A die is thrown twice and the sum of the numbers appearing is observed to be 6. What is the conditional probab

Worked Examples · Example 6

Q. A die is thrown twice and the sum of the numbers appearing is observed to be 6. What is the conditional probability that the number 4 has appeared at least once?

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Given that the sum is 6, we restrict to the 5 equally likely outcomes: (1,5), (2,4), (3,3), (4,2), (5,1). Among these, the outcomes containing a 4 are (2,4) and (4,2). So the conditional probability is $\frac{2}{5}$ .

Why conditional probability works here

When we say “a die is thrown twice and the sum is 6,” we are no longer in the full sample space of 36 outcomes. The condition shrinks the space to only those pairs that add to 6. Conditional probability is simply: count the favourable outcomes within this restricted space, then divide by the total number of outcomes in the restricted space.

The formula is:

P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{\text{number of outcomes in both } A \text{ and } B}{\text{number of outcomes in } B}

where $B$ is “sum = 6” and $A$ is “at least one 4 appears.”

Step-by-step reasoning

List all outcomes where the sum is 6. Since each die shows 1 to 6, the ordered pairs $(x, y)$ with $x + y = 6$ are:

(1,5),\ (2,4),\ (3,3),\ (4,2),\ (5,1)

That’s 5 equally likely outcomes.

Watch out

Do not forget that (3,3) is a single ordered pair — it is not counted twice. Also, (2,4) and (4,2) are distinct because the dice are distinct (first throw and second throw).

Identify which of these contain at least one 4.

Scan the list:
- (1,5) — no 4
- (2,4) — contains a 4
- (3,3) — no 4
- (4,2) — contains a 4
- (5,1) — no 4
So exactly 2 outcomes satisfy “at least one 4”: (2,4) and (4,2). …

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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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Q.A die is thrown twice and the sum of the numbers appearing is observed to be 6. What is the conditional probability that the number 4 has appeared at least once?

Concept understanding — Conditional Probability

Conditional Probability

The Idea: Shrink the Sample Space

The Definition

Two Cautions

Why conditional probability works here

Step-by-step reasoning

Method: Reduced-sample-space counting for two-dice problems

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 die is thrown twice and the sum of the numbers appearing is observed to be 6. What is the conditional probability that the number 4 has appeared at least once?

Why conditional probability works here

Step-by-step reasoning

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