Chapter 14: Probability

The sample space for throwing a die is $$S = \{1, 2, 3, 4, 5, 6\}$$.

Event $$A$$ consists of those outcomes that are numbers less than $$7$$. Every number on a die ($$1$$ to $$6$$) is less than $$7$$, so $$A$$ contains every outcome of $$S$$.

$$A = \{1, 2, 3, 4, 5, 6\}$$

The sample space is $$S = \{1, 2, 3, 4, 5, 6\}$$.

Event $$B$$ consists of those outcomes that are numbers greater than $$7$$. No number on a die is greater than $$7$$, so no outcome of $$S$$ qualifies.

$$B = \varnothing$$

Thus $$B$$ is an impossible event.

The sample space is $$S = \{1, 2, 3, 4, 5, 6\}$$.

Event $$C$$ consists of the outcomes that are multiples of $$3$$. Among $$1, 2, 3, 4, 5, 6$$ the multiples of $$3$$ are $$3$$ and $$6$$.

$$C = \{3, 6\}$$

The sample space is $$S = \{1, 2, 3, 4, 5, 6\}$$.

Event $$D$$. $$D$$ consists of the outcomes that are numbers less than $$4$$, namely $$1$$, $$2$$ and $$3$$.

$$D = \{1, 2, 3\}$$

This question also asks us to find several combinations of the events, so we collect all six events described in parts (i)-(vi):

$$A = \{1, 2, 3, 4, 5, 6\}, \quad B = \varnothing, \quad C = \{3, 6\}$$

$$D = \{1, 2, 3\}, \quad E = \{6\}, \quad F = \{3, 4, 5, 6\}$$

Recall that $$\cup$$ collects all listed outcomes, $$\cap$$ keeps the common ones, and $$X - Y$$ keeps the outcomes of $$X$$ that are not in $$Y$$.

$$A \cup B = \{1, 2, 3, 4, 5, 6\} \cup \varnothing = \{1, 2, 3, 4, 5, 6\}$$

$$A \cap B = \{1, 2, 3, 4, 5, 6\} \cap \varnothing = \varnothing$$

$$B \cup C = \varnothing \cup \{3, 6\} = \{3, 6\}$$

$$E \cap F = \{6\} \cap \{3, 4, 5, 6\} = \{6\}$$

$$D \cap E = \{1, 2, 3\} \cap \{6\} = \varnothing$$

$$A - C = \{1, 2, 3, 4, 5, 6\} - \{3, 6\} = \{1, 2, 4, 5\}$$

$$D - E = \{1, 2, 3\} - \{6\} = \{1, 2, 3\}$$

For the last two we first need $$F'$$, the complement of $$F$$ in $$S$$:

$$F' = S - F = \{1, 2, 3, 4, 5, 6\} - \{3, 4, 5, 6\} = \{1, 2\}$$

$$E \cap F' = \{6\} \cap \{1, 2\} = \varnothing$$

The sample space is $$S = \{1, 2, 3, 4, 5, 6\}$$.

Event $$E$$ consists of outcomes that are both even and greater than $$4$$. The even numbers on a die are $$2, 4, 6$$; of these, only $$6$$ is greater than $$4$$.

$$E = \{6\}$$

The sample space is $$S = \{1, 2, 3, 4, 5, 6\}$$.

'Not less than $$3$$' means $$\ge 3$$. Event $$F$$ therefore consists of the outcomes $$3, 4, 5, 6$$.

$$F = \{3, 4, 5, 6\}$$

When three coins are tossed the sample space is

$$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$$

The four events as sets of outcomes:

$$A = \{HHH\} \qquad \text{(three heads)}$$

$$B = \{HHT, HTH, THH\} \qquad \text{(two heads, one tail)}$$

$$C = \{TTT\} \qquad \text{(three tails)}$$

$$D = \{HHH, HHT, HTH, HTT\} \qquad \text{(head on the first coin)}$$

Two events are mutually exclusive when their intersection is empty. Checking every pair:

$$A \cap B = \varnothing, \quad A \cap C = \varnothing, \quad B \cap C = \varnothing, \quad C \cap D = \varnothing$$

$$A \cap D = \{HHH\} \ne \varnothing, \quad B \cap D = \{HHT, HTH\} \ne \varnothing$$

So the mutually exclusive pairs are $$A$$ & $$B$$, $$A$$ & $$C$$, $$B$$ & $$C$$, and $$C$$ & $$D$$.

