State whether the following statement is True or False: If $A$ and $B$ are two events such that $P(A) > 0$ and $P(A) + P(B) > 1$, then $P(B \mid A) \ge 1 - \dfrac{P(B')}{P(A)}$.

State whether the following statement is True or F – Class 12 Mathematics

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The statement is True. The key idea is to rewrite the conditional probability inequality using the complement rule and the given condition $P(A) + P(B) > 1$ , which ensures the inequality holds.

Why This Approach Works

The problem asks whether $P(B \mid A) \ge 1 - \dfrac{P(B')}{P(A)}$ is always true under the conditions $P(A) > 0$ and $P(A) + P(B) > 1$ .

At first glance, this looks like a conditional probability inequality that might depend on the specific events. But the complement rule gives us a powerful way to simplify: $P(B') = 1 - P(B)$ . So the right-hand side becomes $1 - \dfrac{1 - P(B)}{P(A)}$ .

The trick is to realise that $P(B \mid A) = \dfrac{P(A \cap B)}{P(A)}$ , and we can relate $P(A \cap B)$ to $P(A) + P(B) - 1$ using the inclusion-exclusion principle. The condition $P(A) + P(B) > 1$ guarantees that $P(A \cap B) > 0$ , which is crucial.

Let's work through it step by step.

Step-by-Step Solution

1. Write the target inequality in terms of $P(A \cap B)$ .

We know $P(B \mid A) = \dfrac{P(A \cap B)}{P(A)}$ . The inequality becomes:

\frac{P(A \cap B)}{P(A)} \ge 1 - \frac{P(B')}{P(A)}

Multiply both sides by $P(A) > 0$ (so the inequality direction stays the same):

P(A \cap B) \ge P(A) - P(B')

2. Replace $P(B')$ using the complement rule.

Since $P(B') = 1 - P(B)$ , we get:

P(A \cap B) \ge P(A) - (1 - P(B)) = P(A) + P(B) - 1

So the inequality we need to prove is:

P(A \cap B) \ge P(A) + P(B) - 1

3. Recognise this as a known inequality from inclusion-exclusion.

The inclusion-exclusion principle for two events states:

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

Since $P(A \cup B) \le 1$ (probabilities cannot exceed 1), we have:

P(A) + P(B) - P(A \cap B) \le 1

Rearranging:

P(A \cap B) \ge P(A) + P(B) - 1

This is exactly the inequality we need! It holds for any two events $A$ and $B$ , regardless of the given conditions.

Note

The inequality $P(A \cap B) \ge P(A) + P(B) - 1$ is always true — it's a direct consequence of $P(A \cup B) \le 1$ . No extra conditions are needed for this step.

4. Check the role of the given conditions.

$P(A) > 0$ : This is necessary so that $P(B \mid A)$ is defined (we can't divide by zero). …

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

Showing the 12 most recent of 18 on this concept.

AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
$P(\bar{A} \cup \bar{B}) = 1 - P(A)P\left(\dfrac{B}{A}\right)$
›Reveal solutionSolution
De Morgan's law gives $P(\bar A\cup\bar B)=1-P(A\cap B)$ , and since $P(A\cap B)=P(A)P(B|A)$ always (definition of conditional probability), option (A) is the universally true statement.

Concept and Intuition

$\bar A\cup\bar B$ is the complement of $A\cap B$ (De Morgan's law: $\overline{A\cap B}=\bar A\cup\bar B$ ). So $P(\bar A\cup\bar B)=1-P(A\cap B)$ always holds, regardless of independence. The multiplication rule $P(A\cap B)=P(A)P(B|A)$ is also always true by definition of conditional probability.

Step-by-Step Solution

By De Morgan's law: $\bar A\cup\bar B = \overline{A\cap B}$ , so $P(\bar A\cup\bar B) = 1-P(A\cap B)$ .

By definition of conditional probability, $P(B|A)=\frac{P(A\cap B)}{P(A)}$ , so $P(A\cap B)=P(A)P(B|A)$ .

