Unit 1: Introduction to Probability - Subjective Questions

A set is a well-defined collection of distinct objects. If and , then:

• Union: contains elements belonging to or or both. Thus, .
• Intersection: contains elements common to both sets. Thus, .
• Complement: If the universal set is , then .
• Difference: contains elements in but not in . Thus, .

These operations are used to represent combinations of events in probability.

De Morgan's laws are:

and

Proof of the first law:

• Let . Then .
• Therefore, and .
• Hence, and , so .
• Reversing these steps proves the opposite inclusion.

Thus, . The second law follows similarly by replacing union with intersection and the word "and" with "or".

• A random experiment is a process whose exact outcome cannot be predicted in advance. Tossing two coins is a random experiment.
• An outcome is a possible result of the experiment, such as .
• The sample space is the set of all possible outcomes:

• An event is any subset of . For example, the event of obtaining exactly one head is

The empty set is the impossible event, while itself is the certain event.

A probability measure on a sample space satisfies:

1. Non-negativity: for every event .
2. Normalization: .
3. Countable additivity: For pairwise disjoint events ,

Since and the sets are disjoint,

which gives .

Also, and . Therefore,

so

If a finite sample space contains equally likely outcomes and event contains favorable outcomes, then

For a fair die,

The prime outcomes are

Therefore,

This calculation is valid because all six outcomes are equally likely.

The conditional probability of given that has occurred is

It measures the probability of after the sample space has effectively been restricted to .

Using the given values,

and

Thus, and .

From the definition of conditional probability,

Rearranging gives the multiplication theorem:

Equivalently,

For three events, first treat as one event:

Substituting gives

This is also called the chain rule of probability.

Events and are independent if

Now decompose into the disjoint events and :

Therefore,

Using independence,

Since ,

Hence, and are independent.

Three events , , and are pairwise independent if

They are mutually independent if they are pairwise independent and additionally

Consider two fair coin tosses with . Define

• : the first toss is a head.
• : the second toss is a head.
• : both tosses have the same result.

Each event has probability , and each pair has intersection probability , so they are pairwise independent. However,

whereas

Thus, they are not mutually independent.

If are mutually independent, then

Proof by induction:

• For , the statement is the definition of independence:

• Assume the result holds for events:

• Mutual independence implies that is independent of the intersection of any subcollection of the other events. Hence,

Substitution gives

Addition principle: If one task can be performed in ways and another mutually exclusive task can be performed in ways, then either task can be performed in ways.

For example, choosing one book from 4 mathematics books or 3 statistics books can be done in

ways.

Multiplication principle: If a process has two successive stages with choices for the first stage and choices for the second stage, then the process can be completed in

ways.

For example, choosing one of 3 shirts and one of 2 pairs of trousers can be done in

ways.

A combination is a selection in which order does not matter:

A permutation is an arrangement in which order or position matters:

Because the three positions are distinct, order matters. Therefore,

Alternatively, select 3 students and assign the positions:

Hence, there are 336 ways.

The total number of five-card hands is

To obtain exactly two aces:

• Choose 2 of the 4 aces in ways.
• Choose the remaining 3 cards from the 48 non-aces in ways.

Thus, the number of favorable hands is

Therefore,

Numerically, this probability is approximately .

Each of the 6 tosses has 2 possible outcomes, so the sample space contains

equally likely sequences.

For exactly 4 heads, choose the 4 positions occupied by heads:

Hence,

For at least one head, use the complementary event of obtaining no heads:

Let be mutually exclusive and exhaustive events such that . For an event with , Bayes' theorem states

Derivation: By conditional probability,

The multiplication theorem gives

Because the form a partition,

where the union is disjoint. Thus, by the law of total probability,

Substituting these expressions proves Bayes' theorem.

Let denote having the disease and denote a positive result. The data are

By Bayes' theorem,

Substitution gives

Therefore,

Thus, despite a positive result, the probability of having the disease is approximately , mainly because the disease is rare.

If form a partition of the sample space, then for any event ,

Let be the event that an item is defective. Then

Substituting the values,

Therefore,

Hence, the probability that a randomly selected item is defective is , or .

• Independent events: and are independent if

The occurrence of one does not change the probability of the other.

• Mutually exclusive events: and are mutually exclusive if

so . If both events have positive probability, they cannot be independent because .

• Conditionally independent events: and are conditionally independent given if

provided .

Conditional independence does not necessarily imply ordinary independence, and ordinary independence does not necessarily imply conditional independence.

There are initially 12 balls.

For both balls to be red,

Therefore,

If the first ball is blue, 11 balls remain, of which 5 are red. Hence,

The draws are dependent because the first draw changes the composition of the urn.

Let , , and denote the events that the respective components function. Since the components function independently,

Thus,

The event that at least one component fails is the complement of all components functioning. Therefore,

Hence, the required probabilities are and , respectively.