A simple event is an event that consists of exactly one outcome (a single sample point).

$$A = \{HHH\}$$ — one outcome, so $$A$$ is simple.

$$C = \{TTT\}$$ — one outcome, so $$C$$ is simple.

$$B = \{HHT, HTH, THH\}$$ has three outcomes and $$D = \{HHH, HHT, HTH, HTT\}$$ has four, so they are not simple.

A compound event is an event that consists of more than one outcome.

$$B = \{HHT, HTH, THH\}$$ has three outcomes, so $$B$$ is compound.

$$D = \{HHH, HHT, HTH, HTT\}$$ has four outcomes, so $$D$$ is compound.

($$A$$ and $$C$$ have only one outcome each, so they are simple, not compound.)

When three coins are tossed, the sample space is

$$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$$

(There are many valid answers; one such pair is given.)

Let $$A$$: 'getting at least two heads' and $$B$$: 'getting at least two tails'. Then

$$A = \{HHH, HHT, HTH, THH\}, \qquad B = \{HTT, THT, TTH, TTT\}$$

Since $$A \cap B = \varnothing$$, the events $$A$$ and $$B$$ cannot occur together, so they are mutually exclusive.

Using the sample space $$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$$, take

$$A$$: 'no head appears' $$= \{TTT\}$$

$$B$$: 'exactly one head appears' $$= \{HTT, THT, TTH\}$$

$$C$$: 'at least two heads appear' $$= \{HHH, HHT, HTH, THH\}$$

These are mutually exclusive because they are pairwise disjoint:

$$A \cap B = \varnothing, \quad B \cap C = \varnothing, \quad A \cap C = \varnothing$$

They are exhaustive because their union is the whole sample space:

$$A \cup B \cup C = S$$

Using the sample space $$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$$, take

$$A$$: 'a head on the first coin' $$= \{HHH, HHT, HTH, HTT\}$$

$$B$$: 'a head on the second coin' $$= \{HHH, HHT, THH, THT\}$$

Their intersection is

$$A \cap B = \{HHH, HHT\} \ne \varnothing$$

Since they have outcomes in common, both events can occur together, so $$A$$ and $$B$$ are not mutually exclusive.

Using the sample space $$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$$, take

$$A$$: 'getting three heads' $$= \{HHH\}$$

$$B$$: 'getting three tails' $$= \{TTT\}$$

They are mutually exclusive since $$A \cap B = \varnothing$$.

But they are not exhaustive, because their union

$$A \cup B = \{HHH, TTT\} \ne S$$

misses outcomes such as $$HHT$$.

Using the sample space $$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$$, take

$$A$$: 'getting three heads' $$= \{HHH\}$$

$$B$$: 'getting three tails' $$= \{TTT\}$$

$$C$$: 'getting exactly one head' $$= \{HTT, THT, TTH\}$$

They are mutually exclusive since they are pairwise disjoint:

$$A \cap B = \varnothing, \quad B \cap C = \varnothing, \quad A \cap C = \varnothing$$

They are not exhaustive, because

$$A \cup B \cup C = \{HHH, TTT, HTT, THT, TTH\} \ne S$$

(the outcomes $$HHT, HTH, THH$$ are left out).

When two dice are thrown the sample space has $$36$$ ordered pairs. The events are:

$$A$$: even number on the first die — first die $$\in \{2, 4, 6\}$$;

$$B$$: odd number on the first die — first die $$\in \{1, 3, 5\}$$;

$$C$$: sum of the numbers $$\le 5$$.

$$A'$$ is the complement of $$A$$: it occurs when the first die does not show an even number, i.e. it shows an odd number. So

$$A' = \{(1,y),(3,y),(5,y) : y = 1,2,3,4,5,6\}$$

That is, $$A'$$ is exactly the event 'an odd number appears on the first die', which is $$B$$.