Substituting: $P(\bar A\cup\bar B)=1-P(A)P(B/A)$ — this is option (A), and it holds for ANY events A, B. …
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AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
$Q. Two candidates A and B have attended an interview conducted by a recruitment board for two jobs. If the probability that candidate A will get the job is 0.8 and the probability that candidate B will get the job is 0.7, then the probability that atleast one of them will get the job is (A) 0.96 (B) 0.94 (C) 0.92 (D) 0.9$
›Reveal solutionSolution
This tests the complement rule for "at least one" events with two independent trials. Answer: 0.94.

Concept and Intuition

When two events are independent, the easiest way to find $P(\text{at least one occurs})$ is to compute the probability that neither occurs (multiply the individual "failure" probabilities) and subtract from 1.

Step-by-Step Solution

$P(A\text{ gets job})=0.8\Rightarrow P(A\text{ doesn't})=0.2$ .

$P(B\text{ gets job})=0.7\Rightarrow P(B\text{ doesn't})=0.3$ .

Assuming independence, $P(\text{neither gets the job})=0.2\times0.3=0.06$ .

$P(\text{at least one gets the job})=1-0.06=0.94$ .

Common Mistakes …
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AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
$\dfrac{2}{9}$
›Reveal solutionSolution
This tests the addition rule for probability and the complement rule. The answer is $\dfrac{7}{18}$ .

Concept and Intuition

"Neither" is the complement of "at least one", i.e. $1-P(A\cup B)$ . First find $P(A\cup B)$ using the inclusion-exclusion formula $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ , then subtract from 1.

Step-by-Step Solution

$P(A)=\dfrac29$ (road), $P(B)=\dfrac59$ (building), $P(A\cap B)=\dfrac16$ (both).

$P(A\cup B) = \dfrac29+\dfrac59-\dfrac16 = \dfrac79-\dfrac16$ .

Common denominator 18: $\dfrac79=\dfrac{14}{18}$ , $\dfrac16=\dfrac{3}{18}$ , so $P(A\cup B)=\dfrac{14-3}{18}=\dfrac{11}{18}$ . …
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AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
$\dfrac{1}{4}$
›Reveal solutionSolution
This tests the complement rule for "at least one" probability with two independent events. The probability at least one student qualifies is $\dfrac{11}{20}$ .

Concept and Intuition

When asked for $P(\text{at least one of several independent events occurs})$ , it is almost always easiest to compute the complement — the probability that none occur — and subtract from 1, since "none occur" for independent events is just the product of each event's complement probability.

Step-by-Step Solution

$P(\text{student 1 qualifies})=\tfrac14\Rightarrow P(\text{student 1 fails})=\tfrac34$ .

$P(\text{student 2 qualifies})=\tfrac25\Rightarrow P(\text{student 2 fails})=\tfrac35$ .

Assuming independence, $P(\text{neither qualifies})=\tfrac34\times\tfrac35=\tfrac{9}{20}$ . …
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AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
$\dfrac{3}{20}$
›Reveal solutionSolution
"Contradiction" means exactly one of A, B tells the truth; summing the two mutually exclusive cases gives $\dfrac7{20}$ .

Concept and Intuition

A and B "contradict" each other exactly when one speaks truth and the other lies (if both tell the truth or both lie, they'd agree, not contradict). These are two independent, mutually exclusive scenarios whose probabilities add.

Step-by-Step Solution

$P(A\text{ true})=0.75$ , so $P(A\text{ false})=0.25$ .

$P(B\text{ true})=0.80$ , so $P(B\text{ false})=0.20$ .

Contradiction case 1: A true, B false: $0.75\times0.20=0.15$ .

Contradiction case 2: A false, B true: $0.25\times0.80=0.20$ . …
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AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
$P(A \cap \bar{B}) =$
›Reveal solutionSolution
Split 1–100 by 2-adic valuation (odd / $\equiv2\bmod4$ / divisible by 4); the event 'even but not divisible by 4' happens only for an odd– $(\equiv2\bmod4)$ pair. Probability is $25/99$ .

