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%.

Given:
Total loans = $15 + 5 + 10 + 10 = 40$

Let:
- $L_1$ = Personal Loan, $L_2$ = Education Loan, $L_3$ = Housing Loan, $L_4$ = Car Loan
- $D$ = event that the selected person has defaulted

Prior Probabilities:
$$P(L_1) = \frac{15}{40}, \quad P(L_2) = \frac{5}{40}, \quad P(L_3) = \frac{10}{40}, \quad P(L_4) = \frac{10}{40}$$

Likelihoods (default rates):
$$P(D \mid L_1) = 0.03, \quad P(D \mid L_2) = 0.01, \quad P(D \mid L_3) = 0.02, \quad P(D \mid L_4) = 0.05$$

Total Probability of Default:
$$P(D) = \frac{15}{40}(0.03) + \frac{5}{40}(0.01) + \frac{10}{40}(0.02) + \frac{10}{40}(0.05)$$
$$= \frac{1}{40}(0.45 + 0.05 + 0.20 + 0.50) = \frac{1.20}{40} = 0.03$$

By Bayes' Theorem:
$$P(L_4 \mid D) = \frac{P(L_4)\cdot P(D \mid L_4)}{P(D)} = \frac{\dfrac{10}{40} \times 0.05}{0.03} = \frac{0.0125}{0.03} = \frac{5}{12}$$

Answer: The probability that the defaulted application was for a car loan is $\boxed{\dfrac{5}{12}}$.

Given:
Let $A_1$ = sent by air, $A_2$ = sent by bus/local, $A_3$ = sent by train, $L$ = delivered late.

$$P(A_1) = 0.30, \quad P(A_2) = 0.50, \quad P(A_3) = 0.20$$
$$P(L \mid A_1) = 0.02, \quad P(L \mid A_2) = 0.07, \quad P(L \mid A_3) = 0.05$$

(i) Total Probability of Late Delivery:
$$P(L) = P(A_1)P(L \mid A_1) + P(A_2)P(L \mid A_2) + P(A_3)P(L \mid A_3)$$
$$= 0.30 \times 0.02 + 0.50 \times 0.07 + 0.20 \times 0.05$$
$$= 0.006 + 0.035 + 0.010 = 0.051$$

$$\boxed{P(L) = 0.051 = \frac{51}{1000}}$$

(ii) Probability that parcel was sent by train given it is late (Bayes' Theorem):
$$P(A_3 \mid L) = \frac{P(A_3)\cdot P(L \mid A_3)}{P(L)} = \frac{0.20 \times 0.05}{0.051} = \frac{0.010}{0.051} = \frac{10}{51}$$

$$\boxed{P(A_3 \mid L) = \frac{10}{51}}$$

Given:
Let $C$ = correct setup, $I$ = incorrect setup.
$$P(C) = 0.80, \quad P(I) = 0.20$$

Let $E$ = event that the machine produces an acceptable item followed by an unacceptable item.

If setup is correct: $P(\text{acceptable}) = 0.90$, $P(\text{unacceptable}) = 0.10$
$$P(E \mid C) = 0.90 \times 0.10 = 0.09$$

If setup is incorrect: $P(\text{acceptable}) = 0.50$, $P(\text{unacceptable}) = 0.50$
$$P(E \mid I) = 0.50 \times 0.50 = 0.25$$

Total Probability:
$$P(E) = P(C)\cdot P(E \mid C) + P(I)\cdot P(E \mid I)$$
$$= 0.80 \times 0.09 + 0.20 \times 0.25 = 0.072 + 0.050 = 0.122$$

By Bayes' Theorem:
$$P(I \mid E) = \frac{P(I)\cdot P(E \mid I)}{P(E)} = \frac{0.20 \times 0.25}{0.122} = \frac{0.050}{0.122} = \frac{50}{122} = \frac{25}{61}$$

Answer: The probability that the machine is incorrectly set up is $\boxed{\dfrac{25}{61}}$.

Given:
Let $S$ = scooter driver, $C$ = car driver, $B$ = bus driver insured.

Ratio = 4:5:3, so total parts = 12.
$$P(S) = \frac{4}{12} = \frac{1}{3}, \quad P(C) = \frac{5}{12}, \quad P(B) = \frac{3}{12} = \frac{1}{4}$$

Let $A$ = event of meeting with an accident.
$$P(A \mid S) = 0.007, \quad P(A \mid C) = 0.004, \quad P(A \mid B) = 0.012$$

Total Probability:
$$P(A) = \frac{1}{3}(0.007) + \frac{5}{12}(0.004) + \frac{1}{4}(0.012)$$
$$= 0.00233 + 0.00167 + 0.003 = 0.007$$

By Bayes' Theorem:
$$P(S \mid A) = \frac{P(S)\cdot P(A \mid S)}{P(A)} = \frac{\dfrac{1}{3} \times 0.007}{0.007} = \frac{1}{3}$$

Answer: The probability that the insured person who met with an accident is a scooter driver is $\boxed{\dfrac{1}{3}}$.

