P(A) = 2/3 ; P(B) = 3/5 ; P(A B) = 5/6 Find P(B/A)

Option b: 13/20. Solution: We are given: - P(A) = 2/3 - P(B) = 3/5 - P(A B) = 5/6 First, find the probability of the intersection P(A B) using the addition theorem of probability: P(A B) = P(A) + P(B) - P(A B) 5/6 = 2/3 + 3/5 - P(A B) Rearranging to solve for P(A B) : P(A B) = 2/3 + 3/5 - 5/6 To compute this, find a common denominator (which is 30): P(A B) = 20/30 + 18/30 - 25/30 = 20 + 18 - 25/30 = 13/30 . Now, calculate the conditional probability P(B/A) : P(B/A) = P(A B)/P(A) = 13/30/2/3 = 13/30 × 3/2 = 13/20 . Hence, Option B is the correct answer.

$P(A) = \frac{2}{3}$; $P(B) = \frac{3}{5}$; $P(A \cup B) = \frac{5}{6}$ Find $P(B/A)$

The correct answer is Option b: \frac{13}{20}. We are given: - $P(A) = \frac{2}{3}$ - $P(B) = \frac{3}{5}$ - $P(A \cup B) = \frac{5}{6}$ First, find the probability of the intersection $P(A \cap B)$ using the addition theorem of probability: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ $\frac{5}{6} = \frac{2}{3} + \frac{3}{5} - P(A \cap B)$ Rearranging to solve for $P(A \cap B)$: $P(A \cap B) = \frac{2}{3} + \frac{3}{5} - \frac{5}{6}$ To compute this, find a common denominator (which is 30): $P(A \cap B) = \frac{20}{30} + \frac{18}{30} - \frac{25}{30} = \frac{20 + 18 - 25}{30} = \frac{13}{30}$. Now, calculate the conditional probability $P(B/A)$: $P(B/A) = \frac{P(A \cap B)}{P(A)} = \frac{13/30}{2/3} = \frac{13}{30} \times \frac{3}{2} = \frac{13}{20}$. Hence, **Option B** is the correct answer.

ProbabilityMCQMTP June 24 Series IQuestion 2834 of 295

All Questions

$\displaystyle P(A) = \frac{2}{3}$ ; $\displaystyle P(B) = \frac{3}{5}$ ; $\displaystyle P(A \cup B) = \frac{5}{6}$ Find $\displaystyle P(B/A)$

Options

\displaystyle \frac{11}{20}

\displaystyle \frac{13}{20}

\displaystyle \frac{13}{18}

\displaystyle \frac{15}{20}

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Correct Answer

✅ Option b — $\displaystyle \frac{13}{20}$

All Options:

A $\displaystyle \frac{11}{20}$
B $\displaystyle \frac{13}{20}$
C $\displaystyle \frac{13}{18}$
D $\displaystyle \frac{15}{20}$

Detailed Solution & Explanation

We are given: -

\displaystyle P(A) = \frac{2}{3}

\displaystyle P(B) = \frac{3}{5}

\displaystyle P(A \cup B) = \frac{5}{6}

First, find the probability of the intersection

\displaystyle P(A \cap B)

using the addition theorem of probability:

\displaystyle P(A \cup B) = P(A) + P(B) - P(A \cap B)

\displaystyle \frac{5}{6} = \frac{2}{3} + \frac{3}{5} - P(A \cap B)

Rearranging to solve for

\displaystyle P(A \cap B)

\displaystyle P(A \cap B) = \frac{2}{3} + \frac{3}{5} - \frac{5}{6}

To compute this, find a common denominator (which is 30):

\displaystyle P(A \cap B) = \frac{20}{30} + \frac{18}{30} - \frac{25}{30} = \frac{20 + 18 - 25}{30} = \frac{13}{30}

. Now, calculate the conditional probability

\displaystyle P(B/A)

\displaystyle P(B/A) = \frac{P(A \cap B)}{P(A)} = \frac{13/30}{2/3} = \frac{13}{30} \times \frac{3}{2} = \frac{13}{20}

. Hence, **Option B** is the correct answer.

About This Chapter: Probability

Paper

Paper 3: Quantitative Aptitude

Weightage

5-7 Marks

Key Topics

Probability Operations, Expected Value

A logic-heavy chapter dealing with random experiments, events (mutually exclusive, exhaustive), set theory probability, conditional probability, and Bayes' Theorem. It forms the basis for Theoretical Distributions.

View Official ICAI Syllabus

Exam Strategy Tip

Always draw a quick Venn Diagram or tree when faced with 'At least one' or 'Only A but not B' wording. It saves you from double-counting.

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P(A)=23\displaystyle P(A) = \frac{2}{3}P(A)=32​; P(B)=35\displaystyle P(B) = \frac{3}{5}P(B)=53​; P(A∪B)=56\displaystyle P(A \cup B) = \frac{5}{6}P(A∪B)=65​ Find P(B/A)\displaystyle P(B/A)P(B/A)