Class 12Mathematics · ProbabilityFull chapter

Probability

The whole chapter in one place — read it, then test yourself. Clear notes, formula sheet, a practice quiz, and worked NCERT solutions & PYQs.

In this chapter

Conditional Probability Multiplication Theorem and Independent Events Total Probability Theorem and Bayes' Theorem Random Variables and Probability Distributions Mean and Variance of a Random Variable Bernoulli Trials and Binomial Distribution

Conditional Probability

Quick answer Conditional probability measures the chance of an event occurring given that another event has already happened, and it forms the foundation for every other idea in this chapter.

In many real situations, the occurrence of one event changes what we know about the likelihood of another. Conditional probability captures this idea: it is the probability of an event E occurring, given that another event F has already occurred (with P(F) > 0). Because F has occurred, the sample space effectively shrinks from the whole sample space S to just F, and we ask what fraction of F also satisfies E.

Formally, P(E|F) = P(E∩F)/P(F). This is read as "the probability of E given F". Conditional probabilities obey the same axioms as ordinary probabilities: P(E|F) ≥ 0, P(S|F) = 1, and for mutually exclusive events A and B, P((A∪B)|F) = P(A|F) + P(B|F). From these axioms follow two useful results: P(E′|F) = 1 − P(E|F), and for any two events A and B, P((A∪B)|F) = P(A|F) + P(B|F) − P((A∩B)|F).

Worked Example. A fair die is thrown once. Let E = {2, 4, 6} (die shows an even number) and F = {4, 5, 6} (die shows a number greater than 3). Find P(E|F).

P(F) = 3/6 = 1/2 (outcomes 4, 5, 6 are equally likely).
E∩F = {4, 6}, so P(E∩F) = 2/6 = 1/3.
P(E|F) = P(E∩F)/P(F) = (1/3)/(1/2) = 2/3.

Notice that the unconditional probability P(E) = 3/6 = 1/2, but once we know the die showed a number greater than 3, the probability of an even number rises to 2/3. This is exactly what conditioning does — it updates a probability in light of new information.

Conditional Probability of E given F P(E|F) = P(E∩F) / P(F), provided P(F) > 0 Read as 'probability of E given F'; restricts the sample space to F.

Multiplication Rule (rearranged form) P(E∩F) = P(F) · P(E|F) Direct rearrangement of the conditional probability definition.

Complement Rule under Conditioning P(E′|F) = 1 − P(E|F) The conditional probabilities of an event and its complement given F must add to 1.

Addition Rule under Conditioning P((A∪B)|F) = P(A|F) + P(B|F) − P((A∩B)|F) The ordinary addition theorem applied within the reduced sample space F.

Remember

Conditional probability P(E|F) is defined only when P(F) > 0.
It effectively reduces the sample space from S to the event F.
P(E|F) need not equal P(E); comparing the two tells us whether F carries information about E.
The three probability axioms hold for conditional probabilities, giving addition and complement rules.
Conditional probability underlies the multiplication theorem and Bayes' theorem covered later in the chapter.

Multiplication Theorem and Independent Events

Quick answer The multiplication theorem gives P(A∩B) in terms of conditional probabilities, and independence is the special case where knowing one event tells us nothing about the other.

The definition of conditional probability can be rearranged to give the multiplication theorem on probability: for two events A and B with P(A) > 0 and P(B) > 0, P(A∩B) = P(A)·P(B|A) = P(B)·P(A|B). This lets us compute the probability of two events both occurring by multiplying the probability of the first by the conditional probability of the second given that the first has occurred. The theorem extends to three events A, B, C as P(A∩B∩C) = P(A)·P(B|A)·P(C|A∩B), and similarly for more events.

