Year 10 Probability and counting: revision mind map

1Review of probability

$P(A) = \dfrac{\text{favourable}}{\text{total}}$ and $P(A') = 1 - P(A)$.

52 cards: $P(\text{heart}) = \tfrac{13}{52} = \tfrac14$

$P(\text{not an ace}) = \tfrac{48}{52} = \tfrac{12}{13}$

2Venn diagrams and two-way tables

$A \cap B$: both. $A \cup B$: either. $A'$: not A.

30 people: $n(A \cup B) = 27$, $P(A') = \tfrac{12}{30} = \tfrac25$

3The addition rule

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

$0.5 + 0.4 - 0.2 = 0.7$

Mutually exclusive: $P(A \cap B) = 0$

4Conditional probability

$P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$: only look inside B.

From the Venn in card 2: $P(A \mid B) = \tfrac{6}{15} = \tfrac25$

5Arrays for two steps

List every outcome in a table.

Two dice: 36 outcomes. $P(\text{double}) = \tfrac{6}{36} = \tfrac16$

$P(\text{sum} \ge 10) = \tfrac{6}{36} = \tfrac16$

6Tree diagrams

Multiply along the branches, add the branches you want.

2 red, 3 blue, take 2 without replacement:

$P(\text{one of each}) = \tfrac25 \cdot \tfrac34 + \tfrac35 \cdot \tfrac24 = \tfrac35$

7Independent events

Independent if $P(A \cap B) = P(A) \times P(B)$, so $P(A \mid B) = P(A)$.

Coin and die: $P(\text{H and } 6) = \tfrac12 \times \tfrac16 = \tfrac{1}{12}$

8Counting and factorials

Multiply the number of choices at each step. $n! = n \times (n - 1) \times \cdots \times 1$

3 shirts and 4 pairs of pants: 3 × 4 = 12 outfits

$5! = 120$   $0! = 1$

9Arrangements

Order matters: ${}^nP_r = \dfrac{n!}{(n - r)!}$

Arrange 3 of 5 books: $5 \times 4 \times 3 = 60$

Arrange the letters of MATHS: $5! = 120$

10Selections

Order doesn't matter: ${}^nC_r = \dfrac{n!}{r!(n - r)!}$

Choose 3 from 5: ${}^5C_3 = 10$

A pair from 6 people: ${}^6C_2 = 15$

Watch out: a team is a selection; a 1st, 2nd and 3rd place is an arrangement.

11Counting in probability

$P = \dfrac{\text{favourable selections}}{\text{total selections}}$

Pick 3 from 5 boys and 4 girls: $P(\text{all girls}) = \dfrac{{}^4C_3}{{}^9C_3} = \dfrac{4}{84} = \dfrac{1}{21}$