Basic Probability • BP2

IOE Probability: Basic Problems on Events

Applied practice on mutually exclusive events, exhaustive events, complements, independent events and conditional probability.

Watch Video Solution

This video should explain all basic event problems before Bayes’ theorem.

How to Write Probability Solutions

A good solution should not start directly with numbers. First define the events, then identify the correct formula, then substitute the values.

Exam writing order:
1. Let \(A=\cdots\), \(B=\cdots\).
2. Decide whether the events are mutually exclusive, independent, exhaustive or conditional.
3. Write the formula.
4. Substitute the values.
5. Write the final answer with a short interpretation.

Question 1: Mutually Exclusive Events

Mutually Exclusive

A site engineer checks daily safety reports from a construction site. A report may contain a minor electrical fault or a plumbing fault. From previous records, \(35\%\) of the reports had a minor electrical fault and \(28\%\) had a plumbing fault. These two faults cannot occur together in the same report. Find the probability that a randomly selected report has either a minor electrical fault or a plumbing fault.

First write:
Let \(A=\) report has a minor electrical fault.
Let \(B=\) report has a plumbing fault.
Here, \(P(A)=0.35,\ P(B)=0.28\), and \(A\cap B=\varnothing\).
Correct answer: \(0.63\)
Try this first.

Try once before opening the solution.

Define the events and write the correct formula first. Then compare your work with the full solution.

Full Solution

Since the events are mutually exclusive,

\[P(A\cup B)=P(A)+P(B)\]

\[P(A\cup B)=0.35+0.28=0.63\]

Therefore, the probability that the report has either fault is \(0.63\).

Watch Video Solution

This problem is part of the BP2 basic probability video.

Question 2: General Addition Theorem

Addition Theorem

In an electronics lab, a circuit may have voltage fluctuation or overheating. From inspection records, \(62\%\) of circuits had voltage fluctuation, \(45\%\) had overheating, and \(20\%\) had both problems. Find the probability that a randomly selected circuit has voltage fluctuation or overheating.

First write:
Let \(A=\) circuit has voltage fluctuation.
Let \(B=\) circuit has overheating.
\(P(A)=0.62,\ P(B)=0.45,\ P(A\cap B)=0.20\).
Correct answer: \(0.87\)
Try this first.

Try once before opening the solution.

Define the events and write the correct formula first. Then compare your work with the full solution.

Full Solution

Use the addition theorem:

\[P(A\cup B)=P(A)+P(B)-P(A\cap B)\]

\[P(A\cup B)=0.62+0.45-0.20=0.87\]

Therefore, the required probability is \(0.87\).

Watch Video Solution

This problem is part of the BP2 basic probability video.

Question 3: Complement Rule

Complement

A factory produces sensors for measurement devices. The probability that a randomly selected sensor is defective is \(0.04\). Find the probability that the sensor is not defective.

First write:
Let \(D=\) sensor is defective.
Then \(D'=\) sensor is not defective.
\(P(D)=0.04\).
Correct answer: \(0.96\)
Try this first.

Try once before opening the solution.

Define the events and write the correct formula first. Then compare your work with the full solution.

Full Solution

Using the complement rule,

\[P(D')=1-P(D)\]

\[P(D')=1-0.04=0.96\]

Therefore, the probability that the sensor is not defective is \(0.96\).

Watch Video Solution

This problem is part of the BP2 basic probability video.

Question 4: Independent Events

Independent

In a computer lab, the probability that the projector fails on a given day is \(0.03\). The probability that the network cable fails on the same day is \(0.05\). Assume the two failures are independent. Find the probability that both failures occur on the same day.

First write:
Let \(A=\) projector fails.
Let \(B=\) network cable fails.
\(P(A)=0.03,\ P(B)=0.05\).
Correct answer: \(0.0015\)
Try this first.

Try once before opening the solution.

Define the events and write the correct formula first. Then compare your work with the full solution.

Full Solution

Since the events are independent,

\[P(A\cap B)=P(A)P(B)\]

\[P(A\cap B)=0.03(0.05)=0.0015\]

Therefore, the probability that both fail is \(0.0015\).

Watch Video Solution

This problem is part of the BP2 basic probability video.

Question 5: At Least One Event

Independent + Addition

For the same computer lab, the probability that the projector fails is \(0.03\), and the probability that the network cable fails is \(0.05\). Assume the failures are independent. Find the probability that at least one of the two failures occurs.

First write:
Let \(A=\) projector fails.
Let \(B=\) network cable fails.
\(P(A)=0.03,\ P(B)=0.05\).
Correct answer: \(0.0785\)
Try this first.

Try once before opening the solution.

Define the events and write the correct formula first. Then compare your work with the full solution.

Full Solution

First find the probability that both occur:

\[P(A\cap B)=0.03(0.05)=0.0015\]

Now use addition theorem:

\[P(A\cup B)=0.03+0.05-0.0015=0.0785\]

Therefore, the probability that at least one failure occurs is \(0.0785\).

Watch Video Solution

This problem is part of the BP2 basic probability video.

Question 6: Conditional Probability

Conditional

During building inspection, \(12\%\) of walls were found to have cracks. Among all inspected walls, \(4.5\%\) had both cracks and water leakage. Find the probability that a wall has water leakage given that it has cracks.

First write:
Let \(C=\) wall has cracks.
Let \(L=\) wall has water leakage.
\(P(C)=0.12,\ P(C\cap L)=0.045\).
Correct answer: \(0.375\)
Try this first.

Try once before opening the solution.

Define the events and write the correct formula first. Then compare your work with the full solution.

Full Solution

Use conditional probability:

\[P(L|C)=\frac{P(L\cap C)}{P(C)}\]

\[P(L|C)=\frac{0.045}{0.12}=0.375\]

Therefore, the required probability is \(0.375\).

