A person has undertaken a construction job. The probabilities are 0.65 that there will be a strike, 0.80 that

Worked Examples · Example 15

Q. A person has undertaken a construction job. The probabilities are 0.65 that there will be a strike, 0.80 that the construction job will be completed on time if there is no strike, and 0.32 that the construction job will be completed on time if there is a strike. Determine the probability that the construction job will be completed on time.

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Concept understanding — Conditional Probability

Conditional Probability

Roll a die and ask "what is the chance of an even number?" — that is $3/6$ . But suppose someone tells you the result is greater than 3. Now you are no longer looking at all six faces, only at $\{4,5,6\}$ , and two of those (4 and 6) are even, so the probability becomes $2/3$ . That change — from the probability of $A$ to the probability of $A$ given that $B$ has already occurred — is conditional probability.

The Idea: Shrink the Sample Space

Conditioning on $B$ throws away every outcome where $B$ is false and treats $B$ as the new "whole world." You measure $A$ only against what is still possible.

Note

Think of filtering a table of data: unconditional probability uses every row; conditional probability keeps only the rows where the condition is true.

The Definition

For events $A$ and $B$ with $P(B) > 0$ ,

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

We divide by $P(B)$ to rescale so that $B$ itself has probability $1$ ; the surviving part of $A$ is the overlap $A \cap B$ . Checking the die: $P(A \cap B) = P(\{4,6\}) = \tfrac{2}{6}$ and $P(B) = \tfrac{3}{6}$ , so $P(A \mid B) = \dfrac{2/6}{3/6} = \dfrac{2}{3}$ , matching the intuition.

Tip

Rearranging gives the multiplication rule $P(A \cap B) = P(A \mid B)\,P(B)$ , which is usually the easier way to compute a joint probability when a problem says "given that."

Two Cautions

$P(A \mid B)$ and $P(B \mid A)$ are generally not equal; swapping them is the classic mistake. They are linked by Bayes' theorem, $P(A \mid B) = \dfrac{P(B \mid A)\,P(A)}{P(B)}$ .
If $P(A \mid B) = P(A)$ , then knowing $B$ tells you nothing about $A$ — the events are independent. That is a special case, not the general rule.

Conditional probability is the foundation of the multiplication theorem, independence, and Bayes' theorem — every "given that" question in this chapter rests on it.

Conditional Probability opens the CBSE Class 12 Probability chapter and is foundational for everything that follows in that unit, including Bayes' theorem and the multiplication rule — making "conditional probability formula class 12 with examples" one of the most searched topics in Class 12 Mathematics. It is equally important for JEE Main and CUET, where conditional probability questions are set almost every year.

Using the law of total probability, we split the event "completed on time" into two mutually exclusive cases (strike vs. no strike). The overall probability is $0.65 \times 0.32 + 0.35 \times 0.80 = 0.488$ , or 0.488.

This is a classic law of total probability problem. When an event (here, "job completed on time") can happen under two different conditions (strike or no strike), we cannot just average the conditional probabilities — we must weight each by how likely that condition is.

The key insight: the two conditions (strike, no strike) are mutually exclusive and cover all possibilities. So the total probability of completion is the sum of:

Probability of completion given a strike, times the probability of a strike.
Probability of completion given no strike, times the probability of no strike.

Let’s define events clearly:

$S$ : there is a strike.
$C$ : the construction job is completed on time.

We are given:

$P(S) = 0.65$
$P(C \mid S^c) = 0.80$ (completed on time given no strike)
$P(C \mid S) = 0.32$ (completed on time given strike)

We need $P(C)$ .

Find the probability of no strike. Since $S$ and $S^c$ are complementary:

P(S^c) = 1 - P(S) = 1 - 0.65 = 0.35

Apply the law of total probability. The event $C$ can be written as:

C = (C \cap S) \cup (C \cap S^c)

Since $S$ and $S^c$ are disjoint, the two intersections are also disjoint. Therefore:

P(C) = P(C \cap S) + P(C \cap S^c)

Rewrite each intersection using conditional probability. By definition: $P(C \cap S) = P(S) \cdot P(C \mid S)$ and $P(C \cap S^c) = P(S^c) \cdot P(C \mid S^c)$ . So:

P(C) = P(S) \cdot P(C \mid S) + P(S^c) \cdot P(C \mid S^c)

Substitute the given numbers.

P(C) = (0.65)(0.32) + (0.35)(0.80)

Calculate each term.
- $0.65 \times 0.32 = 0.208$
- $0.35 \times 0.80 = 0.280$
Adding:

P(C) = 0.208 + 0.280 = 0.488

Watch out

A common mistake is to simply average 0.32 and 0.80, getting 0.56. That would be correct only if strike and no strike were equally likely (each 0.5). But here, a strike is more likely (0.65), so the overall probability is pulled closer to 0.32 than to 0.80.

Tip

Think of this as a weighted average: the weights are the probabilities of the conditions. The formula $P(C) = P(S)P(C|S) + P(S^c)P(C|S^c)$ is just a weighted mean of the two conditional probabilities.

✓Final answer

\boxed{0.488}

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

Divide by the given event's probability: since maths is given, the denominator is $P(M)=0.7$ , not $P(S)$ or the total. Dividing the other way ( $0.7/0.5$ ) gives a value above 1, which is impossible for a probability.

Tip

"Given that" tells you the denominator. Here it is "given studying mathematics," so put $P(M)$ on the bottom: $0.5/0.7 = 5/7$ .

✓Final answer
$(D) 5/7$
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.
$P = \frac{3}{6} = \frac{1}{2}.$
Watch out

The condition "sum = 7" shrinks the sample space to those 6 outcomes — divide by 6, not by the full 36. Using $3/36$ gives $1/12$ , which isn't even an option.

Tip

None of the sum-7 pairs are ties, so by symmetry "first > second" and "first < second" split the 6 outcomes evenly — the answer is simply half.

✓Final answer
$(A) 1/2$
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).

Tip

Anchor everything on P(A∩B): both conditionals flow from it via division by the conditioning event's probability.

✓Final answer
$\dfrac{1}{8}$

Note

This answer is verified — independently worked and reviewed by our experienced subject lecturers before we published it.
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.

Tip

Conditional probability always divides the joint probability by the probability of the given (conditioning) event.

✓Final answer
$\dfrac{1}{2}$

Note

This answer is verified — independently worked and reviewed by our experienced subject lecturers before we published it.
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.

Watch out

The trap is to multiply $P(A\cap B)$ by $P(B)$ instead of dividing (giving small fractions like 1/9, 2/9 in options C/D). Always divide by the given/conditioning event.

Tip

$P(A\mid B)$ = joint over the second letter's probability; $P(B\mid A)$ = joint over the first letter's probability.

✓Final answer
$(A) 5/9, 6/11$

Note

This answer is verified — independently worked and reviewed by our experienced subject lecturers before we published it.
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.

Tip

'Given that ...' problems: throw away everyone outside the given category, then take the simple fraction.

✓Final answer
$(B) 0.60$

Note

This answer is verified — independently worked and reviewed by our experienced subject lecturers before we published it.

←Previous section · 13.5.2Theorem of Total Probability →Next question · 13.5.2Example 16 — Bag I contains 3 red and 4 black balls while another Bag II contains 5…

Concept understanding — Conditional Probability

Conditional Probability

The Idea: Shrink the Sample Space

The Definition

Two Cautions

Method: Law of Total Probability (splitting on cases)

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

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