If $P(B) = \dfrac{3}{5}$, $P(A \mid B) = \dfrac{1}{2}$ and $P(A \cup B) = \dfrac{4}{5}$, then $P((A \cup B)') + P(A' \cup B)$ equals (A) $\dfrac{1}{5}$ (B) $\dfrac{4}{5}$ (C) $\dfrac{1}{2}$ (D) $1$

If P(B) = (3)/(5), P(A mid B) = (1)/(2) and P(A ∪ B) = (4)/(5), then P((A ∪ B)') + P(A' ∪ B) equals (A) (1)/(5

NCERT Exemplar · Q40

P(B) = \dfrac{3}{5}

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Appeared in past exams:COMEDK 2023· Set 2023-E· 1mexact

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$P((A\cup B)')=\tfrac15$ and $P(A'\cup B)=\tfrac45$ , so their sum is $1$ — option (D).

Setup

We are given $P(B)=\tfrac35,\ P(A\mid B)=\tfrac12,\ P(A\cup B)=\tfrac45$ . First recover $P(A\cap B)$ and $P(A)$ :

P(A\cap B)=P(A\mid B)\,P(B)=\tfrac12\cdot\tfrac35=\tfrac{3}{10},

P(A)=P(A\cup B)-P(B)+P(A\cap B)=\tfrac45-\tfrac35+\tfrac{3}{10}=\tfrac12.

First term

By the complement rule, $P((A\cup B)')=1-P(A\cup B)=1-\tfrac45=\tfrac15.$

Second term

Use De Morgan's law: $(A'\cup B)'=A\cap B'$ , so $P(A'\cup B)=1-P(A\cap B').$

Since $A$ splits into the parts inside and outside $B$ , $P(A\cap B')=P(A)-P(A\cap B)=\tfrac12-\tfrac{3}{10}=\tfrac15.$ …

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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. …
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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,
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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. $\binom{6}{1} = 6 \quad \text{(namely } \{1\},\{2\},\{3\},\{4\},\{5\},\{6\}\text{)}$ …
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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 …
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←Previous questionQ39 — In Exercise 64 above, P(B mid A') is equal to (A) (1)/(5) (B) (3)/(10)…→Next questionQ41 — Let P(A) = (7)/(13), P(B) = (9)/(13) and P(A ∩ B) = (4)/(13). Then P(A…

Q.If $P(B) = \dfrac{3}{5}$ , $P(A \mid B) = \dfrac{1}{2}$ and $P(A \cup B) = \dfrac{4}{5}$ , then $P((A \cup B)') + P(A' \cup B)$ equals
(A) $\dfrac{1}{5}$
(B) $\dfrac{4}{5}$
(C) $\dfrac{1}{2}$
(D) $1$

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

Setup

First term

Second term

Method: Complement rule and De Morgan for compound expressions

Steps

Common Mistakes

Step 1 — Find the complement probabilities

Step 2 — Multiply the independent probabilities …

Setup

First term

Second term

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