Two events A and B are called independent if the occurrence of one does not affect the probability of the other, i.e. P(A|B) = P(A) and P(B|A) = P(B) whenever these are defined. Substituting into the multiplication theorem gives the standard test for independence: A and B are independent if and only if P(A∩B) = P(A)·P(B). Independence must not be confused with mutually exclusive events — two events of non-zero probability that are mutually exclusive can never be independent, because P(A∩B) = 0 while P(A)·P(B) ≠ 0.

For three or more events, pairwise independence (every pair independent) does not automatically guarantee mutual independence; mutual independence requires the product condition to hold for every combination of two, three, and more events taken together.

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

Let A = "first ball is red" and B = "second ball is red".
P(A) = 5/8.
After one red ball is removed, 4 red and 3 black balls (7 total) remain, so P(B|A) = 4/7.
By the multiplication theorem, P(A∩B) = P(A)·P(B|A) = (5/8)×(4/7) = 20/56 = 5/14.

Multiplication Theorem (two events) P(A∩B) = P(A)·P(B|A) = P(B)·P(A|B) Requires P(A) > 0 and P(B) > 0.

Multiplication Theorem (three events) P(A∩B∩C) = P(A)·P(B|A)·P(C|A∩B) Extends the two-event rule by conditioning on all previously occurred events.

Test for Independence A, B independent ⟺ P(A∩B) = P(A)·P(B) Equivalent to P(A|B) = P(A) and P(B|A) = P(B).

Remember

Multiplication theorem: P(A∩B) = P(A)P(B|A) = P(B)P(A|B), derived directly from the conditional probability definition.
Independent events satisfy P(A∩B) = P(A)P(B); this is the operational test used in problems.
Independence and 'mutually exclusive' are different (and often opposite) ideas for events of positive probability.
Without-replacement draws use the multiplication theorem with a changing conditional probability at each stage.
Pairwise independence of several events does not imply mutual independence.

Total Probability Theorem and Bayes' Theorem

Quick answer When an event can occur through several distinct causes that partition the sample space, the total probability theorem finds its overall probability, and Bayes' theorem reverses this to find the probability of a particular cause given that the event occurred.

A collection of events E1, E2, …, En is called a partition of the sample space S if the events are pairwise mutually exclusive, exhaustive (their union is S), and each has positive probability. Partitions are useful when an event A can occur only "through" one of several distinct, non-overlapping causes or paths.

The Total Probability Theorem states that if E1, E2, …, En form a partition of S and A is any event, then P(A) = P(E1)·P(A|E1) + P(E2)·P(A|E2) + … + P(En)·P(A|En) = Σ P(Eᵢ)·P(A|Eᵢ). This works because A splits into the disjoint pieces A∩E1, A∩E2, …, A∩En, whose probabilities can simply be added.

Bayes' Theorem answers the reverse question: given that A has occurred, what is the probability that it happened via a particular Eᵢ? It states P(Eᵢ|A) = [P(Eᵢ)·P(A|Eᵢ)] / Σⱼ P(Eⱼ)·P(A|Eⱼ), where the denominator is simply P(A) from the total probability theorem. Here P(Eᵢ) is called the prior probability and P(Eᵢ|A) the posterior probability of the "cause" Eᵢ.

Worked Example. Bag I contains 4 red and 4 black balls, and Bag II contains 2 red and 6 black balls. One of the two bags is chosen at random and a ball is drawn from it; the ball turns out to be red. Find the probability that the ball was drawn from Bag I.

Let E1 = "Bag I chosen", E2 = "Bag II chosen", A = "ball drawn is red". P(E1) = P(E2) = 1/2.
P(A|E1) = 4/8 = 1/2 and P(A|E2) = 2/8 = 1/4.
Total probability: P(A) = (1/2)(1/2) + (1/2)(1/4) = 1/4 + 1/8 = 3/8.
By Bayes' theorem, P(E1|A) = [P(E1)P(A|E1)] / P(A) = (1/4)/(3/8) = 2/3.

So having drawn a red ball makes it more likely (probability 2/3) that Bag I was chosen, since Bag I has a higher proportion of red balls.

