Two students Mehul and Rashi are seeking admission in a college. The probability that Mehul is selected is 0.4

Question

Q. Two students Mehul and Rashi are seeking admission in a college. The probability that Mehul is selected is 0.4 and the probability of selection of exactly one of the them is 0.5. Chances of selection of them is independent of each other. Find the chances of selection of Rashi. Also find the probability of selection of at least one of them. 3

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✓ Free question

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.

Solving $0.4(1-p)+0.6p=0.5$ gives Rashi's selection probability $p=0.5$ , and the probability that at least one is selected is $0.7$ .

Let $P(M)=0.4$ (Mehul selected) and $P(R)=p$ (Rashi selected), with the two selections independent.

Exactly one selected = Mehul only, or Rashi only:

P(M)\,(1-p)+\bigl(1-P(M)\bigr)\,p=0.5

0.4(1-p)+0.6p=0.5\ \Rightarrow\ 0.4+0.2p=0.5\ \Rightarrow\ p=0.5.

So Rashi's probability of selection is $0.5$ .

At least one selected $=1-P(\text{none})$ :

P(\text{at least one})=1-(1-0.4)(1-0.5)=1-0.6\times0.5=1-0.3=0.7.

✓Final answer

0.5

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

CBSE 2026Set 65/2/11 markMCQ
$A$
›Reveal solutionSolution
We need to find $P(A'|B')$ . By applying the definition of conditional probability, De Morgan's Law, and the complement rule, we can express this as $\frac{1 - P(A \cup B)}{P(B')}$ , which corresponds to option (D).

Let's break down this problem by first understanding the core concepts involved: conditional probability and the complement rule.

Conditional probability, $P(X|Y)$ , represents the probability of event $X$ occurring given that event $Y$ has already occurred. Its definition is fundamental:
$P(X|Y) = \frac{P(X \cap Y)}{P(Y)}$
The complement rule states that the probability of an event not happening is $1$ minus the probability of it happening. If $X'$ denotes the complement of event $X$ (i.e., $X$ does not occur), then:
$P(X') = 1 - P(X)$
We are asked to find $P(A'|B')$ , which means "the probability that event $A$ does not occur, given that event $B$ does not occur."

Now, let's work through the problem step-by-step.

Apply the definition of conditional probability. Using the formula $P(X|Y) = \frac{P(X \cap Y)}{P(Y)}$ , we replace $X$ with $A'$ and $Y$ with $B'$ .

$P(A'|B') = \frac{P(A' \cap B')}{P(B')}$
The problem states $P(B) \ne 1$. This is important because it implies $P(B') = 1 - P(B) \ne 0$, ensuring that the denominator is not zero and the conditional probability is well-defined.

2. Simplify the numerator using De Morgan's Law.

The term $A' \cap B'$ represents the event where neither $A$ nor $B$ occurs. This is equivalent to the event that $A \cup B$ (either $A$ or $B$ or both occur) does not occur. This is a direct application of De Morgan's Law for sets:
$(A \cup B)' = A' \cap B'$
Therefore, we can rewrite the numerator:
$P(A' \cap B') = P((A \cup B)')$

Apply the complement rule to the numerator. Now we have $P((A \cup B)')$ . Using the complement rule $P(X') = 1 - P(X)$ , where $X$ is the event $(A \cup B)$ :

$P((A \cup B)') = 1 - P(A \cup B)$

Substitute back into the conditional probability formula. Substitute the simplified numerator back into the expression from Step 1:

$P(A'|B') = \frac{1 - P(A \cup B)}{P(B')}$

Compare with the given options.

Let's look at the options:

(A) $1 - P(A/B)$

(B) $1 - P(A'/B)$

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

(D) $\frac{1 - P(A \cup B)}{P(B')}$

Our derived expression matches option (D).

Watch out

A common mistake is to assume $P(A'|B') = 1 - P(A|B)$ . This is generally incorrect. The complement rule applies to the event itself, not necessarily to a conditional probability in this direct manner. For example, $P(A'|B')$ is the probability of $A'$ given $B'$ , while $1 - P(A|B)$ is $1$ minus the probability of $A$ given $B$ . These are different conditions.

The condition $P(A) \ne 0$ is given in the problem but is not directly used in this specific derivation. It might be relevant in other contexts or to ensure $P(A')$ is not 1, but for finding $P(A'|B')$ , it doesn't alter the steps.

✓Final answer
$The correct option is (D) .$
CBSE 2026Set A1 markMCQ
$1 - P(A' \cap B') =$
(a) $P(A \cap B)$
(b) $P(A \cup B)$
(c) $P(A)$
(d) $P(B)$
›Reveal solutionSolution

$1 - P(A'\cap B') = P(A\cup B)$ .

By De Morgan's law,

$A' \cap B' = (A\cup B)'.$

So

$1 - P(A'\cap B') = 1 - P\big((A\cup B)'\big) = P(A\cup B),$

using $P(E') = 1 - P(E)$ .

