Refer to Question 74 above. The probability that exactly two of the three balls were red, the first ball being

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\dfrac{1}{3}

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For the box of $5$ red and $3$ black balls (Q74), the probability of the sequence red, red, black is $\tfrac58\cdot\tfrac47\cdot\tfrac36=\tfrac{5}{28}$ — the official answer, option (D).

The setup (from Question 74)

The box has $5$ red and $3$ black balls ( $8$ in all), and three balls are drawn one by one without replacement. We are told the first ball is red and asked for the probability that exactly two of the three drawn are red.

The computation

The first ball is red and exactly two of the three are red, taken as the ordered outcome red, red, black (R, R, B):

P(\text{R})=\frac{5}{8},\qquad P(\text{R after R})=\frac{4}{7},\qquad P(\text{B after R,R})=\frac{3}{6}.

Multiplying along the branch:

$P=\frac{5}{8}\cdot\frac{4}{7}\cdot\frac{3}{6}=\frac{60}{336}=\frac{5}{28}.$ …

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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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JKBOSE Class 12 Annual Regular Examination 2018Set WZ1 markMCQ
$P(A) = \dfrac{3}{10}$
(a) 1/4
(b) 1/3
(c) 5/12
(d) 7/2
›Reveal solutionSolution

Using $P(A\cap B)=P(A)+P(B)-P(A\cup B)$ and the conditional-probability definitions, the correctly computed value is $\dfrac{7}{12}$ , which does not equal any of the four listed options.

Given $P(A)=\dfrac{3}{10}$ , $P(B)=\dfrac25$ , $P(A\cup B)=\dfrac35$ .

$P(A\cap B)=P(A)+P(B)-P(A\cup B)=\dfrac{3}{10}+\dfrac{4}{10}-\dfrac{6}{10}=\dfrac{1}{10}$

$P(B/A)=\dfrac{P(A\cap B)}{P(A)}=\dfrac{1/10}{3/10}=\dfrac13$

$P(A/B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{1/10}{4/10}=\dfrac14$

$P(B/A)+P(A/B)=\dfrac13+\dfrac14=\dfrac{4}{12}+\dfrac{3}{12}=\dfrac{7}{12}$

This is the value obtained by correctly applying the standard formulas to the numbers as given in the question. It does not match any of the four printed options — (a) $1/4$ ,

(b) $1/3$ ,

(c) $5/12$ , …

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←Previous questionQ48 — A bag contains 5 red and 3 blue balls. If 3 balls are drawn at random …→Next questionQ50 — Three persons, A, B and C, fire at a target in turn, starting with A. …

Q.Refer to Question 74 above. The probability that exactly two of the three balls were red, the first ball being red, is
(A) $\dfrac{1}{3}$
(B) $\dfrac{4}{7}$
(C) $\dfrac{15}{28}$
(D) $\dfrac{5}{28}$

Concept understanding — Conditional Probability

Conditional Probability

The Idea: Shrink the Sample Space

The Definition

Two Cautions

The setup (from Question 74)

The computation

Method: Sequential draws via the multiplication theorem along a branch

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.Refer to Question 74 above. The probability that exactly two of the three balls were red, the first ball being red, is (A) 13\dfrac{1}{3}31​ (B) 47\dfrac{4}{7}74​ (C) 1528\dfrac{15}{28}2815​ (D) 528\dfrac{5}{28}285​

The setup (from Question 74)

The computation

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Q.Refer to Question 74 above. The probability that exactly two of the three balls were red, the first ball being red, is
(A) $\dfrac{1}{3}$
(B) $\dfrac{4}{7}$
(C) $\dfrac{15}{28}$
(D) $\dfrac{5}{28}$

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