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Probability: Class 12 Maths Practice Questions

30 original exam-pattern questions with full answers, matched to the current CBSE Class 12 paper design, including case-based questions. Attempt each question before opening the answer — or start a free 14-day trial ↓ for the full bank.

Q1Case-based4 marks

A factory has three machines M₁, M₂, and M₃ producing bolts. M₁ produces 50%, M₂ produces 30%, and M₃ produces 20% of the total bolts. The defective rates are: M₁ → 2%, M₂ → 3%, M₃ → 4%. A bolt drawn at random is found to be defective.

A factory has three machines — M₁, M₂, and M₃ — that produce bolts. Machine M₁ produces 50% of the total bolts, machine M₂ produces 30%, and machine M₃ produces the remaining 20%. From past records, it is known that 2% of the bolts produced by M₁ are defective, 3% of those by M₂ are defective, and 4% of those by M₃ are defective. A bolt is drawn at random from the day's production and is found to be defective.

(i) What is the probability that the defective bolt was produced by machine M₁?
(ii) What is the probability that the defective bolt was produced by machine M₂?
(iii) Which machine is most likely to have produced the defective bolt? Find the probability for that machine.
OR
(iii) Find the total probability that a randomly selected bolt from the day's production is defective.

Show answer
Let E₁, E₂, E₃ denote the events that the bolt was produced by M₁, M₂, M₃ respectively, and let A denote the event that the bolt is defective.

Then:
P(E₁) = 50/100 = 1/2, P(E₂) = 30/100 = 3/10, P(E₃) = 20/100 = 1/5

P(A|E₁) = 2/100 = 1/50, P(A|E₂) = 3/100, P(A|E₃) = 4/100 = 1/25

─────────────────────────────────────
Total Probability P(A) (needed for all parts):

By the law of total probability,
P(A) = P(E₁)·P(A|E₁) + P(E₂)·P(A|E₂) + P(E₃)·P(A|E₃)

P(A) = (1/2)(1/50) + (3/10)(3/100) + (1/5)(1/25)

P(A) = 1/100 + 9/1000 + 1/125

Converting to a common denominator of 1000:
P(A) = 10/1000 + 9/1000 + 8/1000 = 27/1000

─────────────────────────────────────
(i) Probability that the defective bolt was produced by M₁: [1 mark]

Using Bayes' theorem,
P(E₁|A) = P(E₁)·P(A|E₁) / P(A)

P(E₁|A) = (1/100) / (27/1000)

P(E₁|A) = (1/100) × (1000/27)

∴ P(E₁|A) = 10/27

─────────────────────────────────────
(ii) Probability that the defective bolt was produced by M₂: [1 mark]

Using Bayes' theorem,
P(E₂|A) = P(E₂)·P(A|E₂) / P(A)

P(E₂|A) = (9/1000) / (27/1000)

∴ P(E₂|A) = 9/27 = 1/3

─────────────────────────────────────
(iii) Finding the most likely machine: [2 marks]

Using Bayes' theorem for M₃,
P(E₃|A) = P(E₃)·P(A|E₃) / P(A)

P(E₃|A) = (8/1000) / (27/1000)

P(E₃|A) = 8/27

Comparing the three posterior probabilities:
P(E₁|A) = 10/27, P(E₂|A) = 9/27, P(E₃|A) = 8/27

∵ 10/27 > 9/27 > 8/27,

∴ Machine M₁ is most likely to have produced the defective bolt, and the required probability is 10/27.

─────────────────────────────────────
OR
─────────────────────────────────────
(iii) Total probability that a randomly selected bolt is defective: [2 marks]

By the law of total probability,
P(A) = P(E₁)·P(A|E₁) + P(E₂)·P(A|E₂) + P(E₃)·P(A|E₃)

P(A) = (1/2)(1/50) + (3/10)(3/100) + (1/5)(1/25)

P(A) = 10/1000 + 9/1000 + 8/1000

∴ P(A) = 27/1000
Q2Case-based4 marks

A quality control manager at an electronics factory inspects circuit boards produced by three machines M₁, M₂, and M₃. The machines produce 50%, 30%, and 20% of the total output respectively. The probability that a circuit board produced by M₁ is defective is 0.02, by M₂ is 0.04, and by M₃ is 0.05. A circuit board is selected at random from the day's production and is found to be defective.

A quality control manager at a electronics factory inspects circuit boards produced by three machines M₁, M₂, and M₃. The machines produce 50%, 30%, and 20% of the total output respectively. The probability that a circuit board produced by M₁ is defective is 0.02, by M₂ is 0.04, and by M₃ is 0.05. A circuit board is selected at random from the day's production and is found to be defective.

(i) Find the total probability that a randomly selected circuit board is defective.
(ii) Given that the selected circuit board is defective, find the probability that it was produced by machine M₂.
(iii) The manager wants to identify the machine most likely responsible for the defective board. Which machine has the highest posterior probability of having produced the defective board? Also find the probability that the defective board was NOT produced by M₁.

Show answer
Let E₁, E₂, E₃ denote the events that the board is produced by M₁, M₂, M₃ respectively, and let A denote the event that the selected board is defective.

The prior probabilities and conditional probabilities are:

P(E₁) = 0.50, P(E₂) = 0.30, P(E₃) = 0.20

P(A|E₁) = 0.02, P(A|E₂) = 0.04, P(A|E₃) = 0.05

(i) By the Law of Total Probability:

P(A) = P(E₁)·P(A|E₁) + P(E₂)·P(A|E₂) + P(E₃)·P(A|E₃)

⟹ P(A) = (0.50)(0.02) + (0.30)(0.04) + (0.20)(0.05)

⟹ P(A) = 0.010 + 0.012 + 0.010

∴ P(A) = 0.032

(ii) By Bayes' Theorem:

P(E₂|A) = P(E₂)·P(A|E₂) / P(A)

⟹ P(E₂|A) = (0.30 × 0.04) / 0.032

⟹ P(E₂|A) = 0.012 / 0.032

∴ P(E₂|A) = 3/8

(iii) By Bayes' Theorem, the posterior probabilities for all three machines are:

P(E₁|A) = P(E₁)·P(A|E₁) / P(A) = (0.50 × 0.02) / 0.032 = 0.010 / 0.032 = 5/16

P(E₂|A) = 3/8 = 6/16 [from part (ii)]

P(E₃|A) = P(E₃)·P(A|E₃) / P(A) = (0.20 × 0.05) / 0.032 = 0.010 / 0.032 = 5/16

Comparing: P(E₁|A) = 5/16, P(E₂|A) = 6/16, P(E₃|A) = 5/16

∵ P(E₂|A) > P(E₁|A) and P(E₂|A) > P(E₃|A),

∴ Machine M₂ has the highest posterior probability of having produced the defective board.