Total Probability Theorem P(A) = Σᵢ P(Eᵢ)·P(A|Eᵢ), for i = 1 to n E1, E2, …, En must form a partition of the sample space.

Bayes' Theorem P(Eᵢ|A) = [P(Eᵢ)·P(A|Eᵢ)] / Σⱼ P(Eⱼ)·P(A|Eⱼ) Gives the posterior probability of cause Eᵢ given that A has occurred.

Remember

A partition splits the sample space into mutually exclusive, exhaustive events, each of positive probability.
Total probability theorem sums P(Eᵢ)P(A|Eᵢ) over the partition to give the overall probability of A.
Bayes' theorem 'reverses the conditioning' to find the probability of a cause given an observed effect.
The denominator in Bayes' theorem is always just P(A), computed via the total probability theorem.
Prior probabilities P(Eᵢ) are updated to posterior probabilities P(Eᵢ|A) once evidence A is observed.

Random Variables and Probability Distributions

Quick answer A random variable assigns a number to every outcome of a random experiment, and its probability distribution lists every possible value together with the probability of occurrence.

A random variable is a real-valued function defined on the sample space of a random experiment; it assigns a number to every outcome. For example, if two coins are tossed, the number of heads obtained is a random variable that can take the values 0, 1, or 2 depending on the outcome. This chapter deals with discrete random variables, which take a finite (or countable) set of values.

The probability distribution of a random variable X is the complete list of its possible values x1, x2, …, xn together with the corresponding probabilities p1, p2, …, pn = P(X = xi). A valid probability distribution must satisfy two conditions: each pᵢ ≥ 0, and the probabilities must add up to exactly 1, i.e. Σ pᵢ = 1. These conditions are used both to check whether a claimed distribution is legitimate and to find an unknown probability in a table using the fact that the total must equal 1.

Worked Example. Two fair coins are tossed together. Let X denote the number of heads obtained. Find the probability distribution of X.

Sample space: {HH, HT, TH, TT}, each outcome with probability 1/4.
X = 0 (no heads): only TT, so P(X=0) = 1/4.
X = 1 (one head): HT or TH, so P(X=1) = 2/4 = 1/2.
X = 2 (two heads): only HH, so P(X=2) = 1/4.

Check: 1/4 + 1/2 + 1/4 = 1, so this is a valid probability distribution. This same distribution is used in the next section to compute the mean and variance of X.

Probability Distribution Validity pᵢ ≥ 0 for all i, and Σᵢ pᵢ = 1 Both conditions must hold for a table of values to be a genuine probability distribution.

Remember

A random variable maps each outcome of a random experiment to a real number.
Its probability distribution lists every possible value along with the probability of taking that value.
Every valid probability distribution must have non-negative probabilities that sum to exactly 1.
Probability distributions are usually presented as a table with X-values in one row and probabilities in the next.
The condition Σpᵢ = 1 is commonly used to find a missing probability in an otherwise-given distribution.

Mean and Variance of a Random Variable

Quick answer The mean (expectation) gives the long-run average value of a random variable, while the variance and standard deviation measure how spread out its values typically are.

Just as a data set has a mean and variance, a random variable's probability distribution has a theoretical mean, called its expectation or expected value, denoted E(X) or μ. It is defined as E(X) = Σ xᵢpᵢ, a probability-weighted average of the values X can take. E(X) represents the long-run average value of X if the experiment were repeated a very large number of times.

The variance of X measures how spread out the distribution is around the mean, defined as Var(X) = Σ(xᵢ − μ)²pᵢ, which simplifies to (and is normally computed using) the shortcut formula Var(X) = E(X²) − [E(X)]², where E(X²) = Σxᵢ²pᵢ. The non-negative square root of the variance is the standard deviation, SD(X) = √Var(X), expressed in the same units as X.

Worked Example. Using the probability distribution of X = number of heads in two coin tosses from the previous section (X = 0, 1, 2 with probabilities 1/4, 1/2, 1/4), find the mean and variance of X.

