If $P(A \cap B) = \dfrac{7}{10}$ and $P(B) = \dfrac{17}{20}$, then $P(A \mid B)$ equals (A) $\dfrac{14}{17}$ (B) $\dfrac{17}{20}$ (C) $\dfrac{7}{8}$ (D) $\dfrac{1}{8}$

If P(A ∩ B) = (7)/(10) and P(B) = (17)/(20), then P(A mid B) equals (A) (14)/(17) (B) (17)/(20) (C) (7)/(8) (D

NCERT Exemplar · Q31

P(A \cap B) = \dfrac{7}{10}

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The key idea is the definition of conditional probability: $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$ . Substituting the given values gives $\frac{14}{17}$ , which matches option (A).

Conditional probability answers the question: If we know that event $B$ has happened, what is the chance that $A$ also happened? The sample space shrinks from the whole universe to just $B$ , so we compare the overlap $A \cap B$ to the new total $B$ .

The formula is direct:

P(A \mid B) = \frac{P(A \cap B)}{P(B)}

This is not a theorem to memorise blindly — it’s common sense. If $B$ is certain, the only part of $A$ that can occur is the part inside $B$ , and we measure that against the size of $B$ .

Now apply it.

Identify what’s given.

$P(A \cap B) = \frac{7}{10}$ and $P(B) = \frac{17}{20}$ .
Plug into the formula.

P(A \mid B) = \frac{\frac{7}{10}}{\frac{17}{20}}

Simplify the fraction. Dividing by a fraction: multiply by its reciprocal.

\frac{7}{10} \div \frac{17}{20} = \frac{7}{10} \times \frac{20}{17} = \frac{7 \times 20}{10 \times 17} = \frac{140}{170}

Reduce to lowest terms. Divide numerator and denominator by 10: $\frac{140}{170} = \frac{14}{17}$ …

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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

$P(S \mid M) = \frac{P(M \cap S)}{P(M)} = \frac{0.5}{0.7} = \frac{5}{7}.$ …
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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.

$P = \frac{3}{6} = \frac{1}{2}.$ …
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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). …

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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. …

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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. …
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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. …

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←Previous questionQ30 — If P(A) = (4)/(5) and P(A ∩ B) = (7)/(10), then P(B mid A) is equal to…→Next questionQ32 — If P(A) = (3)/(10), P(B) = (2)/(5) and P(A ∪ B) = (3)/(5), then P(B mi…

Q.If $P(A \cap B) = \dfrac{7}{10}$ and $P(B) = \dfrac{17}{20}$ , then $P(A \mid B)$ equals
(A) $\dfrac{14}{17}$
(B) $\dfrac{17}{20}$
(C) $\dfrac{7}{8}$
(D) $\dfrac{1}{8}$

Concept understanding — Conditional Probability

Conditional Probability

The Idea: Shrink the Sample Space

The Definition

Two Cautions

Method: Conditional probability from a joint 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.If P(A∩B)=710P(A \cap B) = \dfrac{7}{10}P(A∩B)=107​ and P(B)=1720P(B) = \dfrac{17}{20}P(B)=2017​, then P(A∣B)P(A \mid B)P(A∣B) equals (A) 1417\dfrac{14}{17}1714​ (B) 1720\dfrac{17}{20}2017​ (C) 78\dfrac{7}{8}87​ (D) 18\dfrac{1}{8}81​

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Q.If $P(A \cap B) = \dfrac{7}{10}$ and $P(B) = \dfrac{17}{20}$ , then $P(A \mid B)$ equals
(A) $\dfrac{14}{17}$
(B) $\dfrac{17}{20}$
(C) $\dfrac{7}{8}$
(D) $\dfrac{1}{8}$

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