Two events $E$ and $F$ are independent. If $P(E) = 0.3$, $P(E \cup F) = 0.5$, then $P(E \mid F) - P(F \mid E)$ equals (A) $\dfrac{2}{7}$ (B) $\dfrac{3}{35}$ (C) $\dfrac{1}{70}$ (D) $\dfrac{1}{7}$

Two events E and F are independent. If P(E) = 0.3, – Class 12 Mathematics

NCERT Exemplar · Q47

E

Punjab PsebMCQ· 1mImportance★★★★★

Appeared in past exams:WBJEE 2025· Set math-2025· 1mexact

80% · 132/165 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

For independent events, $P(E \mid F) = P(E)$ and $P(F \mid E) = P(F)$ . The difference reduces to $P(E) - P(F)$ . Using $P(E \cup F) = P(E) + P(F) - P(E)P(F)$ we find $P(F) = \frac{2}{7}$ , so the answer is $0.3 - \frac{2}{7} = \frac{1}{70}$ .

The key idea here is that independence simplifies conditional probability drastically. When two events are independent, knowing that one occurred gives you no information about the other — so the conditional probability is just the unconditional probability. That means:

P(E \mid F) = P(E) \quad \text{and} \quad P(F \mid E) = P(F)

Therefore, the expression we need becomes simply:

P(E \mid F) - P(F \mid E) = P(E) - P(F)

We already know $P(E) = 0.3$ . So the whole problem reduces to finding $P(F)$ .

Use the union formula for independent events. For any two events, $P(E \cup F) = P(E) + P(F) - P(E \cap F)$ . Because $E$ and $F$ are independent, $P(E \cap F) = P(E) \cdot P(F)$ . So:

P(E \cup F) = P(E) + P(F) - P(E)P(F)

Plug in the known values. $P(E \cup F) = 0.5$ and $P(E) = 0.3$ . Let $p = P(F)$ . Then:

0.5 = 0.3 + p - (0.3)p

Simplify:

0.5 = 0.3 + p - 0.3p

0.5 = 0.3 + 0.7p

Solve for $p$ . Subtract 0.3 from both sides:

0.2 = 0.7p

So:

$p = \frac{0.2}{0.7} = \frac{2}{7}$ …

Start your 14-day free trial to unlock the full solution →

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

CA Foundation 2026Set jan-20261 markMCQ
$\frac{3}{7}$
›Reveal solutionSolution

Conditional probability $P(S\mid M) = \dfrac{P(M \cap S)}{P(M)}$ .

Step 1 — identify the probabilities

50% study both maths and science, so $P(M \cap S) = 0.5$ ; 70% study maths, so $P(M) = 0.7$ .

Step 2 — apply the conditional-probability formula

$P(S \mid M) = \frac{P(M \cap S)}{P(M)} = \frac{0.5}{0.7} = \frac{5}{7}.$ …
Start your 14-day free trial to see the full solution →
CA Foundation 2026Set jan-20261 markMCQ
$\frac{1}{2}$
›Reveal solutionSolution

Conditional probability on a reduced sample space: $P(A\mid B) = \dfrac{n(A \cap B)}{n(B)}$ .

Step 1 — list the outcomes with sum 7.
$(1,6),(2,5),(3,4),(4,3),(5,2),(6,1) \Rightarrow n(B) = 6.$
Step 2 — count first die greater than second, among those.

$(4,3),(5,2),(6,1)$ → 3 outcomes.

Step 3 — conditional probability.

$P = \frac{3}{6} = \frac{1}{2}.$ …
Start your 14-day free trial to see the full solution →
CA Foundation 2023Set jun-20231 markMCQ
$P(A)=\dfrac{1}{3}, P(B)=\dfrac{1}{4}, P(A/B)=\dfrac{1}{6}$
›Reveal solutionSolution

P(B/A) = P(A∩B)/P(A) = (1/24)/(1/3) = 1/8.

Step 1 — Find the joint probability
$P(A\cap B) = P(A/B)\,P(B) = \frac{1}{6}\times\frac{1}{4} = \frac{1}{24}$
Step 2 — Apply the definition of conditional probability
$P(B/A) = \frac{P(A\cap B)}{P(A)} = \frac{1/24}{1/3} = \frac{3}{24} = \frac{1}{8}$
Watch out

P(A/B) and P(B/A) are not equal — you must recompute the joint probability first, then divide by P(A), not P(B). …

Start your 14-day free trial to see the full solution →
CA Foundation 2022Set dec-20221 markMCQ
$P(A)=\dfrac{1}{3}$
›Reveal solutionSolution

P(A∩B)=1/6, so P(B|A)=(1/6)/(1/3)=1/2.