E(X) = 0×(1/4) + 1×(1/2) + 2×(1/4) = 0 + 1/2 + 1/2 = 1.
E(X²) = 0²×(1/4) + 1²×(1/2) + 2²×(1/4) = 0 + 1/2 + 1 = 3/2.
Var(X) = E(X²) − [E(X)]² = 3/2 − 1² = 3/2 − 1 = 1/2.
SD(X) = √(1/2) = 1/√2 ≈ 0.707.

So on average, two coin tosses give 1 head, with a variance of 0.5 describing how much the actual count typically deviates from this average.

Mean / Expectation E(X) = Σᵢ xᵢ·pᵢ Probability-weighted average of the values of X; also written μ.

Variance (computational form) Var(X) = E(X²) − [E(X)]², where E(X²) = Σᵢ xᵢ²pᵢ Standard shortcut used instead of the direct definition Σ(xᵢ−μ)²pᵢ.

Standard Deviation SD(X) = √Var(X) Same units as X; the positive square root of the variance.

Remember

Mean/Expectation E(X) = Σxᵢpᵢ is the probability-weighted average of a random variable.
Variance measures spread around the mean; the shortcut formula E(X²) − [E(X)]² is easier to compute than the direct definition.
Standard deviation SD(X) = √Var(X) is in the same units as X, unlike variance.
E(X²) is calculated as Σxᵢ²pᵢ, not as [E(X)]² — these are commonly confused.
These two numbers (mean and variance) summarise the location and spread of an entire probability distribution.

Bernoulli Trials and Binomial Distribution

Quick answer Bernoulli trials are repeated, independent trials with exactly two outcomes and a constant success probability; the number of successes in such trials follows the binomial distribution.

Trials of a random experiment are called Bernoulli trials if: (i) there is a finite number of trials, (ii) the trials are independent of one another, (iii) each trial has exactly two possible outcomes, conventionally labelled "success" and "failure", and (iv) the probability of success, denoted p, remains the same in every trial (with q = 1 − p the probability of failure). Repeatedly tossing a coin, or repeatedly checking items from a large batch with a fixed defect rate, are typical examples.

If X denotes the number of successes in n Bernoulli trials, X follows a binomial distribution, and the probability of getting exactly r successes is given by P(X = r) = ⁿCᵣ pʳ qⁿ⁻ʳ, for r = 0, 1, 2, …, n. This is often written as a binomial distribution with parameters n and p, B(n, p). For a binomial distribution, the mean is Mean = np and the variance is Variance = npq; since q < 1, the variance of a binomial distribution is always less than its mean.

Worked Example 1. A fair coin is tossed 6 times. Find the probability of getting exactly 4 heads.

Here n = 6, p = q = 1/2, and r = 4.
P(X=4) = ⁶C₄ (1/2)⁴(1/2)² = 15 × (1/2)⁶ = 15/64.

Worked Example 2. Find the mean and variance of the number of heads in 10 tosses of a fair coin.

n = 10, p = q = 1/2.
Mean = np = 10 × 1/2 = 5.
Variance = npq = 10 × 1/2 × 1/2 = 2.5.

Binomial Probability Formula P(X=r) = ⁿCᵣ · pʳ · qⁿ⁻ʳ, r = 0, 1, …, n q = 1 − p; n is the number of Bernoulli trials.

Mean of Binomial Distribution Mean = n·p Expected number of successes in n trials.

Variance of Binomial Distribution Variance = n·p·q Always less than the mean np, since q < 1.

Remember

Bernoulli trials require a fixed number of independent trials, exactly two outcomes each, and a constant success probability p.
The binomial probability formula is P(X=r) = ⁿCᵣpʳqⁿ⁻ʳ, where q = 1 − p.
Mean of a binomial distribution = np; variance = npq, and variance is always less than the mean since q < 1.
Binomial distribution problems typically involve repeated independent trials such as coin tosses, dice throws, or fixed defect-rate sampling.
The sum of P(X=r) over r = 0 to n always equals 1, matching the binomial expansion of (q+p)ⁿ.