Now, probability that the defective board was NOT produced by M₁:

P(E₁'|A) = 1 − P(E₁|A) = 1 − 5/16

∴ P(E₁'|A) = 11/16
Q3Case-based4 marks

A school organises its annual sports day by randomly selecting one of two events — a relay race or a tug-of-war. The probability of selecting the relay race is 1/3 and that of tug-of-war is 2/3. For the relay race, participants are chosen from Group X (3 girls, 5 boys); for the tug-of-war, from Group Y (6 girls, 4 boys). In either event, exactly two students are selected at random from the respective group.

A school organises its annual sports day by randomly selecting one of two events — a relay race or a tug-of-war. The probability of selecting the relay race is 1/3 and the probability of selecting the tug-of-war is 2/3.

For the relay race, participants are chosen from Group X, which has 3 girls and 5 boys.
For the tug-of-war, participants are chosen from Group Y, which has 6 girls and 4 boys.

In either event, exactly two students are selected at random from the respective group.

(i) Find the probability that both selected students are girls, given that the relay race is chosen.
(ii) Find the probability that both selected students are girls, given that the tug-of-war is chosen.
(iii) Find the total probability that both selected students are girls.
OR
(iii) Given that both selected students are girls, find the probability that the tug-of-war was the event chosen.

Show answer
Let E₁ = event that relay race is chosen, E₂ = event that tug-of-war is chosen, and A = event that both selected students are girls.

Given:
P(E₁) = 1/3, P(E₂) = 2/3

Group X: 3 girls, 5 boys → total 8 students.
Group Y: 6 girls, 4 boys → total 10 students.

(i) P(A|E₁) = probability both are girls, given relay race (Group X)

P(A|E₁) = ³C₂ / ⁸C₂

³C₂ = 3, ⁸C₂ = 28

∴ P(A|E₁) = 3/28 …(1 mark)

(ii) P(A|E₂) = probability both are girls, given tug-of-war (Group Y)

P(A|E₂) = ⁶C₂ / ¹⁰C₂

⁶C₂ = 15, ¹⁰C₂ = 45

∴ P(A|E₂) = 15/45 = 1/3 …(1 mark)

(iii) By the Law of Total Probability,

P(A) = P(E₁)·P(A|E₁) + P(E₂)·P(A|E₂)

P(A) = (1/3)·(3/28) + (2/3)·(1/3)

P(A) = 3/84 + 2/9

P(A) = 1/28 + 2/9

Taking LCM of 28 and 9 = 252,

P(A) = 9/252 + 56/252

∴ P(A) = 65/252 …(2 marks)

∴ Required probability that both selected students are girls = 65/252.

OR

(iii) Using Bayes' Theorem, the required probability is P(E₂|A).

P(E₂|A) = P(E₂)·P(A|E₂) / [P(E₁)·P(A|E₁) + P(E₂)·P(A|E₂)]

Using results from (i) and (ii):

Numerator = P(E₂)·P(A|E₂) = (2/3)·(1/3) = 2/9

Denominator = P(E₁)·P(A|E₁) + P(E₂)·P(A|E₂)
= (1/3)·(3/28) + (2/3)·(1/3)
= 1/28 + 2/9
= 9/252 + 56/252
= 65/252

P(E₂|A) = (2/9) / (65/252)

P(E₂|A) = (2/9) × (252/65)

P(E₂|A) = 504/585

P(E₂|A) = 56/65 …(2 marks)

∴ Required probability that tug-of-war was chosen, given both selected students are girls = 56/65.
Q4Case-based4 marks

A quality control manager at a pharmaceutical company tests tablets from three production lines. Line L₁ produces 50% of the total tablets, Line L₂ produces 30%, and Line L₃ produces 20%. The probability that a tablet produced by L₁ is defective is 0·02, by L₂ is 0·05, and by L₃ is 0·04. A tablet is selected at random from the day's production and is found to be defective.

A quality control manager at a pharmaceutical company tests tablets from three production lines. Line L₁ produces 50% of the total tablets, Line L₂ produces 30%, and Line L₃ produces 20%. The probability that a tablet produced by L₁ is defective is 0·02, by L₂ is 0·05, and by L₃ is 0·04. A tablet is selected at random from the day's production and is found to be defective.

(i) Find the total probability that a randomly selected tablet is defective.

(ii) Given that the selected tablet is defective, find the probability that it came from Line L₂.

(iii) The manager decides to shut down the line most likely responsible for the defective tablet. Which line should be shut down? Also find the probability that the defective tablet came from Line L₁, and verify that the probabilities from all three lines sum to 1.

Show answer
Let E₁, E₂, E₃ denote the events that the tablet is from Line L₁, L₂, L₃ respectively, and let A denote the event that the tablet is defective.

Given:
P(E₁) = 0·50, P(E₂) = 0·30, P(E₃) = 0·20
P(A|E₁) = 0·02, P(A|E₂) = 0·05, P(A|E₃) = 0·04

(i) By the theorem of total probability:

P(A) = P(E₁)·P(A|E₁) + P(E₂)·P(A|E₂) + P(E₃)·P(A|E₃)

P(A) = (0·50)(0·02) + (0·30)(0·05) + (0·20)(0·04)

P(A) = 0·010 + 0·015 + 0·008

∴ P(A) = 0·033

(ii) By Bayes' theorem:

P(E₂|A) = P(E₂)·P(A|E₂) / P(A)

P(E₂|A) = (0·30 × 0·05) / 0·033

P(E₂|A) = 0·015 / 0·033

∴ P(E₂|A) = 15/33 = 5/11

(iii) Using Bayes' theorem for all three lines:

P(E₁|A) = P(E₁)·P(A|E₁) / P(A) = (0·50 × 0·02) / 0·033 = 0·010 / 0·033 = 10/33

P(E₂|A) = 5/11 = 15/33 (computed above)

P(E₃|A) = P(E₃)·P(A|E₃) / P(A) = (0·20 × 0·04) / 0·033 = 0·008 / 0·033 = 8/33

Comparing: P(E₁|A) = 10/33, P(E₂|A) = 15/33, P(E₃|A) = 8/33

∵ P(E₂|A) = 15/33 is the largest among the three,

∴ Line L₂ is most likely responsible for the defective tablet and should be shut down.

