Probability

Probability scale. $0 \le P \le 1$.

• $P = 0$: impossible.
• $P = 1$: certain.
• $P = 0.5$: even chance.

If your calculation gives $P = 1.3$ or $P = -0.2$, something has gone wrong.

Equally likely outcomes. When all outcomes are equally likely: $P(A) = \dfrac{\text{number of favourable outcomes}}{\text{total outcomes}}$

Worked example. Bag has 4 red and 6 blue balls. $P(\text{red}) = 4/10 = 2/5$.

Complement. $P(\text{not } A) = 1 - P(A)$.

Worked example. $P(\text{not red}) = 1 - 2/5 = 3/5$. (Or directly: $6/10 = 3/5$.)

When complement is useful. "At least one" problems become easy: $P(\text{at least one}) = 1 - P(\text{none})$.

Worked example. Roll a die three times. $P(\text{at least one 6})$?

• Direct: messy (could be 1, 2, or 3 sixes).
• Complement: $P(\text{no 6}) = (5/6)^3 = 125/216$.
• $P(\text{at least one}) = 1 - 125/216 = 91/216$.

Edexcel tip. "At least one" → use complement. Saves work.

OR rule (mutually exclusive events).

Mutually exclusive: events that CANNOT happen at the same time.

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

Worked example. Roll a die. $P(\text{2 or 5}) = 1/6 + 1/6 = 1/3$.

AND rule (independent events).

Independent: outcome of one doesn't affect the other.

$P(A \cap B) = P(A) \times P(B)$

Worked example. Two coins. $P(\text{HH}) = 0.5 \times 0.5 = 0.25$.

Worked example. A bag has 4 red and 6 blue. Take a ball, replace it, take another. $P(\text{both red}) = (4/10)(4/10) = 16/100 = 4/25$.

Memory aid. "AND = times. OR = add."

Are events independent?

• Coins, dice, spinners: yes.
• Picking with replacement: yes.
• Picking WITHOUT replacement: NO — second probability changes.

Edexcel tip. Look for keywords:

• "Both" / "and": multiply.
• "Either" / "or": add (if mutually exclusive).
• "Replaced" / "with replacement": independent.
• "Not replaced" / "without replacement": NOT independent.

Without replacement. When you don't put the first item back:

• The TOTAL decreases by 1 for the second pick.
• The favourable count may also decrease (depending on what you picked first).

Worked example. Bag: 3 red, 5 blue (total 8). Pick two WITHOUT replacement. $P(\text{both red})$?

• $P(\text{1st red}) = 3/8$.
• After taking a red: 2 red, 5 blue, total 7. $P(\text{2nd red}) = 2/7$.
• $P(\text{both red}) = (3/8)(2/7) = 6/56 = 3/28$.

Note both numerator (3 → 2) AND denominator (8 → 7) decreased.

Worked example. Same bag. $P(\text{1st red, 2nd blue})$?

• $P(\text{1st red}) = 3/8$.
• After red: 5 blue out of 7. $P(\text{2nd blue}) = 5/7$.
• Product: $(3/8)(5/7) = 15/56$.

Conditional probability $P(B|A)$. "Probability of $B$ GIVEN $A$ has happened."

In the example above, $P(\text{2nd red} | \text{1st red}) = 2/7$.

Tree diagrams are the standard tool — see the Tree Diagrams subtopic.

Edexcel tip. When you see "without replacement", IMMEDIATELY note that the second-pick probability differs. The most common error is using $3/8 \times 3/8$ for both red.

Expected frequency. Expected number of successes in $n$ trials:

$E = n \times p$

Worked example. Roll a die 60 times. Expected number of sixes: $E = 60 \times 1/6 = 10$.

This is the THEORETICAL expectation — actual results vary.

Relative frequency (experimental probability). Estimate probability from data:

$\text{Relative freq.} = \dfrac{\text{number of successes}}{\text{total trials}}$

Worked example. Spinner spun 200 times, lands on red 80 times. Relative frequency of red = $80/200 = 0.4$.

When does relative frequency = theoretical probability? As trials → ∞ (Law of Large Numbers). With small samples, relative frequency is just an estimate.

Worked example. Tossed a coin 4 times, got 3 heads. Relative frequency = 0.75. Doesn't mean coin is biased — small sample.

Edexcel tip. "Estimate the probability" or "use the data" → relative frequency. "Theoretical probability" → use formula or count outcomes.