Step 1 — Intersection via the addition rule
$P(A\cap B)=P(A)+P(B)-P(A\cup B)=\frac13+\frac34-\frac{11}{12}=\frac{4+9-11}{12}=\frac{2}{12}=\frac16$
Step 2 — Apply the conditional-probability formula
$P\!\left(\frac{B}{A}\right)=\frac{P(A\cap B)}{P(A)}=\frac{1/6}{1/3}=\frac12$
Watch out

Option (A) 1/6 is just P(A∩B) — you must still divide by P(A) to get the conditional probability. …

Start your 14-day free trial to see the full solution →
CA Foundation 2021Set dec-20211 markMCQ
$A$
›Reveal solutionSolution

Divide the joint probability by the conditioning event's probability: $P(A\mid B)=5/9$ , $P(B\mid A)=6/11$ .

Step 1 — Apply the conditional probability formula for $P(A\mid B)$
$P(A\mid B) = \frac{P(A\cap B)}{P(B)} = \frac{10/33}{6/11} = \frac{10}{33}\times\frac{11}{6} = \frac{110}{198} = \frac{5}{9}$
Step 2 — Apply it for $P(B\mid A)$
$P(B\mid A) = \frac{P(A\cap B)}{P(A)} = \frac{10/33}{5/9} = \frac{10}{33}\times\frac{9}{5} = \frac{90}{165} = \frac{6}{11}$
Step 3 — Sanity check

Since $P(A)P(B) = (5/9)(6/11) = 10/33 = P(A\cap B)$ , the conditionals collapse to the marginals — consistent with the computed values. …
Start your 14-day free trial to see the full solution →
CA Foundation 2021Set dec-20211 markMCQ
$Q. In a group of 20 males and 15 females, 12 males and 8 females are service holders. What is the probability that a person selected at random from the group is a service holder given that the selected person is a male? (A) 0.40 (B) 0.60 (C) 0.45 (D) 0.55$
›Reveal solutionSolution

Condition on males only: 12 service holders out of 20 males = 0.60.

Step 1 — Identify the reduced sample space

Given the person is male, only the 20 males matter.

Step 2 — Apply the conditional formula
$P(\text{service}\mid \text{male}) = \frac{\text{male service holders}}{\text{total males}} = \frac{12}{20} = 0.60$
Watch out

Do not divide by the full group of 35 — the condition 'given male' shrinks the denominator to 20. …

Start your 14-day free trial to see the full solution →

←Previous questionQ46 — If the events A and B are independent, then P(A ∩ B) is equal to (A) P…→Next questionQ48 — A bag contains 5 red and 3 blue balls. If 3 balls are drawn at random …

Q.Two events $E$ and $F$ are independent. If $P(E) = 0.3$ , $P(E \cup F) = 0.5$ , then $P(E \mid F) - P(F \mid E)$ equals
(A) $\dfrac{2}{7}$
(B) $\dfrac{3}{35}$
(C) $\dfrac{1}{70}$
(D) $\dfrac{1}{7}$

Concept understanding — Conditional Probability

Conditional Probability

The Idea: Shrink the Sample Space

The Definition

Two Cautions

Method: Independence + addition rule to find a missing probability

Steps

Common Mistakes

Step 1 — identify the probabilities

Step 2 — apply the conditional-probability formula

Step 1 — list the outcomes with sum 7.

Step 2 — count first die greater than second, among those.

Step 3 — conditional probability.

Step 1 — Find the joint probability

Step 2 — Apply the definition of conditional probability

Step 1 — Intersection via the addition rule

Step 2 — Apply the conditional-probability formula

Step 1 — Apply the conditional probability formula for $P(A\mid B)$

Step 2 — Apply it for $P(B\mid A)$

Step 3 — Sanity check

Step 1 — Identify the reduced sample space

Step 2 — Apply the conditional formula

Q.Two events EEE and FFF are independent. If P(E)=0.3P(E) = 0.3P(E)=0.3, P(E∪F)=0.5P(E \cup F) = 0.5P(E∪F)=0.5, then P(E∣F)−P(F∣E)P(E \mid F) - P(F \mid E)P(E∣F)−P(F∣E) equals (A) 27\dfrac{2}{7}72​ (B) 335\dfrac{3}{35}353​ (C) 170\dfrac{1}{70}701​ (D) 17\dfrac{1}{7}71​

🎓Unlock everything free for 14 days

Q.Two events $E$ and $F$ are independent. If $P(E) = 0.3$ , $P(E \cup F) = 0.5$ , then $P(E \mid F) - P(F \mid E)$ equals
(A) $\dfrac{2}{7}$
(B) $\dfrac{3}{35}$
(C) $\dfrac{1}{70}$
(D) $\dfrac{1}{7}$

Unlock everything free for 14 days