Complementary events

The complement of an event $E$ is the event "$E$ does not occur". It is denoted $\bar E$ (or $E'$ or $E^c$).

For example:

• If $E$ = "rolling a $6$", then $\bar E$ = "rolling something other than $6$".
• If $E$ = "drawing a king", then $\bar E$ = "drawing a non-king".
• If $E$ = "rain tomorrow", then $\bar E$ = "no rain tomorrow".

The complement rule

The fundamental property is: $P(E) + P(\bar E) = 1$, or equivalently,

$\boxed{P(\bar E) = 1 - P(E).}$

This is because the events $E$ and $\bar E$ together cover all possible outcomes (the entire sample space), and they don't overlap. Their probabilities sum to $1$.

Why this matters

The complement rule is the single most useful shortcut in probability. For events that are hard to count directly but easy to count not happening, it's a huge time-saver.

Example. "What is the probability of rolling at least one $6$ in three rolls of a die?"

Direct count is tedious: enumerate cases with exactly one $6$, exactly two $6$s, exactly three $6$s.

Complement: $P($no $6$ in any roll$) = (5/6)^3 = 125/216$. So $P($at least one 6$$) = 1 - 125/216 = 91/216.

(Technically the $(5/6)^3$ step uses multiplication of independent probabilities, which is beyond Class X , but the reasoning via the complement is standard at this level.)

Standard examples

Worked examples

Example 1. A die is rolled. Find $P($the number is not greater than 4$$).

$P($greater than 4$$) = 2/6 = 1/3. So $P($not$) = 1 - 1/3 = 2/3$.

(Or directly: $\{1, 2, 3, 4\}$ has $4$ outcomes, $4/6 = 2/3$.)

Example 2. A card is drawn from a deck. Find $P($it is neither a king nor a queen$)$.

$P($king or queen$) = (4 + 4)/52 = 8/52 = 2/13$. $P($neither$) = 1 - 2/13 = 11/13$.

Example 3. A bag contains $10$ red, $15$ green, and $5$ blue balls. One is drawn. Find $P($neither red nor blue$)$.

Total: $30$. $P($red or blue$) = 15/30 = 1/2$. $P($neither$) = 1/2$.

(Or: $P($green$) = 15/30 = 1/2$.)

Example 4. Two coins are tossed. Find $P($at least one tail$)$.

$P($no tail$) = P($two heads$) = 1/4$. So $P($at least one tail$) = 1 - 1/4 = 3/4$.

Example 5. A die is rolled. The probability of getting an even number is $1/2$. What is the probability of getting an odd number?

By complement: $1 - 1/2 = 1/2$.

Example 6. Probability of an event is $0.4$. Find $P($complement$)$.

$1 - 0.4 = 0.6$.

Example 7. $P(\bar E) = 5/8$. Find $P(E)$.

$P(E) = 1 - 5/8 = 3/8$.

Example 8. A box has $5$ pink balls and $x$ green balls. The probability of drawing a green ball is $3/4$. Find $x$.

$P($green$) = x/(x + 5) = 3/4 \Rightarrow 4x = 3x + 15 \Rightarrow x = 15$.

Example 9. A box has $24$ balls, some red and some blue. The probability of drawing a red ball is twice the probability of drawing a blue ball. Find the number of red balls.

Let red $= r$. Then $P($red$) = r/24$, $P($blue$) = (24 - r)/24$. Given $r = 2(24 - r) \Rightarrow r = 48 - 2r \Rightarrow 3r = 48 \Rightarrow r = 16$.

Example 10. A die is rolled twice. Find $P($at least one 1$$) using complement.

$P($no $1$ on either$) = (5/6)(5/6) = 25/36$. $P($at least one 1$$) = 1 - 25/36 = 11/36.

(Direct enumeration also works: $11$ pairs have at least one $1$ , verify.)

Try it yourself

1. A coin is tossed. $P($not head$) = ?$
2. A die is rolled. $P($not a multiple of 3$$) = ?
3. A card is drawn. $P($not a heart$) = ?$
4. A bag has $4$ red and $6$ green balls. $P($not red$) = ?$
5. $P(E) = 0.3$. Find $P(\bar E)$.
6. $P(\bar E) = 0.45$. Find $P(E)$.
7. Two coins are tossed. $P($at least one head$) = ?$ (use complement)
8. A card is drawn. $P($neither king nor queen$) = ?$
9. A die is rolled. $P($neither $1$ nor 6$$) = ?
10. A bag has $5$ red, $4$ blue, $1$ yellow ball. $P($not blue$) = ?$
11. A card is drawn. $P($not a face card$) = ?$
12. A box has $20$ cards numbered $1$ to $20$. $P($not divisible by 4$$) = ?

Pitfalls / Insight

(1) $P(\bar E) = 1 - P(E)$ is a complementary relation, not a "subtraction" , both probabilities are between $0$ and $1$.

(2) "Neither A nor B" is the complement of "A or B". To find it, compute $P(A \cup B)$ using inclusion-exclusion, then subtract from $1$.

(3) "At least one X" is usually easier through the complement "no X".

(4) Some students try to compute $P($not A$)$ as $-P(A)$. This is wrong. Probability is non-negative; the correct formula is $1 - P(A)$.

(5) When solving for an unknown (like "find $x$"), set up the probability equation directly and solve. Don't guess.