Concept and Intuition

Whether a product of two numbers is divisible by 4 depends on the total power of 2 across both factors. Splitting the range into three classes by how many factors of 2 each number contributes (0, exactly 1, or $\ge2$ ) turns this into simple counting.

Step-by-Step Solution

Among 1–100: odd numbers (0 factors of 2) — 50 of them. Numbers $\equiv2\pmod4$ (exactly 1 factor of 2, e.g. 2,6,10,...,98) — 25 of them. Numbers divisible by 4 (at least 2 factors of 2) — 25 of them.

Two distinct numbers are chosen (without replacement) from 1–100; total ways $=\binom{100}{2}=4950$ .

Event $A\cap\bar B$ = product is even AND not divisible by 4, i.e. the total power of 2 across the two numbers is exactly 1.

This happens only when one number is odd (0 power) and the other is $\equiv2\pmod4$ (exactly 1 power) — any other combination gives total power 0 (both odd, not in $A$ ) or $\ge2$ (one or both contribute $\ge1$ in a way that sums to $\ge2$ , landing in $B$ ). …
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AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
$(0.9)^9$
›Reveal solutionSolution
This is a binomial "exactly one success" computation; the factor of $10\times0.1$ conveniently equals $1$ , leaving a clean power of $0.9$ .

Concept and Intuition

With $X\sim B(10,0.1)$ counting left-handed people in the group, $P(X=1)=\binom{10}{1}(0.1)(0.9)^9$ . The combinatorial factor $10$ exactly cancels the $0.1$ , collapsing the whole expression to a pure power of $0.9$ — which is exactly the pattern the answer choices are built around.

Step-by-Step Solution

$X\sim B(n=10,p=0.1)$ .

$P(X=1)=\binom{10}{1}\,(0.1)^1\,(0.9)^{9}$ . …
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AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
$1/3$
›Reveal solutionSolution
"Neither red nor green" simply means "blue"; the probability is just the blue count over the total count.

Concept and Intuition

For a simple random draw from a finite set with equally likely outcomes, the probability of an event equals the count of favourable outcomes divided by the total count.

Step-by-Step Solution

Total balls $= 8\ (\text{red}) + 7\ (\text{blue}) + 6\ (\text{green}) = 21$ .

The complement of "red or green" among these three categories is simply "blue" (since every ball is one of the three colours).

Number of blue balls $=7$ .

$P(\text{blue}) = \dfrac{7}{21} = \dfrac13$ .

Common Mistakes …
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AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
$1-(1.02)e^{0.2}$
›Reveal solutionSolution
With Poisson mean $\lambda=0.2$ per day, $P(X\ge3)$ is found as the complement of $P(0)+P(1)+P(2)$ , giving $1-1.22\,e^{-0.2}$ .

Concept and Intuition

For a Poisson-distributed random variable with mean $\lambda$ , $P(X=k)=\dfrac{e^{-\lambda}\lambda^k}{k!}$ . Here the average accidents per day is estimated from the given data: 10 accidents over 50 days gives $\lambda = \dfrac{10}{50}=0.2$ accidents/day. "3 or more" is most easily computed as $1$ minus the sum of the (easy, low- $k$ ) probabilities for 0, 1, and 2 accidents.

Step-by-Step Solution

$\lambda = \dfrac{10\text{ accidents}}{50\text{ days}} = 0.2$ per day.

$P(X=0) = e^{-0.2}$ .

$P(X=1) = 0.2\,e^{-0.2}$ .

$P(X=2) = \dfrac{(0.2)^2}{2}e^{-0.2} = 0.02\,e^{-0.2}$ .