'not $$B$$' is the complement $$B'$$. The event $$B$$ is 'an odd number on the first die', so $$B'$$ occurs when the first die does not show an odd number, i.e. it shows an even number.

$$B' = \{(2,y),(4,y),(6,y) : y = 1,2,3,4,5,6\}$$

This is exactly the event 'an even number appears on the first die', which is $$A$$.

'$$A$$ or $$B$$' is the union $$A \cup B$$. Event $$A$$ occurs when the first die is even and $$B$$ occurs when the first die is odd.

Since the number on the first die is always either even or odd, every outcome of the experiment belongs to $$A$$ or to $$B$$. Therefore

$$A \cup B = S$$

i.e. $$A \cup B$$ is the whole sample space (all $$36$$ outcomes) — a sure event.

'$$A$$ and $$B$$' is the intersection $$A \cap B$$. Event $$A$$ needs an even number on the first die and $$B$$ needs an odd number on the first die.

The first die cannot show a number that is both even and odd, so no outcome can belong to both events.

$$A \cap B = \varnothing$$

Hence $$A$$ and $$B$$ are mutually exclusive.

'$$A$$ but not $$C$$' is the difference $$A - C = A \cap C'$$: an even number on the first die and a sum greater than $$5$$.

First list $$C$$ (sum $$\le 5$$):

$$C = \{(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(3,1),(3,2),(4,1)\}$$

The outcomes of $$C$$ that have an even first die are $$(2,1),(2,2),(2,3),(4,1)$$. Removing these from $$A$$ leaves the outcomes with an even first die and sum $$> 5$$:

$$A - C = \{(2,4),(2,5),(2,6),(4,2),(4,3),(4,4),(4,5),(4,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}$$

This event has $$14$$ outcomes.

'$$B$$ or $$C$$' is the union $$B \cup C$$: an odd number on the first die or a sum $$\le 5$$.

$$B$$ contains all $$18$$ outcomes with an odd first die. $$C$$ adds the outcomes with sum $$\le 5$$; those with an odd first die are already in $$B$$, so we only add the ones from $$C$$ with an even first die, namely $$(2,1),(2,2),(2,3),(4,1)$$.

$$B \cup C = \{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\}$$

This event has $$18 + 4 = 22$$ outcomes.

'$$B$$ and $$C$$' is the intersection $$B \cap C$$: an odd number on the first die and a sum $$\le 5$$.

From $$C = \{(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(3,1),(3,2),(4,1)\}$$, keep only the outcomes whose first die is odd ($$1$$, $$3$$ or $$5$$):

$$B \cap C = \{(1,1),(1,2),(1,3),(1,4),(3,1),(3,2)\}$$

This event has $$6$$ outcomes.

From part (ii), $$B' = A$$ (an even number on the first die). Therefore

$$A \cap B' \cap C' = A \cap A \cap C' = A \cap C'$$

which is just '$$A$$ but not $$C$$' — an even number on the first die and a sum greater than $$5$$. From part (v),

$$A \cap B' \cap C' = \{(2,4),(2,5),(2,6),(4,2),(4,3),(4,4),(4,5),(4,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}$$

This event has $$14$$ outcomes.

Recall $$A$$: even number on the first die, $$B$$: odd number on the first die.

The first die cannot be both even and odd, so $$A$$ and $$B$$ have no common outcome:

$$A \cap B = \varnothing$$

Hence $$A$$ and $$B$$ are mutually exclusive. The statement is true.

From part (i), $$A \cap B = \varnothing$$, so $$A$$ and $$B$$ are mutually exclusive.

Also, the first die is always either even or odd, so every outcome lies in $$A$$ or in $$B$$:

$$A \cup B = S$$

Since the events are mutually exclusive and their union is the whole sample space, they are also exhaustive. The statement is true.

$$B'$$ is the complement of $$B$$ (odd number on the first die), so $$B'$$ occurs when the first die is not odd, i.e. when it is even.

But 'an even number on the first die' is exactly the event $$A$$. Hence

$$B' = A$$

The statement is true.

$$A$$: even number on the first die; $$C$$: sum $$\le 5$$.