The formula sheet

Every formula in this chapter, in one place — screenshot it before your exam.

P(E|F) = P(E∩F) / P(F), provided P(F) > 0

Conditional Probability of E given F

P(E∩F) = P(F) · P(E|F)

Multiplication Rule (rearranged form)

P(E′|F) = 1 − P(E|F)

Complement Rule under Conditioning

P((A∪B)|F) = P(A|F) + P(B|F) − P((A∩B)|F)

Addition Rule under Conditioning

P(A∩B) = P(A)·P(B|A) = P(B)·P(A|B)

Multiplication Theorem (two events)

P(A∩B∩C) = P(A)·P(B|A)·P(C|A∩B)

Multiplication Theorem (three events)

A, B independent ⟺ P(A∩B) = P(A)·P(B)

Test for Independence

P(A) = Σᵢ P(Eᵢ)·P(A|Eᵢ), for i = 1 to n

Total Probability Theorem

P(Eᵢ|A) = [P(Eᵢ)·P(A|Eᵢ)] / Σⱼ P(Eⱼ)·P(A|Eⱼ)

Bayes' Theorem

pᵢ ≥ 0 for all i, and Σᵢ pᵢ = 1

Probability Distribution Validity

E(X) = Σᵢ xᵢ·pᵢ

Mean / Expectation

Var(X) = E(X²) − [E(X)]², where E(X²) = Σᵢ xᵢ²pᵢ

Variance (computational form)

SD(X) = √Var(X)

Standard Deviation

P(X=r) = ⁿCᵣ · pʳ · qⁿ⁻ʳ, r = 0, 1, …, n

Binomial Probability Formula

Mean = n·p

Mean of Binomial Distribution

Variance = n·p·q

Variance of Binomial Distribution

Test yourself

Tap an answer to check it instantly — you'll see why it's right, and what to revise if it isn't.

0 correct · 0/12 answered

Q1 Conditional Probability easy

What is the correct definition of conditional probability P(A|B)?

Q2 Independent Events easy

If A and B are independent events, then P(A∩B) equals

Q3 Random Variable & Probability Distribution easy

For any valid discrete probability distribution of a random variable X, the sum of all the probabilities pᵢ must equal

Q4 Conditional Probability medium

A fair die is thrown once. Let E = {2, 4, 6} and F = {4, 5, 6}. What is P(E|F)?

Q5 Conditional Probability medium

Two dice are thrown. What is the probability that the sum of the numbers is 8, given that at least one die shows a 5?

Q6 Multiplication Theorem medium

A bag has 5 red and 3 black balls. Two balls are drawn one after another without replacement. What is the probability that both are red?

Q7 Conditional Probability medium

A card is drawn from a well-shuffled deck of 52 cards. Given that the card drawn is a face card, what is the probability that it is a king?

Q8 Bayes' Theorem hard

Bag I has 4 red and 4 black balls; Bag II has 2 red and 6 black balls. A bag is chosen at random and a ball drawn from it turns out to be red. What is the probability that it came from Bag I?

Q9 Mean of a Random Variable medium

Two fair coins are tossed. If X denotes the number of heads, what is E(X)?

Q10 Variance of a Random Variable hard

For the same distribution (X = number of heads in two coin tosses), what is Var(X)?

Q11 Binomial Distribution hard

A fair coin is tossed 6 times. What is the probability of getting exactly 4 heads?

Q12 Binomial Distribution hard

For a binomial distribution with n = 10 and p = 0.4, what is the variance?

NCERT solutions & previous-year questions

Step-by-step model answers — tap a question to reveal the full solution.

