NCERT Class 12 Mathematics — Probability, Example 15.
Problem
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.
Key insight. “On time” can happen through two entirely separate paths — with a strike, or without one — and these two paths are mutually exclusive and cover every possibility. The Law of Total Probability says the overall probability is just the weighted sum of “on time” across both paths, weighted by how likely each path is in the first place.
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Approach
Identify the two mutually exclusive scenarios — “strike” and “no strike” — and their probabilities. Multiply each scenario’s probability by the conditional probability of completing on time given that scenario, then add the two products together (the Law of Total Probability).
Solution
Step 1 — Identify the given probabilities
$$P(\text{strike}) = 0.65, \qquad P(\text{no strike}) = 1-0.65 = 0.35$$
$$P(\text{on time}\mid\text{no strike}) = 0.80, \qquad P(\text{on time}\mid\text{strike}) = 0.32$$
Step 2 — Apply the Law of Total Probability
$$P(\text{on time}) = P(\text{strike})\times P(\text{on time}\mid\text{strike}) + P(\text{no strike})\times P(\text{on time}\mid\text{no strike})$$
Step 3 — Substitute the values
$$P(\text{on time}) = (0.65)(0.32) + (0.35)(0.80)$$
Step 4 — Compute each product
$$0.65\times0.32 = 0.208$$
$$0.35\times0.80 = 0.28$$
Step 5 — Add the results
$$P(\text{on time}) = 0.208+0.28 = 0.488$$
Answer
$$0.488$$
Common mistakes
- Forgetting to find $P(\text{no strike})$ as the complement of $P(\text{strike})$. Since only the probability of a strike is given directly, the “no strike” probability must be computed as $1-0.65$ before it can be used.
- Multiplying the wrong conditional probability with the wrong scenario. It’s easy to accidentally pair $P(\text{strike})$ with $P(\text{on time}\mid\text{no strike})$ instead of $P(\text{on time}\mid\text{strike})$ — always double-check that each weight is paired with its matching condition.
Practise next
- A factory has a $0.4$ probability of a machine breakdown on any given day. If there’s a breakdown, the probability of meeting the daily production target is $0.25$; if not, it’s $0.9$. Find the overall probability of meeting the target, using the same total probability approach.
Show answer
$\dfrac{16}{25}=0.64$. The two cases — breakdown or not — partition the day, so the law of total probability applies directly.
$$P(\text{target})=0.4\times0.25+0.6\times0.9=0.10+0.54=0.64.$$
Note it sits between the two conditional probabilities $0.25$ and $0.9$, and nearer the larger one because the no-breakdown case carries more weight — a useful sanity check on any total-probability answer.

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