✓Final answer
$P(A\cup B)$
CBSE 2025Set 65/2/11 markMCQ
$P(E) > 0$
›Reveal solutionSolution
Use the complement rule and conditional probability definition: $P(E' \mid F') = \frac{P(E' \cap F')}{P(F')}$ , then recognize that $E' \cap F' = (E \cup F)'$ to arrive at option (D): $\dfrac{1 - P(E \cup F)}{P(F')}$ .

The heart of this problem is understanding what conditional probability means when both events are complemented, and how set operations interact with complements.

Conditional probability $P(A \mid B)$ asks: "Given that $B$ has occurred, what is the probability of $A$ ?" The formula is always
$P(A \mid B) = \frac{P(A \cap B)}{P(B)}.$
When we want $P(E' \mid F')$ , we're asking: "Given that $F$ did not occur, what is the probability that $E$ also did not occur?" So we need the intersection $E' \cap F'$ in the numerator.

The key insight is recognizing what $E' \cap F'$ represents. By De Morgan's law, the region where neither $E$ nor $F$ occurs is precisely the complement of their union:
$E' \cap F' = (E \cup F)'.$
This transforms our problem into something we can express in terms of $P(E \cup F)$ .

Step-by-step derivation:

Write the definition of conditional probability for $P(E' \mid F')$ :

$P(E' \mid F') = \frac{P(E' \cap F')}{P(F')}.$

Apply De Morgan's law to the numerator: The event "not $E$ and not $F$ " is the same as "not ( $E$ or $F$ )":

$E' \cap F' = (E \cup F)'.$

Express the complement in terms of probability:

$P(E' \cap F') = P\bigl((E \cup F)'\bigr) = 1 - P(E \cup F).$

Substitute back into the conditional probability formula:

$P(E' \mid F') = \frac{1 - P(E \cup F)}{P(F')}.$
This matches option (D) exactly.

Watch out

A common mistake is to think $P(E' \mid F') = 1 - P(E \mid F')$ or $1 - P(E \mid F)$ . The complement rule $P(A') = 1 - P(A)$ applies to unconditional probabilities or when the conditioning event is the same. Here, we're conditioning on $F'$ , not $F$ , so neither option (B) nor (C) follows directly.

Tip

Whenever you see intersections of complements, think De Morgan: $A' \cap B' = (A \cup B)'$ and $A' \cup B' = (A \cap B)'$ . These identities are the bridge between conditional probabilities involving complements and union/intersection probabilities.

Why the other options don't work:

(A) $\frac{P(E')}{P(F')}$ ignores the intersection entirely; it treats $E'$ and $F'$ as if they were independent, which is not given.

(B) $1 - P(E' \mid F)$ would equal $P(E \mid F)$ , which conditions on $F$ , not $F'$ .

(C) $1 - P(E \mid F)$ equals $P(E' \mid F)$ , again conditioning on $F$ instead of $F'$ .

✓Final answer
$\dfrac{1 - P(E \cup F)}{P(F')}$
CBSE 2025Set ANNUAL1 markMCQ
$Q. A problem is given to three students whose chances of solving it are 1/4, 1/5 and 1/6 respectively. Then, the probability that the problem is solved is -$
(i) 1/120
(ii) 1/4
(iii) 1/2
(iv) 3/4
›Reveal solutionSolution

It's easier to find the probability that NONE of the three solve it, then subtract from 1.

Chances of solving: $1/4$ , $1/5$ , $1/6$ , so chances of NOT solving: $3/4$ , $4/5$ , $5/6$ respectively (independent students).

$P(\text{none solves}) = \dfrac34\times\dfrac45\times\dfrac56 = \dfrac{60}{120}=\dfrac12$ .

$P(\text{problem is solved}) = 1-P(\text{none solves}) = 1-\dfrac12=\dfrac12$ .

✓Final answer
$P(\text{problem is solved}) = \dfrac12$
CBSE 2024Set EX1 mark
$A$
›Reveal solutionSolution

In one race only one person can win, so the events " $A$ wins" and " $B$ wins" are mutually exclusive. $P(A\text{ or }B)=\tfrac13+\tfrac14=\tfrac{7}{12}$ ; neither $=1-\tfrac{7}{12}=\dfrac{5}{12}$ .

Concept. For mutually exclusive events, $P(A\cup B)=P(A)+P(B)$ . "Neither wins" is the complement of " $A$ or $B$ wins".

Add the winning chances.

$P(A\cup B)=P(A)+P(B)=\frac13+\frac14=\frac{4+3}{12}=\frac{7}{12}.$

Complement.

$P(\text{neither wins})=1-\frac{7}{12}=\frac{5}{12}.$

✓Final answer
$P(\text{neither }A\text{ nor }B\text{ wins})=\dfrac{5}{12}$
CBSE 2023Set M1 mark
$A$
›Reveal solutionSolution

Tests De Morgan's law with the addition rule; $P(A'\cap B')=\tfrac38$ .

