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.

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.

Check the others: (B) $1-P(A\cup B)=P(\bar A\cap\bar B)$ , not $P(\bar A\cup\bar B)$ in general — false. (C) and (D) only hold in special cases (e.g. mutual exclusivity), not generally.

Common Mistakes

Confusing $P(\bar A\cup\bar B)$ (complement of the intersection) with $P(\bar A\cap\bar B)=1-P(A\cup B)$ (complement of the union) — these are different De Morgan pairs.

Assuming independence when none is stated, which would wrongly validate option (D).

✓Final answer
$P(\bar{A} \cup \bar{B}) = 1 - P(A)P\left(\dfrac{B}{A}\right)$

ANSWER: A
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

Simply adding $0.8+0.7-0.8\times0.7$ incorrectly or making an arithmetic slip in the multiplication step.

Forgetting the independence assumption is what allows multiplying the individual "not getting the job" probabilities.

✓Final answer
$The correct option is (B) — 0.94.$

ANSWER: B
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 $P(A\cup B)=\dfrac{14-3}{18}=\dfrac{11}{18}$ .

$P(\text{neither}) = 1 - P(A\cup B) = 1-\dfrac{11}{18} = \dfrac{7}{18}$ .

Common Mistakes

Forgetting to subtract $P(A\cap B)$ (double-counting the overlap).

Arithmetic slip converting to the common denominator 18.

✓Final answer
$\dfrac{7}{18}$

ANSWER: C
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, $P(\text{neither qualifies})=\tfrac34\times\tfrac35=\tfrac{9}{20}$ .

$P(\text{at least one qualifies})=1-\tfrac{9}{20}=\tfrac{11}{20}$ .

Common Mistakes

Adding the two probabilities directly ( $\tfrac14+\tfrac25$ ) without subtracting the overlap — this overcounts the case both qualify.

Forgetting to take the complement correctly (mixing up qualify/fail probabilities).

✓Final answer
$\dfrac{11}{20}$

ANSWER: D
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

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

$P(X=1)=\binom{10}{1}\,(0.1)^1\,(0.9)^{9}$ .

$\binom{10}{1}=10$ , and $10\times0.1=1$ .

So $P(X=1)=1\times(0.9)^9=(0.9)^9$ .

Common Mistakes

Computing $P(X\ge1)=1-(0.9)^{10}$ instead, which does not match any option and is not what the answer choices are built for.

✓Final answer
$(0.9)^9$

ANSWER: A
AP EAPCET 2022Set eng-2022-07-04-FN1 markMCQ
$P(X \geq 3) =$
›Reveal solutionSolution
This tests the complement rule for a Poisson tail probability: compute the first three terms directly and subtract from 1.

Concept and Intuition

A Poisson random variable with mean $\lambda$ has $P(X=k) = \dfrac{e^{-\lambda}\lambda^k}{k!}$ . Since summing $P(X\geq 3)$ directly is an infinite sum, it is far easier to use the complement: $P(X\geq 3) = 1 - P(X\le 2) = 1 - [P(0)+P(1)+P(2)]$ .

Step-by-Step Solution

Here $\lambda = 6$ .

$P(0) = e^{-6}\dfrac{6^0}{0!} = e^{-6}$ .

$P(1) = e^{-6}\dfrac{6^1}{1!} = 6e^{-6}$ .

$P(2) = e^{-6}\dfrac{6^2}{2!} = \dfrac{36}{2}e^{-6} = 18e^{-6}$ .

Sum: $P(0)+P(1)+P(2) = (1+6+18)e^{-6} = 25e^{-6}$ .

$P(X\geq 3) = 1 - 25e^{-6} = 1 - \dfrac{25}{e^6}$ .

Common Mistakes

Forgetting the complement rule and trying to sum an infinite series directly.

Mis-computing $2! = 2$ , giving a wrong coefficient for $P(2)$ .

✓Final answer
$1 - 25/e^6$

ANSWER: A
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: $0.25\times0.80=0.20$ .

Total (mutually exclusive, so add): $0.15+0.20=0.35=\dfrac{35}{100}=\dfrac7{20}$ .

Common Mistakes

Including the "both true" or "both false" cases, which are agreement, not contradiction.

Forgetting independence and trying to combine the probabilities in some other (incorrect) way.

✓Final answer
$\dfrac7{20}$

ANSWER: C
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

Overcomplicating with inclusion-exclusion when the categories are mutually exclusive and exhaustive — direct counting of the complementary category (blue) is enough.

Forgetting to reduce the fraction $\frac{7}{21}$ to its simplest form $\frac13$ .

✓Final answer
$1/3$

ANSWER: A
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 $B$ ).

Number of such pairs $=50\times25=1250$ (one from each class, automatically distinct numbers).

$P(A\cap\bar B) = \dfrac{1250}{4950} = \dfrac{125}{495}=\dfrac{25}{99}$ .

Common Mistakes

Treating the selection as with replacement (would change the denominator to $100^2$ or similar) when the answer choices' denominators (99, 198) signal a without-replacement / combinatorial selection from 100.

Forgetting that a number divisible by 4 paired with anything even keeps the product divisible by 4, so those pairs must NOT be counted in $A\cap\bar B$ .

✓Final answer
$\dfrac{25}{99}$

ANSWER: C
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:

$P(\text{not black}) = \sum_i P(U_i)P(\text{not black}\mid U_i) = \frac13\left(\frac{8}{10}+\frac{8}{10}+\frac{7}{10}\right)=\frac13\times\frac{23}{10}=\frac{23}{30}$
Common Mistakes

Computing $P(\text{black})$ instead of $P(\text{not black})$ and forgetting to complement.

Pooling all balls from all urns together (30+ total) instead of correctly weighting per-urn probabilities by $\tfrac13$ each.

✓Final answer
$\dfrac{23}{30}$

ANSWER: B
AP EAPCET 2022Set eng-2022-07-04-FN1 markMCQ
$1/10$
›Reveal solutionSolution
This tests basic classical probability: favourable outcomes over total outcomes for a single random draw. The probability of not winning is $\dfrac57$ .

Concept and Intuition

For a single random draw from a finite set with equally likely outcomes, the probability of an event is simply the count of outcomes satisfying that event divided by the total count. Here the event is "drawing a non-prize ticket".

Step-by-Step Solution

Total tickets $=35$ ; prize-bearing tickets $=10$ .

Non-prize tickets $=35-10=25$ .

$P(\text{not getting a prize}) = \dfrac{25}{35}$ .

Simplify: $\dfrac{25}{35}=\dfrac{5}{7}$ (dividing numerator and denominator by 5).

Common Mistakes

Computing the probability of WINNING ( $10/35=2/7$ ) instead of the probability of NOT winning, which is what's asked.

Failing to simplify the fraction to lowest terms.

✓Final answer
$5/7$

ANSWER: D
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: $P(0)+P(1)+P(2) = e^{-0.2}(1+0.2+0.02) = 1.22\,e^{-0.2}$ .

$P(X\ge3) = 1 - 1.22\,e^{-0.2}$ .

Common Mistakes

Using the total count (10) as $\lambda$ instead of the per-day rate (0.2) — the Poisson parameter must match the unit of interest ("in a day").

Forgetting the $\dfrac{\lambda^2}{2!}$ factor (dividing by $2!=2$ ) when computing $P(X=2)$ .

✓Final answer
$1-(1.22)e^{-0.2}$

ANSWER: B

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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 $P(A') + P(B')$ .

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