A bag contains $5$ red marbles and $3$ black marbles. Three marbles are drawn one by one without replacement. What is the probability that at least one of the three marbles drawn be black, if the first marble is red?

A bag contains 5 red marbles and 3 black marbles. Three marbles are drawn one by one without replacement. What

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Given the first marble is red, the bag holds $4$ red and $3$ black. The chance the next two are both red is $\tfrac{4}{7}\cdot\tfrac{3}{6}=\tfrac{2}{7}$ , so the probability of at least one black is $1-\tfrac{2}{7}=\dfrac{5}{7}$ .

Since the first marble drawn is red, the remaining bag contains

5-1=4\ \text{red},\qquad 3\ \text{black},\qquad \text{total }7,

and two more marbles are drawn without replacement. "At least one of the three is black" is now the same as "at least one of the next two is black," so use the complement.

1. Both of the next two are red (no black).

Second marble red: $\dfrac{4}{7}$ .
Then $3$ red remain out of $6$ , so third marble red: $\dfrac{3}{6}=\dfrac{1}{2}$ .

P(\text{both red})=\frac{4}{7}\times\frac{1}{2}=\frac{2}{7}.

2. At least one black. …

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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
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›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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Q.A bag contains $5$ red marbles and $3$ black marbles. Three marbles are drawn one by one without replacement. What is the probability that at least one of the three marbles drawn be black, if the first marble is red?

Concept understanding — Conditional Probability

Conditional Probability

The Idea: Shrink the Sample Space

The Definition

Two Cautions

Method: Conditional Probability with an Updated Sample Space and the "At Least One" Complement

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.A bag contains 555 red marbles and 333 black marbles. Three marbles are drawn one by one without replacement. What is the probability that at least one of the three marbles drawn be black, if the first marble is red?

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Q.A bag contains $5$ red marbles and $3$ black marbles. Three marbles are drawn one by one without replacement. What is the probability that at least one of the three marbles drawn be black, if the first marble is red?

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