Verification:

P(E₁|A) + P(E₂|A) + P(E₃|A) = 10/33 + 15/33 + 8/33 = 33/33 = 1 ✓

∴ Required probability that the defective tablet came from Line L₁ = 10/33, and the sum of all three posterior probabilities = 1. Hence verified.
Q5Case-based4 marks

A school health survey classifies students into three groups based on their daily screen time: Low (less than 2 hours), Moderate (2–4 hours), and High (more than 4 hours). The proportion of students in each group is 30%, 50%, and 20% respectively. The probability that a student reports eye strain given their screen-time group is: 10% for Low, 25% for Moderate, and 60% for High.

A school health survey classifies students into three groups based on their daily screen time: Low (less than 2 hours), Moderate (2–4 hours), and High (more than 4 hours). From the survey data, the proportion of students in each group is 30%, 50%, and 20% respectively. The probability that a student reports eye strain given their screen-time group is as follows: 10% for Low, 25% for Moderate, and 60% for High.

Based on this information, answer the following:
(i) What is the total probability that a randomly selected student reports eye strain?
(ii) Given that a student reports eye strain, what is the probability that they belong to the High screen-time group?
(iii) Given that a student reports eye strain, what is the probability that they belong to the Low screen-time group? OR
(iii) Are the events 'student belongs to the Moderate screen-time group' and 'student reports eye strain' independent? Justify.

Show answer
Let E₁, E₂, E₃ denote the events that a student belongs to the Low, Moderate, and High screen-time group respectively, and let A denote the event that a student reports eye strain.

Given:
P(E₁) = 0·30, P(E₂) = 0·50, P(E₃) = 0·20
P(A|E₁) = 0·10, P(A|E₂) = 0·25, P(A|E₃) = 0·60

(i) By the Law of Total Probability,
P(A) = P(E₁)·P(A|E₁) + P(E₂)·P(A|E₂) + P(E₃)·P(A|E₃)
= (0·30)(0·10) + (0·50)(0·25) + (0·20)(0·60)
= 0·030 + 0·125 + 0·120
∴ P(A) = 0·275

(ii) By Bayes' Theorem,
P(E₃|A) = P(E₃)·P(A|E₃) / P(A)
= (0·20 × 0·60) / 0·275
= 0·120 / 0·275
= 12/27·5
= 24/55
∴ P(E₃|A) = 24/55

(iii) By Bayes' Theorem,
P(E₁|A) = P(E₁)·P(A|E₁) / P(A)
= (0·30 × 0·10) / 0·275
= 0·030 / 0·275
= 3/27·5
= 6/55
∴ P(E₁|A) = 6/55

OR

(iii) For independence of E₂ and A, we require P(E₂ ∩ A) = P(E₂)·P(A).
Now,
P(E₂ ∩ A) = P(E₂)·P(A|E₂) = 0·50 × 0·25 = 0·125
P(E₂)·P(A) = 0·50 × 0·275 = 0·1375
∵ P(E₂ ∩ A) = 0·125 ≠ 0·1375 = P(E₂)·P(A)
∴ The events 'student belongs to the Moderate screen-time group' and 'student reports eye strain' are NOT independent.
Q6Case-based4 marks

A tech company uses an AI-based screening tool to shortlist job applicants. From past data, it is known that 70% of applicants are genuinely qualified (event Q) and 30% are not qualified (event Q'). The screening tool correctly identifies a qualified applicant (gives a 'Pass') with probability 0.90, and incorrectly passes an unqualified applicant with probability 0.20.

A tech company uses an AI-based screening tool to shortlist job applicants. From past data, it is known that 70% of applicants are genuinely qualified (event Q) and 30% are not qualified (event Q'). The screening tool correctly identifies a qualified applicant (gives a 'Pass') with probability 0.90, and incorrectly passes an unqualified applicant with probability 0.20.

Based on this information, answer the following:

(i) What is the probability that a randomly selected applicant receives a 'Pass' from the screening tool?

(ii) Given that an applicant received a 'Pass', what is the probability that the applicant is actually qualified? (Use Bayes' Theorem.)

(iii) The company decides to interview only those who receive a 'Pass'. An interviewer randomly picks one 'Pass' holder. What is the probability that this person is NOT qualified? Also, if the company screens 1000 applicants, how many 'Pass' holders are expected to be unqualified?

Show answer
Let Q = event that the applicant is qualified, Q' = event that the applicant is not qualified, and P = event that the applicant receives a 'Pass'.

Given:
P(Q) = 0.70, P(Q') = 0.30
P(P|Q) = 0.90, P(P|Q') = 0.20

(i) Finding P(P) using the Total Probability Theorem:

P(P) = P(Q)·P(P|Q) + P(Q')·P(P|Q')

⟹ P(P) = (0.70)(0.90) + (0.30)(0.20)

⟹ P(P) = 0.630 + 0.060

∴ P(P) = 0.69

[1 mark]

(ii) Finding P(Q|P) using Bayes' Theorem:

P(Q|P) = P(Q)·P(P|Q) / P(P)

⟹ P(Q|P) = (0.70 × 0.90) / 0.69

⟹ P(Q|P) = 0.630 / 0.69

⟹ P(Q|P) = 63/69 = 21/23

∴ P(Q|P) = 21/23

[1 mark]

(iii) Probability that a 'Pass' holder is NOT qualified:

P(Q'|P) = 1 − P(Q|P)

⟹ P(Q'|P) = 1 − 21/23

∴ P(Q'|P) = 2/23

Alternatively, directly by Bayes' Theorem:
P(Q'|P) = P(Q')·P(P|Q') / P(P) = (0.30 × 0.20) / 0.69 = 0.060/0.69 = 6/69 = 2/23 ✓

Expected number of unqualified 'Pass' holders out of 1000 applicants:

Expected number of 'Pass' holders = 1000 × P(P) = 1000 × 0.69 = 690

Expected number of unqualified among 'Pass' holders = 690 × P(Q'|P)
= 690 × (2/23)
= 60

∴ P(Q'|P) = 2/23, and out of 1000 applicants screened, 60 'Pass' holders are expected to be unqualified.

[2 marks]
Q7Case-based4 marks

A quality control inspector at an electronics factory tests smartphones coming off two assembly lines, Line A and Line B. Line A produces 60% of the total smartphones and Line B produces 40%. From past records, 5% of smartphones from Line A are defective and 3% of smartphones from Line B are defective.

A quality control inspector at a electronics factory tests smartphones coming off two assembly lines, Line A and Line B. Line A produces 60% of the total smartphones and Line B produces 40%. From past records, 5% of smartphones from Line A are defective and 3% of smartphones from Line B are defective. The inspector picks one smartphone at random from the day's production.

(i) What is the probability that the smartphone is defective?
(ii) Given that the smartphone is found to be defective, what is the probability that it came from Line A?
(iii) Let X be a random variable denoting the number of defective smartphones found when 2 smartphones are picked at random (with replacement) from the combined day's production. Find E(X).