Given:
A standard deck has 52 cards: 13 spades and 39 non-spades.

Let the events be:
- $A_0$ = both lost cards are non-spades
- $A_1$ = exactly one lost card is a spade
- $A_2$ = both lost cards are spades
- $E$ = the card drawn from remaining 50 cards is a spade

Prior Probabilities:
$$P(A_0) = \frac{\binom{39}{2}}{\binom{52}{2}} = \frac{741}{1326}, \quad P(A_1) = \frac{\binom{13}{1}\binom{39}{1}}{\binom{52}{2}} = \frac{507}{1326}, \quad P(A_2) = \frac{\binom{13}{2}}{\binom{52}{2}} = \frac{78}{1326}$$

Likelihoods:
- If $A_0$: 13 spades remain in 50 cards $\Rightarrow P(E \mid A_0) = \dfrac{13}{50}$
- If $A_1$: 12 spades remain in 50 cards $\Rightarrow P(E \mid A_1) = \dfrac{12}{50}$
- If $A_2$: 11 spades remain in 50 cards $\Rightarrow P(E \mid A_2) = \dfrac{11}{50}$

Total Probability:
$$P(E) = \frac{741}{1326}\cdot\frac{13}{50} + \frac{507}{1326}\cdot\frac{12}{50} + \frac{78}{1326}\cdot\frac{11}{50}$$
$$= \frac{1}{1326 \times 50}(741 \times 13 + 507 \times 12 + 78 \times 11)$$
$$= \frac{9633 + 6084 + 858}{66300} = \frac{16575}{66300} = \frac{1}{4}$$

By Bayes' Theorem:
$$P(A_2 \mid E) = \frac{P(A_2)\cdot P(E \mid A_2)}{P(E)} = \frac{\dfrac{78}{1326} \times \dfrac{11}{50}}{\dfrac{1}{4}} = \frac{78 \times 11 \times 4}{1326 \times 50} = \frac{3432}{66300} \approx 0.0518$$

Answer: The probability that both lost cards are spades is approximately $\boxed{0.051}$.

Given:
Let $D$ = person has the disease, $D'$ = person does not have the disease, $T$ = test result is positive.

$$P(D) = 0.002, \quad P(D') = 0.998$$
$$P(T \mid D) = 0.99 \quad (\text{test detects disease correctly})$$
$$P(T \mid D') = 0.005 \quad (\text{false positive rate})$$

Total Probability of Positive Test:
$$P(T) = P(D)\cdot P(T \mid D) + P(D')\cdot P(T \mid D')$$
$$= 0.002 \times 0.99 + 0.998 \times 0.005$$
$$= 0.00198 + 0.00499 = 0.00697$$

By Bayes' Theorem:
$$P(D \mid T) = \frac{P(D)\cdot P(T \mid D)}{P(T)} = \frac{0.002 \times 0.99}{0.00697} = \frac{0.00198}{0.00697} = \frac{198}{697}$$

Answer: The probability that a person has the disease given a positive test result is $\boxed{\dfrac{198}{697}} \approx 0.284$ or about 28.4%.

Given:
Let $B_1, B_2, B_3$ = events that the bulb is manufactured by units A, B, C respectively.
Let $E$ = event that the bulb is defective.

$$P(B_1) = 0.25, \quad P(B_2) = 0.35, \quad P(B_3) = 0.40$$
$$P(E \mid B_1) = 0.05, \quad P(E \mid B_2) = 0.04, \quad P(E \mid B_3) = 0.02$$

Total Probability of Defective Bulb:
$$P(E) = P(B_1)P(E \mid B_1) + P(B_2)P(E \mid B_2) + P(B_3)P(E \mid B_3)$$
$$= 0.25 \times 0.05 + 0.35 \times 0.04 + 0.40 \times 0.02$$
$$= 0.0125 + 0.0140 + 0.0080 = 0.0345$$

By Bayes' Theorem:
$$P(B_2 \mid E) = \frac{P(B_2)\cdot P(E \mid B_2)}{P(E)} = \frac{0.35 \times 0.04}{0.0345} = \frac{0.014}{0.0345} = \frac{28}{69}$$

Answer: The probability that the defective bulb was manufactured by unit B is $\boxed{\dfrac{28}{69}}$.