
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
220.5k+ 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
Geometry of Complex Numbers Explained

Electricity and Magnetism Explained: Key Concepts & Applications

JEE Energetics Important Concepts and Tips for Exam Preparation

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

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding Centrifugal Force in Physics

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

NCERT Solutions for Class 11 Maths Chapter 10 Conic Sections

NCERT Solutions for Class 11 Maths Chapter 9 Straight Lines

NCERT Solutions For Class 11 Maths Chapter 8 Sequences And Series

NCERT Solutions For Class 11 Maths Chapter 12 Limits And Derivatives

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