Show answer
Let E₁ = event that smartphone is from Line A, E₂ = event that smartphone is from Line B, and D = event that smartphone is defective.

Given:
P(E₁) = 0.6 = 3/5, P(E₂) = 0.4 = 2/5
P(D|E₁) = 0.05 = 1/20, P(D|E₂) = 0.03 = 3/100

(i) By the Total Probability theorem,
P(D) = P(E₁)·P(D|E₁) + P(E₂)·P(D|E₂)
⟹ P(D) = (3/5)·(1/20) + (2/5)·(3/100)
⟹ P(D) = 3/100 + 6/500
⟹ P(D) = 15/500 + 6/500
⟹ P(D) = 21/500

∴ The probability that the smartphone is defective = 21/500.

(ii) By Bayes' theorem,
P(E₁|D) = P(E₁)·P(D|E₁) / P(D)
⟹ P(E₁|D) = (3/5 × 1/20) / (21/500)
⟹ P(E₁|D) = (3/100) / (21/500)
⟹ P(E₁|D) = (3/100) × (500/21)
⟹ P(E₁|D) = 1500/2100 = 5/7

∴ The probability that the defective smartphone came from Line A = 5/7.

(iii) From part (i), the probability that any one smartphone picked at random is defective is
p = P(D) = 21/500.

Let X = number of defective smartphones when 2 smartphones are picked with replacement.
∵ Each pick is independent with probability of success p = 21/500, X follows a Binomial distribution with n = 2 and p = 21/500.

For a Binomial distribution, E(X) = np.

∴ E(X) = 2 × (21/500) = 42/500 = 21/250.

∴ E(X) = 21/250.
Q8Case-based4 marks

A school's annual health survey records that 60% of students travel by school bus, 30% travel by personal vehicle, and the remaining 10% walk to school. Past data shows that a student travelling by school bus is late on 5% of school days, a student travelling by personal vehicle is late on 10% of school days, and a student who walks is late on 20% of school days.

A school's annual health survey records that 60% of students travel by school bus, 30% travel by personal vehicle, and the remaining 10% walk to school. Past data shows that a student travelling by school bus is late on 5% of school days, a student travelling by personal vehicle is late on 10% of school days, and a student who walks is late on 20% of school days.

Based on the above information, answer the following:

(i) What is the total probability that a randomly selected student is late on a given school day?

(ii) Given that a student is late, what is the probability that they travel by school bus?

(iii) A teacher observes that a student is NOT late on a particular day. Find the probability that the student walks to school.

OR

(iii) Let X denote the number of days (out of 2 randomly chosen school days) on which a student who walks to school is late. Find the probability distribution of X and hence find E(X).

Show answer
Let E₁, E₂, E₃ denote the events that a student travels by school bus, personal vehicle, and walks, respectively, and let A denote the event that a student is late.

Given:
P(E₁) = 0.60, P(E₂) = 0.30, P(E₃) = 0.10
P(A|E₁) = 0.05, P(A|E₂) = 0.10, P(A|E₃) = 0.20

(i) By the Total Probability Theorem,
P(A) = P(E₁)·P(A|E₁) + P(E₂)·P(A|E₂) + P(E₃)·P(A|E₃)
P(A) = (0.60)(0.05) + (0.30)(0.10) + (0.10)(0.20)
P(A) = 0.030 + 0.030 + 0.020
∴ P(A) = 0.08

(ii) By Bayes' Theorem,
P(E₁|A) = P(E₁)·P(A|E₁) / P(A)
P(E₁|A) = (0.60 × 0.05) / 0.08
P(E₁|A) = 0.030 / 0.08
∴ P(E₁|A) = 3/8

(iii) We require P(E₃|A'), where A' is the event that the student is NOT late.
P(A') = 1 − P(A) = 1 − 0.08 = 0.92
P(A'|E₃) = 1 − P(A|E₃) = 1 − 0.20 = 0.80
By Bayes' Theorem,
P(E₃|A') = P(E₃)·P(A'|E₃) / P(A')
P(E₃|A') = (0.10 × 0.80) / 0.92
P(E₃|A') = 0.080 / 0.92
∴ P(E₃|A') = 2/23

OR

(iii) A student who walks is late on any given day with probability p = 0.20, and not late with probability q = 0.80.
X = number of days (out of 2) on which the student is late.
∴ X can take values 0, 1, 2.

P(X = 0) = ²C₀·(0.20)⁰·(0.80)² = 1 × 1 × 0.64 = 0.64 = 16/25
P(X = 1) = ²C₁·(0.20)¹·(0.80)¹ = 2 × 0.20 × 0.80 = 0.32 = 8/25
P(X = 2) = ²C₂·(0.20)²·(0.80)⁰ = 1 × 0.04 × 1 = 0.04 = 1/25

Verification: 16/25 + 8/25 + 1/25 = 25/25 = 1 ✓

Probability Distribution Table:

X | 0 | 1 | 2
P(X) | 16/25 | 8/25 | 1/25

E(X) = Σ x·P(X = x)
E(X) = 0·(16/25) + 1·(8/25) + 2·(1/25)
E(X) = 0 + 8/25 + 2/25
∴ E(X) = 10/25 = 2/5
Q9MCQ1 mark

If A and B are two independent events such that P(A) = 1/3 and P(B) = 1/4, then P(A' ∩ B') is equal to:

Show answer
Option (a) is correct.

Explanation: Since A and B are independent events, A' and B' are also independent. ∴ P(A' ∩ B') = P(A') · P(B') = (1 − 1/3)(1 − 1/4) = (2/3)(3/4) = 1/2.
Q10MCQ1 mark

A and B are two events such that P(A) = 1/3, P(B) = 1/4 and P(A ∩ B) = 1/12. Find P(A | B).

Show answer
Option (B) is correct.

Explanation: Using the conditional probability formula, P(A | B) = P(A ∩ B) / P(B) = (1/12) / (1/4) = 1/3.
Q11Short Answer1 mark

Assertion (A): If A and B are two events such that P(A) = 0.4, P(B) = 0.5 and P(A ∩ B) = 0.2, then P(A|B) = 0.4.

Reason (R): For any two events A and B, P(A|B) = P(A ∩ B) / P(B), provided P(B) ≠ 0.

Show answer
Option (a) is correct.

Explanation: The conditional probability formula states P(A|B) = P(A ∩ B) / P(B), provided P(B) ≠ 0. This is exactly what R states, so R is true.

Applying R to verify A: P(A|B) = P(A ∩ B) / P(B) = 0.2 / 0.5 = 0.4.