NCERT questions 6

1 If P(A) = 0.6, P(B) = 0.3 and P(A∩B) = 0.2, find P(A|B) and P(B|A).Conditional Probability

We use the definition of conditional probability, P(X|Y) = P(X∩Y)/P(Y).

Finding P(A|B):

P(A|B) = P(A∩B)/P(B) = 0.2/0.3 = 2/3 ≈ 0.667.

Finding P(B|A):

P(B|A) = P(A∩B)/P(A) = 0.2/0.6 = 1/3 ≈ 0.333.

2 A die is thrown three times. E is the event '4 appears on the third throw' and F is the event '6 and 5 appear respectively on the first two throws'. Find P(E|F).Conditional Probability

F consists of all outcomes where the first throw is 6, the second throw is 5, and the third throw can be anything from 1 to 6. So F has 6 equally likely outcomes: (6,5,1), (6,5,2), (6,5,3), (6,5,4), (6,5,5), (6,5,6).

E∩F is the outcome where, additionally, the third throw is 4, i.e. only (6,5,4) — exactly 1 outcome out of the 6 in F.

P(E|F) = n(E∩F)/n(F) = 1/6.

This matches the unconditional probability of getting a 4 on the third throw, which makes sense because the three throws of a die are independent — what happened on the first two throws does not affect the third.

3 A pair of dice is thrown. Find the probability of getting a sum of 9, given that the numbers appearing on the two dice are different.Conditional Probability

Let F = "the two numbers appearing are different" and E = "the sum of the numbers is 9".

Total outcomes = 36. Outcomes where both dice show the same number: (1,1), (2,2), …, (6,6) — 6 outcomes. So outcomes with different numbers, n(F) = 36 − 6 = 30, giving P(F) = 30/36 = 5/6.

Outcomes with sum 9: (3,6), (6,3), (4,5), (5,4) — 4 outcomes, all of which already have different numbers on the two dice, so n(E∩F) = 4 and P(E∩F) = 4/36 = 1/9.

P(E|F) = P(E∩F)/P(F) = (1/9)/(5/6) = 6/45 = 2/15.

4 Three machines E1, E2, E3 in a factory produce 30%, 45% and 25% of the total items respectively. Of their outputs, 2%, 3% and 2% respectively are defective. An item is drawn at random from the total output and is found defective. Find the probability that it was produced by machine E2.Bayes' Theorem

Let D = "item is defective". Given: P(E1) = 0.30, P(E2) = 0.45, P(E3) = 0.25, and P(D|E1) = 0.02, P(D|E2) = 0.03, P(D|E3) = 0.02.

Step 1: Total probability of D.

P(D) = P(E1)P(D|E1) + P(E2)P(D|E2) + P(E3)P(D|E3)
= (0.30×0.02) + (0.45×0.03) + (0.25×0.02)
= 0.0060 + 0.0135 + 0.0050 = 0.0245.

Step 2: Apply Bayes' theorem.

P(E2|D) = [P(E2)P(D|E2)] / P(D) = 0.0135 / 0.0245 = 135/245 = 27/49 ≈ 0.551.

So there is roughly a 55.1% chance that a randomly selected defective item came from machine E2.

5 Find the probability distribution of the number of heads in two tosses of a fair coin.Random Variable & Probability Distribution

Let X = number of heads. The sample space of two coin tosses is {HH, HT, TH, TT}, each outcome equally likely with probability 1/4.

X = 0 corresponds to TT only: P(X=0) = 1/4.
X = 1 corresponds to HT or TH: P(X=1) = 2/4 = 1/2.
X = 2 corresponds to HH only: P(X=2) = 1/4.

The probability distribution table is:

X : 0, 1, 2
P(X) : 1/4, 1/2, 1/4

Check: 1/4 + 1/2 + 1/4 = 1, confirming this is a valid probability distribution.