Sum: $P(0)+P(1)+P(2) = e^{-0.2}(1+0.2+0.02) = 1.22\,e^{-0.2}$ . …
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AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
$U_1, U_2, U_3$
›Reveal solutionSolution
Use the Law of Total Probability, averaging the "not black" probability over the three equally likely urns. Answer: $\dfrac{23}{30}$ .

Concept and Intuition

This is a two-stage random experiment: first an urn is picked (uniformly at random among the three), then a ball is drawn from that urn. The overall probability of an event is the weighted average of its probability conditional on each urn, weighted by the probability of picking that urn — this is exactly the Law of Total Probability.

Step-by-Step Solution

Each urn is equally likely to be chosen: $P(U_1)=P(U_2)=P(U_3)=\dfrac13$ .

Totals per urn: $U_1$ has $5+3+2=10$ balls, $U_2$ has $4+4+2=10$ balls, $U_3$ has $3+4+3=10$ balls.

Probability of NOT drawing black from each urn:

$P(\text{not black}\mid U_1)=\frac{5+3}{10}=\frac{8}{10}$ $P(\text{not black}\mid U_2)=\frac{4+4}{10}=\frac{8}{10}$ $P(\text{not black}\mid U_3)=\frac{3+4}{10}=\frac{7}{10}$

By the Law of Total Probability: …
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AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
$\dfrac{8}{35}$
›Reveal solutionSolution
This tests complementary counting for "at least one" events. Answer: $27/35$ .

Concept and Intuition

"At least one correct pair" is hard to count directly (could be exactly 1 pair, or 2 pairs), but its complement — "no correct pair at all" — is very restrictive and easy to count. With exactly 4 pairs and 4 shoes drawn, avoiding any complete pair forces one shoe from every single pair (there's no pair left over to skip).

Step-by-Step Solution

Total ways to choose 4 shoes out of 8: $\binom{8}{4} = 70$ .

Count the complement: no pair among the 4 drawn is a matched (left+right) pair. Since there are exactly 4 pairs and we draw exactly 4 shoes, "no complete pair" means each of the 4 pairs contributes exactly one shoe (if any pair contributed 0 shoes, another would have to contribute 2, forming a complete pair — contradiction).

For each of the 4 pairs, there are 2 choices (which shoe of that pair to include), giving $2^4 = 16$ ways with no complete pair. …
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AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
$\dfrac{1}{5}$
›Reveal solutionSolution
Complement of "all letters correctly placed" (only 1 of the 120 arrangements) is the required event.

Concept and Intuition

"At least one wrongly placed" is the complement of "every letter in its own envelope" — and there's exactly one arrangement (the identity permutation) out of $5!$ equally likely arrangements where all are correct.

Step-by-Step Solution

Total arrangements of 5 letters into 5 envelopes $=5!=120$ .

Exactly one arrangement has every letter correctly placed.

$P(\text{all correct}) = \dfrac{1}{120}$ . …
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Q.State whether the following statement is True or False: If $A$ and $B$ are two events such that $P(A) > 0$ and $P(A) + P(B) > 1$ , then $P(B \mid A) \ge 1 - \dfrac{P(B')}{P(A)}$ .

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

Why This Approach Works

Step-by-Step Solution

Method: Verifying a conditional-probability inequality

Steps

Common Mistakes

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

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Concept and Intuition

Step-by-Step Solution

Q.State whether the following statement is True or False: If AAA and BBB are two events such that P(A)>0P(A) > 0P(A)>0 and P(A)+P(B)>1P(A) + P(B) > 1P(A)+P(B)>1, then P(B∣A)≥1−P(B′)P(A)P(B \mid A) \ge 1 - \dfrac{P(B')}{P(A)}P(B∣A)≥1−P(A)P(B′)​.

Why This Approach Works

Step-by-Step Solution

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Q.State whether the following statement is True or False: If $A$ and $B$ are two events such that $P(A) > 0$ and $P(A) + P(B) > 1$ , then $P(B \mid A) \ge 1 - \dfrac{P(B')}{P(A)}$ .

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