∴ A is true, and R is the correct explanation of A.
Q12MCQ1 mark

A fair coin is tossed twice. Find the probability of getting at least one head.

Show answer
Option (C) is correct.

Explanation: The sample space for tossing a fair coin twice is S = {HH, HT, TH, TT}, so n(S) = 4.

Event A = getting at least one head = {HH, HT, TH}, so n(A) = 3.

∴ P(at least one head) = 3/4.
Q13MCQ1 mark

If P(A) = 1/3, P(B) = 1/4, and A and B are independent events, then P(A ∩ B) is:

Show answer
Option (B) is correct.

Explanation: For independent events A and B, P(A ∩ B) = P(A) · P(B).

∴ P(A ∩ B) = (1/3) × (1/4) = 1/12.
Q14MCQ1 mark

A card is drawn at random from a well-shuffled deck of 52 cards. What is the probability that the card drawn is a king?

Show answer
Option (C) is correct.

Explanation: P(E) = (number of favourable outcomes) / (total number of outcomes). A deck of 52 cards contains 4 kings. ∴ P(drawing a king) = 4/52 = 1/13.
Q15Short Answer2 marks

The probability that a student passes a Mathematics test is 2/3 and the probability that the same student passes a Physics test is 3/5. If the two events are independent, find the probability that the student passes at least one of the two tests.

Show answer
Let M = event that the student passes Mathematics, P = event that the student passes Physics.

Given: P(M) = 2/3, P(P) = 3/5, and M, P are independent events.

P(at least one test passed) = 1 − P(M' ∩ P')

Since M and P are independent, M' and P' are also independent.

∴ P(M' ∩ P') = P(M') × P(P') = (1 − 2/3)(1 − 3/5) = (1/3)(2/5) = 2/15

∴ P(at least one test passed) = 1 − 2/15 = 13/15
Q16Short Answer2 marks

Events A and B are such that P(A) = 1/3, P(B) = 1/4 and P(A ∩ B) = 1/12. Find whether the events A and B are independent or not.

Show answer
For independence of events A and B, we check whether P(A ∩ B) = P(A) · P(B).

P(A) · P(B) = 1/3 × 1/4 = 1/12

Since P(A ∩ B) = 1/12 = P(A) · P(B),

∴ Events A and B are independent.
Q17Short Answer2 marks

A card is drawn at random from a well-shuffled deck of 52 cards. Given that the card drawn is a face card (Jack, Queen, or King), find the probability that it is a King.

Show answer
Let E = event that the card drawn is a King, and F = event that the card drawn is a face card.

In a deck of 52 cards, there are 12 face cards (4 Jacks, 4 Queens, 4 Kings) and 4 Kings.

∴ P(F) = 12/52 = 3/13, and P(E ∩ F) = 4/52 = 1/13.

Using the conditional probability formula,

P(E|F) = P(E ∩ F) / P(F) = (1/13) / (3/13) = 1/3.

∴ Required probability = 1/3.
Q18Short Answer2 marks

A bag contains 5 red and 3 black balls. Two balls are drawn one by one without replacement. Find the probability that both balls drawn are red.

Show answer
Let E₁ = event that the first ball drawn is red, and E₂ = event that the second ball drawn is red.

Total balls in the bag = 5 + 3 = 8.

P(E₁) = 5/8

After drawing one red ball, balls remaining = 7, of which red balls = 4.

∴ P(E₂ | E₁) = 4/7

Using the multiplication rule of probability,

P(both balls are red) = P(E₁) × P(E₂ | E₁)

= 5/8 × 4/7

= 20/56

∴ Required probability = 5/14
Q19Short Answer3 marks

A bag contains 4 red and 6 black balls. Two balls are drawn one after the other without replacement. Let event E: the first ball drawn is red, and event F: the second ball drawn is red. Find P(E), P(F|E), and P(E ∩ F). Also verify that P(E ∩ F) = P(E) · P(F|E).

Show answer
The bag contains 4 red and 6 black balls, so total balls = 10.

Finding P(E):
P(E) = P(first ball is red) = 4/10 = 2/5

Finding P(F|E):
Given that the first ball drawn is red, 9 balls remain in the bag, of which 3 are red.

∴ P(F|E) = P(second ball is red | first ball is red) = 3/9 = 1/3

Finding P(E ∩ F):
Using the multiplication rule,
P(E ∩ F) = P(E) · P(F|E)

= 2/5 × 1/3

∴ P(E ∩ F) = 2/15

Verification:
Alternatively, P(E ∩ F) = P(both balls are red) = (⁴C₂)/(¹⁰C₂) = 6/45 = 2/15 ...(i)

Also, P(E) · P(F|E) = 2/5 × 1/3 = 2/15 ...(ii)

∵ (i) = (ii),

∴ P(E ∩ F) = P(E) · P(F|E). Hence verified.
Q20Short Answer3 marks

E and F are two independent events such that P(Ē) = 0·4 and P(E ∪ F) = 0·7. Find P(F) and P(Ē ∩ F̄).

Show answer
Given: P(Ē) = 0·4 and P(E ∪ F) = 0·7.

∴ P(E) = 1 − P(Ē) = 1 − 0·4 = 0·6

Using the addition rule,
P(E ∪ F) = P(E) + P(F) − P(E ∩ F) ...(i)

Since E and F are independent events,
P(E ∩ F) = P(E) · P(F) ...(ii)

Substituting (ii) in (i),
0·7 = 0·6 + P(F) − 0·6 · P(F)

⟹ 0·7 − 0·6 = P(F)(1 − 0·6)

⟹ 0·1 = P(F) × 0·4

∴ P(F) = 0·1 / 0·4 = 1/4

Now, for P(Ē ∩ F̄):

Using De Morgan's law, Ē ∩ F̄ = (E ∪ F)̄

∴ P(Ē ∩ F̄) = 1 − P(E ∪ F) = 1 − 0·7

∴ P(Ē ∩ F̄) = 0·3
Q21Short Answer3 marks

A bag contains 4 red and 6 black balls. Two balls are drawn one by one without replacement. Let X denote the number of red balls drawn. Write the probability distribution of X and find its mean.

Show answer
The bag contains 4 red and 6 black balls (total = 10). Two balls are drawn without replacement.

X can take values 0, 1, 2.