Watch Video Solution

This problem is part of the BP2 basic probability video.

Question 7: Conditional Probability from Counts

Counts

A programming instructor checked 200 student submissions. Out of these, 50 submissions were late. Among the late submissions, 20 were affected by server issues. Find the probability that a submission had server issues given that it was late.

First write:
Let \(L=\) submission is late.
Let \(S=\) submission had server issues.
\(n(L)=50,\ n(S\cap L)=20\).
Correct answer: \(0.40\)
Try this first.

Try once before opening the solution.

Define the events and write the correct formula first. Then compare your work with the full solution.

Full Solution

Using conditional probability from counts:

\[P(S|L)=\frac{n(S\cap L)}{n(L)}\]

\[P(S|L)=\frac{20}{50}=0.40\]

Therefore, the required probability is \(0.40\).

Watch Video Solution

This problem is part of the BP2 basic probability video.

Question 8: Exhaustive Events

Exhaustive

A machine part is classified as having low tolerance error, medium tolerance error or high tolerance error. These three classes are mutually exclusive and exhaustive. The probability of low error is \(0.52\), and the probability of medium error is \(0.31\). Find the probability of high error.

First write:
Let \(L=\) low error, \(M=\) medium error and \(H=\) high error.
Since the classes are exhaustive, \(P(L)+P(M)+P(H)=1\).
Correct answer: \(0.17\)
Try this first.

Try once before opening the solution.

Define the events and write the correct formula first. Then compare your work with the full solution.

Full Solution

Since the events are exhaustive,

\[P(L)+P(M)+P(H)=1\]

\[P(H)=1-0.52-0.31=0.17\]

Therefore, the probability of high tolerance error is \(0.17\).

Watch Video Solution

This problem is part of the BP2 basic probability video.

Question 9: Neither Event Occurs

Complement + Mutually Exclusive

A component failure is classified as rust failure or overload failure. These two failure modes are mutually exclusive. The probability of rust failure is \(0.18\), and the probability of overload failure is \(0.12\). Find the probability that neither failure occurs.

First write:
Let \(R=\) rust failure.
Let \(O=\) overload failure.
\(P(R)=0.18,\ P(O)=0.12,\ R\cap O=\varnothing\).
Correct answer: \(0.70\)
Try this first.

Try once before opening the solution.

Define the events and write the correct formula first. Then compare your work with the full solution.

Full Solution

Since the events are mutually exclusive,

\[P(R\cup O)=0.18+0.12=0.30\]

Therefore,

\[P(\text{neither})=1-P(R\cup O)=1-0.30=0.70\]

Therefore, the probability that neither failure occurs is \(0.70\).

Watch Video Solution

This problem is part of the BP2 basic probability video.

Question 10: Independent Events and Conditional Probability

Independent Conditional

The probability of rain on a day is \(0.30\). The probability of a small internet outage is \(0.10\). Assume rain and internet outage are independent. Find the probability that there is an internet outage given that it rains.

First write:
Let \(R=\) rain occurs.
Let \(O=\) internet outage occurs.
Since \(R\) and \(O\) are independent, occurrence of rain does not change the probability of outage.
Correct answer: \(0.10\)
Try this first.

Try once before opening the solution.

Define the events and write the correct formula first. Then compare your work with the full solution.

Full Solution

For independent events,

\[P(O|R)=P(O)\]

\[P(O|R)=0.10\]

Therefore, the required probability is \(0.10\).

Watch Video Solution

This problem is part of the BP2 basic probability video.

Question 11: Multiplication Theorem

Multiplication

In a road survey, \(25\%\) of roads are old. Among old roads, \(40\%\) need urgent repair. Find the probability that a randomly selected road is old and needs urgent repair.

First write:
Let \(O=\) road is old.
Let \(R=\) road needs urgent repair.
\(P(O)=0.25,\ P(R|O)=0.40\).
Correct answer: \(0.10\)
Try this first.

Try once before opening the solution.

Define the events and write the correct formula first. Then compare your work with the full solution.

Full Solution

Use multiplication theorem:

\[P(O\cap R)=P(O)P(R|O)\]

\[P(O\cap R)=0.25(0.40)=0.10\]

Therefore, the probability that a road is old and needs urgent repair is \(0.10\).

Watch Video Solution

This problem is part of the BP2 basic probability video.

Question 12: Dice Application

Basic Set Probability

A die is thrown once. Let \(A\) be the event of getting an even number and \(B\) be the event of getting a number greater than 4. Find \(P(A)\), \(P(B)\), \(P(A\cap B)\), and \(P(A\cup B)\).

First write:
Sample space: \(S=\{1,2,3,4,5,6\}\).
\(A=\{2,4,6\}\), \(B=\{5,6\}\).
Correct answer: \(\frac12\)
Try this first.
Correct answer: \(\frac13\)
Try this first.
Correct answer: \(\frac16\)
Try this first.
Correct answer: \(\frac23\)
Try this first.

Try once before opening the solution.

Define the events and write the correct formula first. Then compare your work with the full solution.

Full Solution

\[P(A)=\frac{3}{6}=\frac{1}{2}\]

\[P(B)=\frac{2}{6}=\frac{1}{3}\]

\[A\cap B=\{6\},\quad P(A\cap B)=\frac{1}{6}\]

\[P(A\cup B)=\frac{1}{2}+\frac{1}{3}-\frac{1}{6}=\frac{2}{3}\]

Therefore, \(P(A)=\frac12,\ P(B)=\frac13,\ P(A\cap B)=\frac16,\ P(A\cup B)=\frac23\).

Watch Video Solution

This problem is part of the BP2 basic probability video.

Formulae

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