The probability that at least one of the two events A and B occurs is 0.6. If A and B occur simultaneously wit

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Concept understanding — Probability Complement Rule

The Probability Complement Rule

Every event $A$ splits the sample space in two: outcomes where $A$ happens, and outcomes where it does not. The second group is the complement of $A$ , written $A'$ (also $A^c$ or $\bar{A}$ ). Because exactly one of the two must occur, their probabilities together fill the whole space:

P(A) + P(A') = 1 \quad\Longrightarrow\quad P(A') = 1 - P(A).

That is the complement rule: the probability that $A$ does not happen is $1$ minus the probability that it does.

Why It Holds

$A$ and $A'$ are mutually exclusive (no outcome lies in both) and exhaustive (together they are the entire sample space $S$ , with $P(S) = 1$ ). So $P(A) + P(A') = P(S) = 1$ , and rearranging gives the rule.

A Simple Example

For a fair die, $P(\text{six}) = \tfrac{1}{6}$ , so $P(\text{not six}) = 1 - \tfrac{1}{6} = \tfrac{5}{6}$ .

Why It Is So Useful: the "At Least One" Trick

Counting "at least one" directly often means adding many separate cases, while its complement, "none," is a single easy case. For instance, the probability of at least one head in three tosses of a fair coin:

P(\text{at least one head}) = 1 - P(\text{no heads}) = 1 - \left(\tfrac{1}{2}\right)^3 = 1 - \tfrac{1}{8} = \tfrac{7}{8}.

Computing "no heads" once is far quicker than summing the one-head, two-head and three-head cases separately.

Tip

Whenever a question asks for the probability of "at least one," pause and try $1 - P(\text{none})$ first — it usually turns a long sum into a one-line calculation.

Note

The rule combines with others too: $P(A' \cap B') = 1 - P(A \cup B)$ , which is how De Morgan's laws appear in probability.

The complement rule and its "at least one" shortcut are staples of the NCERT Class 12 Probability chapter, tested constantly in CBSE boards, JEE Main and state CETs wherever a question asks for P(at least one). Students searching "probability of at least one event formula" will find this trick turns some of the hardest-looking probability questions into one-line calculations.

The key idea is to use the complement rule: $P(A') + P(B') = 2 - [P(A) + P(B)]$ . From the given data, $P(A \cup B) = 0.6$ and $P(A \cap B) = 0.3$ , so $P(A) + P(B) = P(A \cup B) + P(A \cap B) = 0.9$ . Thus $P(A') + P(B') = 2 - 0.9 = 1.1$ .

The problem asks for $P(A') + P(B')$ , the sum of the probabilities of the complements of two events. A direct approach would require knowing $P(A)$ and $P(B)$ individually, but we are not given those. Instead, we are given two pieces of information:

$P(A \cup B) = 0.6$ — the probability that at least one occurs.
$P(A \cap B) = 0.3$ — the probability that both occur simultaneously.

The complement rule tells us that $P(A') = 1 - P(A)$ and $P(B') = 1 - P(B)$ . So:

P(A') + P(B') = (1 - P(A)) + (1 - P(B)) = 2 - [P(A) + P(B)].

The problem reduces to finding $P(A) + P(B)$ from the given union and intersection. This is where the addition rule of probability comes in.

A

Rearranging:

P(A) + P(B) = P(A \cup B) + P(A \cap B).

Now substitute the given values:

$P(A \cup B) = 0.6$
$P(A \cap B) = 0.3$

So:

P(A) + P(B) = 0.6 + 0.3 = 0.9.

Therefore:

P(A') + P(B') = 2 - 0.9 = 1.1.

Watch out

A common mistake is to think $P(A') + P(B') = 1 - P(A \cup B)$ or something similar. But complements don't combine that way — you must go through $P(A) + P(B)$ .

Tip

Notice that we never needed $P(A)$ or $P(B)$ individually. The sum $P(A) + P(B)$ was enough. This is a neat trick: whenever you see $P(A') + P(B')$ , think $2 - [P(A) + P(B)]$ , and use the addition rule to get the sum.