P(X = 0) = P(both balls are black)
= (⁶C₂)/(¹⁰C₂) = 15/45 = 1/3

P(X = 1) = P(exactly one red and one black)
= (⁴C₁ × ⁶C₁)/(¹⁰C₂) = 24/45 = 8/15

P(X = 2) = P(both balls are red)
= (⁴C₂)/(¹⁰C₂) = 6/45 = 2/15

Verification: 1/3 + 8/15 + 2/15 = 5/15 + 8/15 + 2/15 = 15/15 = 1 ✓

Probability Distribution of X:

| X | 0 | 1 | 2 |
|---|---|---|---|
| P(X) | 1/3 | 8/15 | 2/15 |

Mean E(X) = Σ x · P(X = x)

E(X) = 0 × (1/3) + 1 × (8/15) + 2 × (2/15)

= 0 + 8/15 + 4/15

= 12/15

∴ Mean E(X) = 4/5
Q22Short Answer3 marks

A bag contains 4 red and 6 black balls. Two balls are drawn one by one without replacement. Find the probability that the first ball drawn is black, given that the second ball drawn is red.

Show answer
Let the events be defined as:
E₁ : First ball drawn is black
E₂ : Second ball drawn is red

We need to find P(E₁ | E₂).

By Bayes' theorem,

P(E₁ | E₂) = P(E₁) · P(E₂ | E₁) / P(E₂) ...(i)

Now, total balls = 4 + 6 = 10.

P(E₁) = P(first ball is black) = 6/10 = 3/5

P(E₁') = P(first ball is red) = 4/10 = 2/5

P(E₂ | E₁) = P(second ball is red | first ball was black)
= 4/9 [4 red balls remain out of 9]

P(E₂ | E₁') = P(second ball is red | first ball was red)
= 3/9 = 1/3 [3 red balls remain out of 9]

By the theorem of total probability,
P(E₂) = P(E₁) · P(E₂ | E₁) + P(E₁') · P(E₂ | E₁')

∴ P(E₂) = (3/5)(4/9) + (2/5)(1/3)

= 12/45 + 2/15

= 12/45 + 6/45

= 18/45

= 2/5

Substituting in (i),

P(E₁ | E₂) = P(E₁) · P(E₂ | E₁) / P(E₂)

= [(3/5)(4/9)] / (2/5)

= [12/45] / [2/5]

= (12/45) × (5/2)

= 60/90

= 2/3

∴ Required probability = 2/3
Q23Short Answer3 marks

A bag contains 3 red and 5 black balls. Two balls are drawn one by one without replacement. Let X denote the number of red balls drawn. Find the probability distribution of X and hence find the mean (expected value) of X.

Show answer
The bag contains 3 red and 5 black balls. Two balls are drawn without replacement.

X can take values 0, 1, 2.

P(X = 0) = P(both balls are black)
= (⁵C₂)/(⁸C₂) = (5×4)/(8×7) = 20/56 = 5/14

P(X = 1) = P(one red, one black)
= (³C₁ × ⁵C₁)/(⁸C₂) = (3×5)/28 = 15/28

P(X = 2) = P(both balls are red)
= (³C₂)/(⁸C₂) = 3/28

Verification: 5/14 + 15/28 + 3/28 = 10/28 + 15/28 + 3/28 = 28/28 = 1 ✓

Probability Distribution Table:

| X | 0 | 1 | 2 |
|---|---|---|---|
| P(X) | 5/14 | 15/28 | 3/28 |

Mean E(X) = Σ xᵢ P(xᵢ)

∴ E(X) = 0 × (5/14) + 1 × (15/28) + 2 × (3/28)
= 0 + 15/28 + 6/28
= 21/28
= 3/4

∴ Required probability distribution is as above and E(X) = 3/4.
Q24Short Answer3 marks

A bag contains 4 red and 6 black balls. Two balls are drawn one by one without replacement. Let event E: the first ball drawn is red, and event F: the second ball drawn is red. Find P(E), P(F) and P(E|F).

Show answer
Total balls in the bag = 4 red + 6 black = 10.

Finding P(E):

P(E) = P(first ball is red) = 4/10 = 2/5.

Finding P(F):

Using the law of total probability, let E' denote the event that the first ball drawn is black.

P(F) = P(first red) · P(second red | first red) + P(first black) · P(second red | first black)

⟹ P(F) = (4/10) · (3/9) + (6/10) · (4/9)

⟹ P(F) = 12/90 + 24/90 = 36/90 = 2/5.

Finding P(E|F):

By the definition of conditional probability,

P(E|F) = P(E ∩ F) / P(F).

Now, P(E ∩ F) = P(both balls drawn are red)

⟹ P(E ∩ F) = (4/10) · (3/9) = 12/90 = 2/15.

Also, P(F) = 2/5 (calculated above).

∴ P(E|F) = (2/15) / (2/5) = (2/15) · (5/2) = 10/30 = 1/3.

∴ P(E) = 2/5, P(F) = 2/5, and P(E|F) = 1/3.
Q25Case-based4 marks

A quality control inspector at a factory checks items from two production lines, Line X and Line Y. Line X produces 60% of the total items and Line Y produces the remaining 40%. From past records, it is known that 5% of items from Line X are defective and 3% of items from Line Y are defective.

A quality control inspector at a factory checks items from two production lines, Line X and Line Y. Line X produces 60% of the total items and Line Y produces the remaining 40%. From past records, it is known that 5% of items from Line X are defective and 3% of items from Line Y are defective. An item is picked at random from the combined output.

(i) Find the probability that the selected item is defective.
(ii) Given that the selected item is found to be defective, find the probability that it came from Line Y.

Show answer
Let E₁ denote the event that the item is from Line X, E₂ denote the event that the item is from Line Y, and A denote the event that the item is defective.

From the given data:

P(E₁) = 60/100 = 3/5, P(E₂) = 40/100 = 2/5

P(A|E₁) = 5/100 = 1/20, P(A|E₂) = 3/100 = 3/100

(i) Using the Total Probability Theorem,

P(A) = P(E₁)·P(A|E₁) + P(E₂)·P(A|E₂)

⟹ P(A) = (3/5)·(1/20) + (2/5)·(3/100)

⟹ P(A) = 3/100 + 6/500

⟹ P(A) = 15/500 + 6/500

⟹ P(A) = 21/500

∴ The probability that the selected item is defective = 21/500.

(ii) Using Bayes' Theorem,

P(E₂|A) = P(E₂)·P(A|E₂) / P(A)

⟹ P(E₂|A) = [(2/5)·(3/100)] / (21/500)

⟹ P(E₂|A) = [6/500] / [21/500]

⟹ P(E₂|A) = 6/21 = 2/7

∴ The probability that the defective item came from Line Y = 2/7.
Q26Case-based4 marks

A quality control manager at a pharmaceutical company inspects tablets produced by two machines, Machine A and Machine B. Machine A produces 60% of the total tablets and Machine B produces the remaining 40%. From past records, it is known that 3% of tablets produced by Machine A are defective, while 5% of tablets produced by Machine B are defective.