By De Morgan's law, $P(\text{not }A\text{ and not }B) = P(A'\cap B') = P\big((A\cup B)'\big) = 1 - P(A\cup B)$ .

Using the addition rule:
$P(A\cup B) = P(A)+P(B)-P(A\cap B) = \frac14 + \frac12 - \frac18 = \frac{2+4-1}{8} = \frac{5}{8}.$
Therefore
$P(A'\cap B') = 1 - \frac{5}{8} = \frac{3}{8}.$
✓Final answer
$\dfrac{3}{8}$
CBSE 2021Set I1 markMCQ
$P(A) + P(A') =$
(a) $0$
(b) $1$
(c) $-1$
(d) $P(S)$
›Reveal solutionSolution

Complementary probabilities sum to $1$ .

An event $A$ and its complement $A'$ together cover the whole sample space $S$ and are mutually exclusive.

So $P(A)+P(A')=P(S)=1$ .

✓Final answer
$1$
CBSE 2021Set I1 markMCQ
$1 - P(A' \cap B') =$
(a) $P(A \cap B)$
(b) $P(A \cup B)$
(c) $P(A)$
(d) $P(B)$
›Reveal solutionSolution

$A'\cap B'$ is the complement of $A\cup B$ ; $1$ minus its probability gives $P(A\cup B)$ .

By De Morgan's law, $A'\cap B'=(A\cup B)'$ .

Using the complement rule $P(E')=1-P(E)$ :

$1-P(A'\cap B')=1-P\big((A\cup B)'\big)=1-\big(1-P(A\cup B)\big)=P(A\cup B)$ .

✓Final answer
$P(A\cup B)$
CBSE 2020Set 65/3/11 mark
$Q. An unbiased coin is tossed 4 times. Find the probability of getting at least one head.$
›Reveal solutionSolution
The complement approach is cleanest: find the probability of no heads (all tails), then subtract from 1. The probability of at least one head is $\boxed{\frac{15}{16}}$ .

Why the complement works

When a problem asks for "at least one" of something, you face the tedious task of adding up many cases: exactly one head, exactly two heads, exactly three heads, or all four heads. The complement sidesteps this entirely. There's only one way to fail the "at least one head" condition: get zero heads, meaning all four tosses land tails. Calculate that single probability, subtract from 1, and you're done.

Solution

Identify the complement event

"At least one head" means one or more heads. The opposite is "no heads at all," which is the same as getting tails on every single toss.

Find the probability of all tails

Each toss is independent, and for an unbiased coin $P(\text{Tail}) = \frac{1}{2}$ . The probability of tails on all four tosses is:

$P(\text{TTTT}) = \left(\frac{1}{2}\right)^4 = \frac{1}{16}$

Apply the complement rule The probability of at least one head is:

$P(\text{at least one head}) = 1 - P(\text{no heads}) = 1 - \frac{1}{16} = \frac{15}{16}$
Tip

Whenever you see "at least one," think complement first. It almost always saves work.

✓Final answer
$\boxed{\frac{15}{16}}$
CBSE 2019Set ANNUAL1 markMCQ
$P(A) + P(A') = ?$
(a) $0$
(b) $1$
(c) $-1$
(d) $P(E)$
›Reveal solutionSolution

$P(A)+P(A')=1$ .

The complement $A'$ consists of all outcomes not in $A$ , so $A$ and $A'$ are mutually exclusive and exhaustive. Hence $P(A)+P(A')=P(S)=1$ .

✓Final answer
$1$
CBSE 2018Set ANNUAL1 mark
$Q. Match the Column-A item 'A pair of coins are thrown simultaneously. What is the total chance of getting at least one head?' with the correct entry from Column-B. Column-B options (as printed, unordered):$
(1) 1;
(2) 6;
(3) 3;
(4) 5;
(5) 4.
›Reveal solutionSolution

List all equally-likely outcomes of two coin tosses; 'at least one head' excludes only the all-tails case.

Sample space for two coins tossed together: $\{HH, HT, TH, TT\}$ — 4 equally likely outcomes.

'At least one head' is satisfied by $HH, HT, TH$ — that's 3 favourable outcomes.

$P(\text{at least one head}) = \dfrac{3}{4}$

✓Final answer
$3$
CBSE 2018Set ANNUAL1 markMCQ
$Q. Three coins are tossed. The probability of at least one head is$
(a) $\dfrac{1}{8}$
(b) $\dfrac{3}{8}$
(c) $\dfrac{6}{8}$
(d) $\dfrac{7}{8}$
›Reveal solutionSolution

Use the complement: at least one head is the opposite of no heads at all.

$P(\text{no head in 3 tosses}) = \left(\dfrac12\right)^3 = \dfrac18$

$P(\text{at least one head}) = 1-P(\text{no head}) = 1-\dfrac18 = \dfrac78$

✓Final answer
$P(\text{at least one head})=\dfrac{7}{8}$

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