Probability

Given: We need an experiment whose sample space is the set of all non-negative integers $S = \{0, 1, 2, \ldots\}$.

Concept: The sample space lists all possible outcomes of a random experiment.

Answer: Observing the number of people who voted in a constituency is one such experiment. The number of voters can be 0, 1, 2, 3, … (any non-negative integer), so the sample space is $S = \{0, 1, 2, \ldots\}$.

Other valid examples: number of calls received at a call centre in a day, number of accidents on a highway in a week, etc.

Given: 3 bulbs are selected; each is classified as D (defective) or N (non-defective).

Concept: The sample space is the set of all possible ordered outcomes when each of the 3 bulbs is tested.

Working: Each bulb has 2 possible outcomes (D or N), so the total number of outcomes = $2^3 = 8$.

Listing all outcomes systematically:

$$S = \{DDD,\ DDN,\ DND,\ NDD,\ DNN,\ NDN,\ NND,\ NNN\}$$

Answer: The sample space has 8 elements as listed above.

Independent Events:

Let:
- $A$ = Event that a person has black hair
- $B$ = Event that a person works in an MNC

The occurrence of $A$ does not affect the probability of $B$ and vice versa. Hence $A$ and $B$ are independent events.

Dependent Events:

Let:
- $A$ = Event of heavy traffic on a road
- $B$ = Event of a road accident

Heavy traffic increases the likelihood of an accident, so $B$ depends on $A$. Hence $A$ and $B$ are dependent events.

Impossible Event: Getting the sun to rise in the west on any given day. The probability of this event is 0, so it is an impossible event.

Sure Event: Getting a sum of numbers $\leq 12$ when a pair of dice is rolled. Since the maximum sum is $6 + 6 = 12$, this always happens. Its probability is 1, making it a sure event.

Exhaustive Events: When a coin is tossed once:
- $A$ = Getting a Head
- $B$ = Getting a Tail

$A \cup B = S$ (the entire sample space), so $A$ and $B$ together cover all possible outcomes. Hence $A$ and $B$ are exhaustive events.

Mutually Exclusive Events: When a person is running:
- $A$ = The person is running forward
- $B$ = The person is running backward

A person cannot run forward and backward at the same time, so $A \cap B = \emptyset$. Hence $A$ and $B$ are mutually exclusive events.

Bonus — Mutually Exclusive and Exhaustive Events: When a die is thrown once:
- $A$ = Getting an even number $\{2, 4, 6\}$
- $B$ = Getting an odd number $\{1, 3, 5\}$

$A \cap B = \emptyset$ (mutually exclusive) and $A \cup B = S$ (exhaustive).

Given:
- Total buyers = 100
- Buyers who bought alarm system: $n(A) = 40 \Rightarrow P(A) = 0.40$
- Buyers who bought bucket seats: $n(B) = 30 \Rightarrow P(B) = 0.30$
- Buyers who bought both: $n(A \cap B) = 20 \Rightarrow P(A \cap B) = 0.20$

Formula (Conditional Probability):
$$P(B \mid A) = \frac{P(A \cap B)}{P(A)}$$

Calculation:
$$P(B \mid A) = \frac{0.20}{0.40} = 0.5$$

Answer: The probability that a buyer also bought bucket seats, given they purchased an alarm system, is $\mathbf{0.5}$ or 50%.

Given:
- Let $M$ = event that the person is male
- Let $P$ = event that the person owns a pet

From the table:
$$P(M \cap P) = 0.41$$
$$P(P) = 0.86$$

Formula (Conditional Probability):
$$P(M \mid P) = \frac{P(M \cap P)}{P(P)}$$

Calculation:
$$P(M \mid P) = \frac{0.41}{0.86} \approx 0.477$$

Answer: The probability that a randomly selected person is male, given that they own a pet, is approximately $\mathbf{0.477}$ or 47.7%.

Given:
- Let $D$ = event that a person sees the doctor regularly
- Let $H$ = event that a person has no health problems in the following year

$$P(D) = 0.80, \quad P(D') = 0.20$$
$$P(H \mid D) = 0.35, \quad P(H \mid D') = 0.05$$

Formula (Total Probability Theorem):
$$P(H) = P(D)\cdot P(H \mid D) + P(D')\cdot P(H \mid D')$$

Calculation:
$$P(H) = 0.80 \times 0.35 + 0.20 \times 0.05$$
$$= 0.28 + 0.01 = 0.29$$

Answer: The probability that a randomly selected person will have no health problems in the following year is $\mathbf{0.29}$ or 29%.