✓Final answer

P(A') + P(B')

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

KCET 2021Set A-11 markMCQ
$P(A) = 0.59$
›Reveal solutionSolution
Recognise $A' \cap B'$ as the complement of $A \cup B$ (De Morgan), find $P(A\cup B)$ by the addition rule, and subtract from 1.

Step 1 — Rewrite the event using De Morgan's law.

De Morgan's law for sets says
$A' \cap B' = (A \cup B)'$
Why: "neither $A$ nor $B$ occurs" is the same event as "it is not the case that at least one of $A$ , $B$ occurs."

Therefore, by the complement rule,
$P(A' \cap B') = P\big((A \cup B)'\big) = 1 - P(A \cup B)$
Step 2 — Find $P(A \cup B)$ using the addition theorem.

The general addition rule (inclusion–exclusion) is
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
The subtraction of $P(A \cap B)$ is essential — the overlap would otherwise be counted twice. (Note we may not use $P(A)+P(B)$ alone, since $P(A\cap B) = 0.21 \ne 0$ , so the events are not mutually exclusive.)

Substituting the given values:
$P(A \cup B) = 0.59 + 0.30 - 0.21 = 0.89 - 0.21 = 0.68$
Step 3 — Complete the calculation.
$P(A' \cap B') = 1 - 0.68 = \mathbf{0.32}$
Step 4 — Sanity check.

Split the sample space into four disjoint pieces:

only $A$ : $0.59 - 0.21 = 0.38$

only $B$ : $0.30 - 0.21 = 0.09$

both: $0.21$

neither ( $A' \cap B'$ ): $0.32$

Sum $= 0.38 + 0.09 + 0.21 + 0.32 = 1.00$ ✓ — everything is consistent, and all four pieces are non-negative, so the data are valid.

(Distractor note: option (B) $0.38$ is the probability of " $A$ only", i.e. $P(A \cap B')$ — a natural slip.)

✓Final answer
$The correct option is (C) — 0.32.$

ANSWER: C
COMEDK 2025Set 2025-M1 markMCQ
$A$
›Reveal solutionSolution
The problem states that one of A or B must occur, so P(A) + P(B) = 1. Given P(A) = (2/3)P(B), we solve to find P(B) = 3/5, so odds in favour of B are P(B) : P(not B) = 3/5 : 2/5 = 3:2. The correct option is (C).

Concept and intuition

When we say “one of the two events must occur,” it means the events are mutually exclusive and exhaustive — together they cover all possibilities. That gives us the key equation: P(A) + P(B) = 1. The problem also gives a relationship between their probabilities: P(A) is two-thirds of P(B). This is a classic system of two equations in two unknowns. Once we find P(B), the odds in favour of B are simply the ratio of the probability that B happens to the probability that it does not happen (i.e., P(B) : P(not B)).

Step-by-step solution

Set up the equations from the problem statement. Since one of A or B must occur, they are complementary in the sense that their probabilities sum to 1:

$P(A) + P(B) = 1.$
Also, the chance of A is $\frac{2}{3}$ the chance of B:
$P(A) = \frac{2}{3} P(B).$

Substitute to solve for P(B). Replace $P(A)$ in the first equation:

$\frac{2}{3} P(B) + P(B) = 1.$
Combine like terms:
$\frac{5}{3} P(B) = 1.$
Multiply both sides by $\frac{3}{5}$ :
$P(B) = \frac{3}{5}.$

Find the probability that B does not occur. Since B either happens or doesn’t,

$P(\text{not } B) = 1 - P(B) = 1 - \frac{3}{5} = \frac{2}{5}.$

Express odds in favour of B. Odds in favour of an event are the ratio of the probability it occurs to the probability it does not occur:

$\text{Odds in favour of } B = P(B) : P(\text{not } B) = \frac{3}{5} : \frac{2}{5}.$
Multiply both sides by 5 to clear denominators:
$3 : 2.$
Tip

A common mistake is to confuse “odds in favour” with “probability.” Probability is a fraction between 0 and 1; odds are a ratio of two probabilities. Here, probability of B is 3/5, but odds are 3:2.

