ability

Formulas (2)

• Favourable over total for one pick · Classical probability
  $$P(E) = \dfrac{n(E)}{n(S)}$$
• Three rules: not, or for exclusive events, and for independent events · The three rules
  $$P(\text{not } A) = 1 - P(A); \quad P(A \text{ or } B) = P(A) + P(B) \ (\text{exclusive}); \quad P(A \text{ and } B) = P(A)\,P(B) \ (\text{independent})$$

Watch out for (4)

• 1 is a perfect square
  → Favourable over total for one pick
• Read which ends are included
  → Favourable over total for one pick
• Adding when you should multiply
  → Three rules: not, or for exclusive events, and for independent events
• Exclusive is not the same as independent
  → Three rules: not, or for exclusive events, and for independent events

Formulas (5)

• Listing the outcomes for up to three coins · Outcomes for n coins
  $$n(S) = 2^n, \qquad P(E) = \dfrac{n(E)}{2^n}$$
• Exactly k heads from n coins: choose the positions · Exactly k heads in n tosses
  $$P(k \text{ heads}) = \dfrac{^nC_k}{2^n}$$
• At least one: one minus none · At least one
  $$P(\text{at least one}) = 1 - P(\text{none})$$
• Order matters: given tosses, a first head, patterns and a biased coin · One sequence of n tosses
  $$P(\text{a given sequence}) = \left(\tfrac12\right)^n$$
• Given that: count inside the condition · Conditional probability
  $$P(A \mid B) = \dfrac{n(A \cap B)}{n(B)}$$

Watch out for (9)

• HT and TH are two outcomes
  → Listing the outcomes for up to three coins
• Write the list in order
  → Listing the outcomes for up to three coins
• Exactly is not at least
  → Exactly k heads from n coins: choose the positions
• The total is 2 to the power n
  → Exactly k heads from n coins: choose the positions
• None is one outcome
  → At least one: one minus none
• On every go means multiply the goes
  → At least one: one minus none
• Positions are not counts
  → Order matters: given tosses, a first head, patterns and a biased coin
• The pattern must not finish early
  → Order matters: given tosses, a first head, patterns and a biased coin
• Divide by the condition, not by everything
  → Given that: count inside the condition

Formulas (5)

• One die: six equally likely faces · One die
  $$P = \dfrac{\text{faces that fit}}{6}$$
• Sums on two dice · Ways to make a sum
  $$n(\text{sum } s) = 6 - |s - 7|, \quad 2 \le s \le 12$$
• Products, differences and doublets · Pairs with difference d
  $$n(|a - b| = d) = 2(6 - d), \quad 1 \le d \le 5$$
• Die by die: multiply, and one minus none · At least one six from n dice
  $$P(\text{at least one six}) = 1 - \left(\tfrac56\right)^n$$
• Three dice: 216 ordered triples · Exactly k sixes on three dice
  $$P = {}^3C_k \left(\tfrac16\right)^k \left(\tfrac56\right)^{3 - k}$$

Watch out for (10)

• Greater than or at least
  → One die: six equally likely faces
• 2 is prime
  → One die: six equally likely faces
• Count both orders
  → Sums on two dice
• The 11 sums are not equally likely
  → Sums on two dice
• A square product counts once
  → Products, differences and doublets
• Differ by means either order
  → Products, differences and doublets
• Do not add for at least one
  → Die by die: multiply, and one minus none
• Exactly one has two orders
  → Die by die: multiply, and one minus none
• Choose which dice
  → Three dice: 216 ordered triples
• A sum of 4 has three triples
  → Three dice: 216 ordered triples

Formulas (5)

• Ranges of values, and two conditions on one card · Cards in some ranks
  $$P = \dfrac{4 \times \text{ranks that fit}}{52}$$
• Or: add, then take off the overlap · Or with an overlap
  $$P(A \text{ or } B) = \dfrac{n(A) + n(B) - n(A \text{ and } B)}{52}$$
• Given that, or a deck with cards removed · Given that
  $$P(A \mid B) = \dfrac{n(A \cap B)}{n(B)}$$
• Two draws with replacement · With replacement
  $$P(A \text{ then } B) = P(A) \times P(B)$$
• Two or three cards without replacement · Without replacement
  $$P(A \text{ then } B) = P(A) \times P(B \mid A)$$

Reference tables (1)

Watch out for (10)

• Twelve face cards, not sixteen
  → The 52-card deck
• Number cards are 2 to 10
  → The 52-card deck
• Ace high or Ace low
  → Ranges of values, and two conditions on one card
• Inclusive ranges
  → Ranges of values, and two conditions on one card
• The overlap is counted twice
  → Or: add, then take off the overlap
• Groups that cannot overlap
  → Or: add, then take off the overlap
• Removed cards change both counts
  → Given that, or a deck with cards removed
• Replacement keeps the totals
  → Two draws with replacement
• Same suit as the first card
  → Two or three cards without replacement
• Shrink both counts
  → Two or three cards without replacement

Formulas (3)

• One ball: favourable over total · One ball
  $$P(\text{colour}) = \dfrac{\text{balls of that colour}}{\text{all balls}}$$
• One after another, without replacement · Without replacement
  $$P(A \text{ then } B) = \dfrac{a}{n} \times \dfrac{b}{n - 1}$$
• With replacement: the bag resets · j of k draws, with replacement
  $$P(j \text{ of } k) = {}^kC_j\, p^j (1 - p)^{k - j}$$

Watch out for (5)

• The total is every ball
  → One ball: favourable over total
• Neither is not either
  → One ball: favourable over total
• Change both numbers
  → One after another, without replacement
• Order is part of the event
  → One after another, without replacement
• With replacement nothing shrinks
  → With replacement: the bag resets

Formulas (4)

• A given mix of colours: choose from each colour and multiply · A mix of colours
  $$P = \dfrac{{}^aC_x \times {}^bC_y}{{}^nC_r}$$
• Same colour: add over every colour · Same colour
  $$P(\text{same}) = \dfrac{{}^aC_r + {}^bC_r + {}^cC_r}{{}^nC_r}$$
• Different colours: one minus same, or one of each · Different colours
  $$P(\text{different}) = 1 - P(\text{same})$$
• At least: add the exact cases, or take away the rest · At least one of a colour
  $$P(\text{at least one}) = 1 - \dfrac{{}^{n - a}C_r}{{}^nC_r}$$

Watch out for (7)

• Every ball is in the total
  → A given mix of colours: choose from each colour and multiply
• Exactly two red needs the rest non-red
  → A given mix of colours: choose from each colour and multiply
• Do not stop at one colour
  → Same colour: add over every colour
• One of each is a product, not a sum
  → Different colours: one minus same, or one of each
• More colours than balls
  → Different colours: one minus same, or one of each
• At least includes the top case
  → At least: add the exact cases, or take away the rest
• The opposite of at least one
  → At least: add the exact cases, or take away the rest