A quality control manager at a pharmaceutical company inspects tablets produced by two machines, Machine A and Machine B. Machine A produces 60% of the total tablets and Machine B produces the remaining 40%. From past records, it is known that 3% of tablets produced by Machine A are defective, while 5% of tablets produced by Machine B are defective.

Based on the above information, answer the following questions:

(i) Find the probability that a randomly selected tablet is produced by Machine A and is defective.

(ii) Find the total probability that a randomly selected tablet is defective.

(iii) If a randomly selected tablet is found to be defective, find the probability that it was produced by Machine B.

OR

(iii) If a randomly selected tablet is found to be non-defective, find the probability that it was produced by Machine A.

Show answer
Let E₁ = event that tablet is produced by Machine A,
E₂ = event that tablet is produced by Machine B,
D = event that a randomly selected tablet is defective.

Then:
P(E₁) = 60/100 = 3/5, P(E₂) = 40/100 = 2/5
P(D|E₁) = 3/100, P(D|E₂) = 5/100 = 1/20

(i) P(E₁ ∩ D) = P(E₁) · P(D|E₁)

∴ P(E₁ ∩ D) = (3/5) × (3/100) = 9/500

(ii) Using the Total Probability Theorem,
P(D) = P(E₁)·P(D|E₁) + P(E₂)·P(D|E₂)

P(D) = (3/5)×(3/100) + (2/5)×(5/100)

P(D) = 9/500 + 10/500

∴ P(D) = 19/500

(iii) Using Bayes' Theorem,
P(E₂|D) = P(E₂)·P(D|E₂) / P(D)

P(E₂|D) = (2/5 × 1/20) / (19/500)

P(E₂|D) = (10/500) / (19/500)

∴ P(E₂|D) = 10/19

OR

(iii) P(D') = 1 − P(D) = 1 − 19/500 = 481/500

P(D'|E₁) = 1 − P(D|E₁) = 1 − 3/100 = 97/100

Using Bayes' Theorem,
P(E₁|D') = P(E₁)·P(D'|E₁) / P(D')

P(E₁|D') = (3/5 × 97/100) / (481/500)

P(E₁|D') = (291/500) / (481/500)

∴ P(E₁|D') = 291/481
Q27Case-based4 marks

A school library has three sections: Fiction (F), Science (S), and History (H). The proportion of books in these sections is 5 : 3 : 2, respectively. A student picks a book at random. The probability that a book is damaged, given it is from Fiction, Science, or History section is 0.10, 0.20, and 0.35 respectively.

A school library has three sections: Fiction (F), Science (S), and History (H). The proportion of books in these sections is 5 : 3 : 2, respectively. A student picks a book at random. The probability that a book is damaged, given it is from Fiction, Science, or History section is 0.10, 0.20, and 0.35 respectively.

Based on the above information, answer the following questions:

(i) What is the probability that a randomly selected book is from the Science section?

(ii) What is the probability that a randomly selected book is damaged and from the History section?

(iii) A book is found to be damaged. What is the probability that it came from the Fiction section?

OR

(iii) A book is found to be NOT damaged. What is the probability that it came from the Science section?

Show answer
Let E₁, E₂, E₃ denote the events that the selected book is from Fiction, Science, and History section respectively, and let A denote the event that the selected book is damaged.

From the given proportions 5 : 3 : 2,

P(E₁) = 5/10 = 1/2, P(E₂) = 3/10, P(E₃) = 2/10 = 1/5

The conditional probabilities of a book being damaged are:

P(A|E₁) = 0.10 = 1/10, P(A|E₂) = 0.20 = 1/5, P(A|E₃) = 0.35 = 7/20

─────────────────────────────────
(i) Probability that the selected book is from the Science section:

∴ P(E₂) = 3/10

─────────────────────────────────
(ii) Probability that the selected book is damaged AND from the History section:

P(E₃ ∩ A) = P(E₃) × P(A|E₃)

⟹ P(E₃ ∩ A) = (1/5) × (7/20) = 7/100

∴ Required probability = 7/100

─────────────────────────────────
(iii) A book is found to be damaged. Find P(E₁|A).

By the theorem of total probability:

P(A) = P(E₁)·P(A|E₁) + P(E₂)·P(A|E₂) + P(E₃)·P(A|E₃)

⟹ P(A) = (1/2)(1/10) + (3/10)(1/5) + (1/5)(7/20)

⟹ P(A) = 1/20 + 3/50 + 7/100

⟹ P(A) = 5/100 + 6/100 + 7/100 = 18/100 = 9/50

By Bayes' theorem:

P(E₁|A) = P(E₁)·P(A|E₁) / P(A)

⟹ P(E₁|A) = (1/2 × 1/10) / (9/50)

⟹ P(E₁|A) = (1/20) / (9/50) = (1/20) × (50/9) = 50/180 = 5/18

∴ Required probability = 5/18

─────────────────────────────────
OR

(iii) A book is found to be NOT damaged. Find P(E₂|A').

P(A') = 1 − P(A) = 1 − 9/50 = 41/50

P(A'|E₂) = 1 − P(A|E₂) = 1 − 1/5 = 4/5

By Bayes' theorem:

P(E₂|A') = P(E₂)·P(A'|E₂) / P(A')

⟹ P(E₂|A') = (3/10 × 4/5) / (41/50)

⟹ P(E₂|A') = (12/50) / (41/50) = 12/41

∴ Required probability = 12/41
Q28Case-based4 marks

A cybersecurity firm analyses login attempts on its server. Based on past data, 55% of all login attempts are made by registered users, 30% are made by bots, and the remaining 15% are made by hackers. The probability that a login attempt is flagged as suspicious, given it is from a registered user, is 0·05. The probability that a login attempt is flagged as suspicious, given it is from a bot, is 0·70. The probability that a login attempt is flagged as suspicious, given it is from a hacker, is 0·90.

A cybersecurity firm analyses login attempts on its server. Based on past data, 55% of all login attempts are made by registered users, 30% are made by bots, and the remaining 15% are made by hackers. The probability that a login attempt is flagged as suspicious, given it is from a registered user, is 0·05. The probability that a login attempt is flagged as suspicious, given it is from a bot, is 0·70. The probability that a login attempt is flagged as suspicious, given it is from a hacker, is 0·90.

Let:
E₁ : event that the login attempt is by a registered user
E₂ : event that the login attempt is by a bot
E₃ : event that the login attempt is by a hacker
A : event that a login attempt is flagged as suspicious

Based on the above information, answer the following:
(i) Find P(E₁), P(E₂) and P(E₃).
(ii) Find the probability that a randomly chosen login attempt is flagged as suspicious, i.e., find P(A).
(iii) Given that a login attempt has been flagged as suspicious, find the probability that it was made by a hacker.
OR
(iii) Given that a login attempt has been flagged as suspicious, find the probability that it was NOT made by a registered user.