6 Find the probability of getting the number 5 exactly twice in 6 throws of a fair die.Bernoulli Trials & Binomial Distribution

Getting a 5 on a single throw is a "success" with p = 1/6, and "failure" (not getting a 5) has q = 1 − 1/6 = 5/6. The 6 throws are independent Bernoulli trials, so X = number of times 5 appears follows a binomial distribution with n = 6.

We need P(X = 2):

P(X=2) = ⁶C₂ (1/6)² (5/6)⁴

⁶C₂ = 15, (1/6)² = 1/36, (5/6)⁴ = 625/1296.

P(X=2) = 15 × (1/36) × (625/1296) = 9375/46656 = 3125/15552 ≈ 0.201.

Previous-year board questions 4

Q1 If A and B are two events such that P(A) = 0.4, P(B) = 0.8 and P(B|A) = 0.6, find P(A|B) and P(A∪B). 2023 2 marks

By the multiplication theorem, P(A∩B) = P(A)×P(B|A) = 0.4 × 0.6 = 0.24.

Finding P(A|B): P(A|B) = P(A∩B)/P(B) = 0.24/0.8 = 0.3.

Finding P(A∪B): P(A∪B) = P(A) + P(B) − P(A∩B) = 0.4 + 0.8 − 0.24 = 0.96.

Q2 Two cards are drawn successively, without replacement, from a well-shuffled deck of 52 playing cards. Find the probability distribution of the number of aces obtained, and hence find the mean of the distribution. 2022 5 marks

Let X = number of aces in the 2 cards drawn; X can be 0, 1, or 2. Total ways to draw 2 cards from 52 = C(52,2) = 1326.

P(X=0): both cards non-ace, chosen from 48 non-ace cards: C(48,2)/C(52,2) = 1128/1326 = 188/221.

P(X=1): one ace (from 4) and one non-ace (from 48): [C(4,1)×C(48,1)]/C(52,2) = (4×48)/1326 = 192/1326 = 32/221.

P(X=2): both cards aces: C(4,2)/C(52,2) = 6/1326 = 1/221.

Check: 188/221 + 32/221 + 1/221 = 221/221 = 1.

Probability distribution:

X : 0, 1, 2
P(X) : 188/221, 32/221, 1/221

Mean: E(X) = 0×(188/221) + 1×(32/221) + 2×(1/221) = (32+2)/221 = 34/221 = 2/13.

Q3 A bag contains 4 white and 6 black balls. Two balls are drawn one after the other without replacement. Find the probability that the two balls drawn are of different colours. 2023 3 marks

Let the two balls be drawn in order. "Different colours" happens in exactly two mutually exclusive ways: (white then black) or (black then white).

P(White then Black): P(W₁) = 4/10; after removing 1 white ball, 6 black balls remain out of 9, so P(B₂|W₁) = 6/9. By the multiplication theorem: (4/10)×(6/9) = 24/90.

P(Black then White): P(B₁) = 6/10; after removing 1 black ball, 4 white balls remain out of 9, so P(W₂|B₁) = 4/9. By the multiplication theorem: (6/10)×(4/9) = 24/90.

Since these two cases are mutually exclusive, add them: P(different colours) = 24/90 + 24/90 = 48/90 = 8/15.

Q4 A person has undertaken a construction job. The probability that there will be a strike is 0.65. The probability that the construction job will be completed on time if there is no strike is 0.80, and the probability of completion on time if there is a strike is 0.32. Determine the probability that the construction job will be completed on time. 2020 5 marks

Let A = "there is a strike", so P(A) = 0.65 and P(A′) = 1 − 0.65 = 0.35. Let T = "job completed on time". Given P(T|A′) = 0.80 and P(T|A) = 0.32.

Since A and A′ form a partition of the sample space, by the Total Probability Theorem:

P(T) = P(A)×P(T|A) + P(A′)×P(T|A′)
= (0.65 × 0.32) + (0.35 × 0.80)
= 0.208 + 0.280 = 0.488.

So the probability that the construction job will be completed on time is 0.488.

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