
If \[4P(A) = 6P(B) = 10P(A \cap B) = 1\] then $P(\frac{B}{A}) = $ -------.
A.$\frac{2}{5}$
B.$\frac{3}{5}$
C.$\frac{7}{{10}}$
D.$\frac{{19}}{{60}}$
Answer
225.3k+ views
Hint: Here, to solve the given problem we use the conditional probability concept.
Given,
\[4P(A) = 6P(B) = 10P(A \cap B) = 1 \to (1)\]
Now, from equation 1, let us find ‘$P(A)$’, ‘$P(B)$’and ‘$P(A \cap B)$’ values.
$4P(A) = 1 \Rightarrow P(A) = \frac{1}{4}$
$6P(B) = 1 \Rightarrow P(B) = \frac{1}{6}$
$10P(A \cap B) = 1 \Rightarrow P(A \cap B) = \frac{1}{{10}}$
Here, we need to find the value of $P(B/A)$ i.e.., the probability of the event B after the
occurrence of event A.
So, to find the $P(B/A)$ let us consider the concept of conditional probability i.e..,
$P(B/A) = \frac{{P(A \cap B)}}{{P(A)}} \to (2)$
Let us substitute the obtained values of $P(A \cap B)$ and $P(A)$ in equation 2, we get
$
\Rightarrow P(B/A) = \frac{{P(A \cap B)}}{{P(A)}} \\
\Rightarrow P(B/A) = \frac{{\frac{1}{{10}}}}{{\frac{1}{4}}} \\
\Rightarrow P(B/A) = \frac{4}{{10}} \\
\Rightarrow P(B/A) = \frac{2}{5} \\
$
Hence, the obtained value of $P(B/A)$ is$\frac{2}{5}$.
Hence the correct option for the given question is ‘A’.
Note: As, to find the conditional probability of $P(B/A) = \frac{{P(A \cap B)}}{{P(A)}}$i.e.., the
probability of the event B after the occurrence of event A .The probability is defined only after the occurrence of event A i.e.., $P(A)$ should be greater than zero.
Given,
\[4P(A) = 6P(B) = 10P(A \cap B) = 1 \to (1)\]
Now, from equation 1, let us find ‘$P(A)$’, ‘$P(B)$’and ‘$P(A \cap B)$’ values.
$4P(A) = 1 \Rightarrow P(A) = \frac{1}{4}$
$6P(B) = 1 \Rightarrow P(B) = \frac{1}{6}$
$10P(A \cap B) = 1 \Rightarrow P(A \cap B) = \frac{1}{{10}}$
Here, we need to find the value of $P(B/A)$ i.e.., the probability of the event B after the
occurrence of event A.
So, to find the $P(B/A)$ let us consider the concept of conditional probability i.e..,
$P(B/A) = \frac{{P(A \cap B)}}{{P(A)}} \to (2)$
Let us substitute the obtained values of $P(A \cap B)$ and $P(A)$ in equation 2, we get
$
\Rightarrow P(B/A) = \frac{{P(A \cap B)}}{{P(A)}} \\
\Rightarrow P(B/A) = \frac{{\frac{1}{{10}}}}{{\frac{1}{4}}} \\
\Rightarrow P(B/A) = \frac{4}{{10}} \\
\Rightarrow P(B/A) = \frac{2}{5} \\
$
Hence, the obtained value of $P(B/A)$ is$\frac{2}{5}$.
Hence the correct option for the given question is ‘A’.
Note: As, to find the conditional probability of $P(B/A) = \frac{{P(A \cap B)}}{{P(A)}}$i.e.., the
probability of the event B after the occurrence of event A .The probability is defined only after the occurrence of event A i.e.., $P(A)$ should be greater than zero.
Recently Updated Pages
JEE Main 2026 Session 1 Correction Window Started: Check Dates, Edit Link & Fees

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Isoelectronic Definition in Chemistry: Meaning, Examples & Trends

Ionisation Energy and Ionisation Potential Explained

Iodoform Reactions - Important Concepts and Tips for JEE

Introduction to Dimensions: Understanding the Basics

Trending doubts
JEE Main 2026: City Intimation Slip and Exam Dates Released, Application Form Closed, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

NCERT Solutions for Class 11 Maths Chapter 10 Conic Sections

NCERT Solutions For Class 11 Maths Chapter 12 Limits And Derivatives

Understanding Atomic Structure for Beginners