✓Final answer
$The correct option is (C). ANSWER: C$
KCET 2020Set A-11 markMCQ
$A = \{1, 2, 3, 4, 5, 6\}$
›Reveal solutionSolution
Total subsets $2^6 = 64$ , minus the $1$ empty subset and the $6$ single-element subsets, gives $57$ .

Step 1 — Total number of subsets.

For a set with $n$ elements, each element is independently either in or out of a subset — two choices each — so the number of subsets is $2^n$ . Here $n = |A| = 6$ :
$\text{Total subsets} = 2^6 = 64$
Step 2 — Use the complement (much faster than adding up).

"At least two elements" is the complement of "fewer than two elements", i.e. of subsets with 0 elements or 1 element:
$N(\ge 2) = \text{Total} - N(0) - N(1)$
Step 3 — Count the unwanted subsets.

Subsets with 0 elements: only the empty set $\varnothing$ .

$\binom{6}{0} = 1$

Subsets with 1 element: choose 1 of the 6 elements.

$\binom{6}{1} = 6 \quad \text{(namely } \{1\},\{2\},\{3\},\{4\},\{5\},\{6\}\text{)}$
Step 4 — Subtract.
$N(\ge 2) = 64 - 1 - 6 = 57$
Step 5 — Cross-check by direct summation.
$\sum_{r=2}^{6}\binom{6}{r} = \binom62 + \binom63 + \binom64 + \binom65 + \binom66 = 15 + 20 + 15 + 6 + 1 = 57 \;✓$
Both routes agree. (Option (B) 63 is the trap of subtracting only the empty set; option (A) 64 is the raw total.)

✓Final answer
$The correct option is (C) — 57.$

ANSWER: C
KCET 2026Set UNKNOWN1 markMCQ
$\frac{1}{4}$
›Reveal solutionSolution

Find the complement (non-survival) probability for each person, then multiply them since the two events are independent.

Step 1 — Find the complement probabilities

Given $P(\text{man lives}) = \frac{1}{4}$ and $P(\text{wife lives}) = \frac{1}{3}$ , the probabilities that each does not live after 20 years are:
$P(\text{man does not live}) = 1 - \frac{1}{4} = \frac{3}{4},$ $P(\text{wife does not live}) = 1 - \frac{1}{3} = \frac{2}{3}.$
Step 2 — Multiply the independent probabilities

"Neither the man nor his wife live" means both non-survival events happen together. Treating the two lifespans as independent:
$P(\text{neither lives}) = \frac{3}{4}\times\frac{2}{3} = \frac{6}{12} = \frac{1}{2}.$
✓Final answer
$\frac{1}{2}$

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Q.The probability that at least one of the two events $A$ and $B$ occurs is $0.6$ . If $A$ and $B$ occur simultaneously with probability $0.3$ , evaluate $P(A') + P(B')$ .

Concept understanding — Probability Complement Rule

The Probability Complement Rule

Why It Holds

A Simple Example

Why It Is So Useful: the "At Least One" Trick

Method: Relating Complement Sums to the Addition Rule

Steps

Common Mistakes

Step 1 — Find the complement probabilities

Step 2 — Multiply the independent probabilities

Q.The probability that at least one of the two events AAA and BBB occurs is 0.60.60.6. If AAA and BBB occur simultaneously with probability 0.30.30.3, evaluate P(A′)+P(B′)P(A') + P(B')P(A′)+P(B′).

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Q.The probability that at least one of the two events $A$ and $B$ occurs is $0.6$ . If $A$ and $B$ occur simultaneously with probability $0.3$ , evaluate $P(A') + P(B')$ .

Unlock everything free for 14 days