Show answer
(i) From the given data:

P(E₁) = 55/100 = 11/20, P(E₂) = 30/100 = 3/10, P(E₃) = 15/100 = 3/20

∴ P(E₁) = 11/20, P(E₂) = 3/10, P(E₃) = 3/20

(ii) Using the theorem of total probability:

P(A) = P(E₁)·P(A|E₁) + P(E₂)·P(A|E₂) + P(E₃)·P(A|E₃)

Given:
P(A|E₁) = 0·05 = 1/20, P(A|E₂) = 0·70 = 7/10, P(A|E₃) = 0·90 = 9/10

Substituting,

P(A) = (11/20)·(1/20) + (3/10)·(7/10) + (3/20)·(9/10)

= 11/400 + 21/100 + 27/200

= 11/400 + 84/400 + 54/400

= 149/400

∴ P(A) = 149/400

(iii) Using Bayes' theorem:

P(E₃|A) = P(E₃)·P(A|E₃) / P(A)

= [(3/20)·(9/10)] / (149/400)

= (27/200) / (149/400)

= (27/200) × (400/149)

= 54/149

∴ Required probability = 54/149

OR

(iii) P(login NOT by registered user | flagged suspicious)

= P(E₂ ∪ E₃ | A)

= [P(E₂)·P(A|E₂) + P(E₃)·P(A|E₃)] / P(A)

= [(3/10)·(7/10) + (3/20)·(9/10)] / (149/400)

= [21/100 + 27/200] / (149/400)

= [42/200 + 27/200] / (149/400)

= (69/200) / (149/400)

= (69/200) × (400/149)

= 138/149

∴ Required probability = 138/149
Q29Case-based4 marks

A quality control inspector at a factory checks items from three production lines L₁, L₂, and L₃. The lines produce 50%, 30%, and 20% of the total output respectively. The probability that an item from L₁ is defective is 0·02, from L₂ is 0·04, and from L₃ is 0·05.

A quality control inspector at a factory checks items from three production lines L₁, L₂, and L₃. The lines produce 50%, 30%, and 20% of the total output respectively. The probability that an item from L₁ is defective is 0·02, from L₂ is 0·04, and from L₃ is 0·05.

(i) Find the total probability that a randomly selected item is defective.
(ii) If a randomly selected item is found to be defective, find the probability that it came from production line L₂.
(iii) If a randomly selected item is found to be defective, find the probability that it came from production line L₃.

OR

(iii) The inspector selects 2 items at random (with replacement) from the combined output. Find the probability that at least one item is defective.

Show answer
Let E₁, E₂, E₃ denote the events that a selected item comes from lines L₁, L₂, L₃ respectively, and let A denote the event that the selected item is defective.

Given:
P(E₁) = 0·50, P(E₂) = 0·30, P(E₃) = 0·20
P(A|E₁) = 0·02, P(A|E₂) = 0·04, P(A|E₃) = 0·05

(i) By the theorem of total probability,

P(A) = P(E₁)·P(A|E₁) + P(E₂)·P(A|E₂) + P(E₃)·P(A|E₃)

P(A) = (0·50)(0·02) + (0·30)(0·04) + (0·20)(0·05)
⟹ P(A) = 0·010 + 0·012 + 0·010
⟹ P(A) = 0·032

∴ The total probability that a randomly selected item is defective = 0·032.

(ii) By Bayes' theorem,

P(E₂|A) = P(E₂)·P(A|E₂) / P(A)

P(E₂|A) = (0·30 × 0·04) / 0·032
⟹ P(E₂|A) = 0·012 / 0·032
⟹ P(E₂|A) = 12/32 = 3/8

∴ Required probability that the defective item came from L₂ = 3/8.

(iii) By Bayes' theorem,

P(E₃|A) = P(E₃)·P(A|E₃) / P(A)

P(E₃|A) = (0·20 × 0·05) / 0·032
⟹ P(E₃|A) = 0·010 / 0·032
⟹ P(E₃|A) = 10/32 = 5/16

∴ Required probability that the defective item came from L₃ = 5/16.

OR

(iii) From part (i), P(A) = 0·032 = 4/125 (probability that any one item is defective).

P(item is not defective) = 1 − 0·032 = 0·968

Let X = number of defective items in 2 draws (with replacement).

P(at least one defective) = 1 − P(no defective item)

P(no defective in 2 draws) = (0·968)² = 0·937024

∴ P(at least one defective) = 1 − 0·937024 = 0·062976

∴ Required probability that at least one item is defective = 0·062976 (or equivalently 1 − (0·968)²).
Q30Case-based4 marks

A quality control inspector at a factory checks electronic chips produced on two assembly lines. Line L₁ produces 60% of the total chips and Line L₂ produces the remaining 40%. From past records, it is known that 5% of chips from L₁ are defective and 8% of chips from L₂ are defective. One chip is selected at random from the combined output.

A quality control inspector at a factory checks electronic chips produced on two assembly lines. Line L₁ produces 60% of the total chips and Line L₂ produces the remaining 40%. From past records, it is known that 5% of chips from L₁ are defective and 8% of chips from L₂ are defective. One chip is selected at random from the combined output.

(i) What is the probability that the selected chip is defective?
(ii) Given that the selected chip is found to be defective, what is the probability that it came from Line L₁?

Show answer
Let E₁ = event that the chip came from Line L₁,
E₂ = event that the chip came from Line L₂,
and A = event that the chip is defective.

Then:
P(E₁) = 0.6 = 3/5, P(E₂) = 0.4 = 2/5
P(A|E₁) = 0.05 = 1/20, P(A|E₂) = 0.08 = 2/25

(i) By the law of total probability:
P(A) = P(E₁)·P(A|E₁) + P(E₂)·P(A|E₂)

⟹ P(A) = (3/5)·(1/20) + (2/5)·(2/25)

⟹ P(A) = 3/100 + 4/125

⟹ P(A) = 15/500 + 16/500

⟹ P(A) = 31/500

∴ Required probability that the selected chip is defective = 31/500

(ii) By Bayes' theorem:
P(E₁|A) = P(E₁)·P(A|E₁) / P(A)

⟹ P(E₁|A) = (3/5 × 1/20) / (31/500)

⟹ P(E₁|A) = (3/100) / (31/500)

⟹ P(E₁|A) = (3/100) × (500/31)

⟹ P(E₁|A) = 1500/3100

⟹ P(E₁|A) = 15/31

∴ Required probability that the defective chip came from Line L₁ = 15/31

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