Consider the outcome $$(2, 1)$$: the first die shows $$2$$ (even), so it is in $$A$$; and the sum is $$3 \le 5$$, so it is in $$C$$. In fact

$$A \cap C = \{(2,1),(2,2),(2,3),(4,1)\} \ne \varnothing$$

Since the events have common outcomes, they are not mutually exclusive. The statement is false.

From part (iii), $$B' = A$$. Therefore

$$A \cap B' = A \cap A = A \ne \varnothing$$

An event always shares outcomes with itself, so $$A$$ and $$B'$$ are not mutually exclusive. The statement is false.

Here $$A' = B$$ (odd first die) and $$B' = A$$ (even first die), so the three events are really $$B$$, $$A$$ and $$C$$.

For all three to be mutually exclusive, every pair must be disjoint. But

$$A' \cap C = B \cap C = \{(1,1),(1,2),(1,3),(1,4),(3,1),(3,2)\} \ne \varnothing$$

Since $$A'$$ and $$C$$ are not disjoint, the three events are not mutually exclusive. (For instance the outcome $$(1,1)$$ lies in both $$A'$$ and $$C$$.) The statement is false.

The sample space for throwing a die is $$S = \{1, 2, 3, 4, 5, 6\}$$, with $$6$$ equally likely outcomes.

The prime numbers among $$1$$ to $$6$$ are $$2, 3, 5$$, so the event is

$$E = \{2, 3, 5\} \quad (3 \text{ outcomes})$$

$$P(E) = \frac{3}{6} = \frac{1}{2}$$

From $$S = \{1, 2, 3, 4, 5, 6\}$$, the numbers that are $$\ge 3$$ are $$3, 4, 5, 6$$:

$$E = \{3, 4, 5, 6\} \quad (4 \text{ outcomes})$$

$$P(E) = \frac{4}{6} = \frac{2}{3}$$

From $$S = \{1, 2, 3, 4, 5, 6\}$$, the only number that is $$\le 1$$ is $$1$$:

$$E = \{1\} \quad (1 \text{ outcome})$$

$$P(E) = \frac{1}{6}$$

From $$S = \{1, 2, 3, 4, 5, 6\}$$, no number is more than $$6$$, so the event has no favourable outcome:

$$E = \varnothing \quad (0 \text{ outcomes})$$

$$P(E) = \frac{0}{6} = 0$$

This is an impossible event.

From $$S = \{1, 2, 3, 4, 5, 6\}$$, the numbers less than $$6$$ are $$1, 2, 3, 4, 5$$:

$$E = \{1, 2, 3, 4, 5\} \quad (5 \text{ outcomes})$$

$$P(E) = \frac{5}{6}$$

One card is selected from a pack of $$52$$ cards. Each card is a possible outcome, and selecting any one of them is a distinct sample point.

Hence the number of points (outcomes) in the sample space is $$52$$.

There are $$52$$ equally likely cards, and exactly one of them is the ace of spades.

$$P(\text{ace of spades}) = \frac{1}{52}$$

(i) An ace. A deck contains $$4$$ aces (one in each suit).

$$P(\text{an ace}) = \frac{4}{52} = \frac{1}{13}$$

(ii) A black card. The black cards are the $$13$$ clubs and the $$13$$ spades, giving $$13 + 13 = 26$$ black cards.

$$P(\text{black card}) = \frac{26}{52} = \frac{1}{2}$$

When three coins are tossed, the sample space is

$$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$$

with $$8$$ equally likely outcomes.

The event '$$3$$ heads' is $$\{HHH\}$$, with $$1$$ favourable outcome.

$$P(3 \text{ heads}) = \frac{1}{8}$$

From $$S$$ (the $$8$$ outcomes of tossing three coins), the outcomes with exactly $$2$$ heads are

$$\{HHT, HTH, THH\} \quad (3 \text{ outcomes})$$

$$P(2 \text{ heads}) = \frac{3}{8}$$

'At least $$2$$ heads' means $$2$$ heads or $$3$$ heads. From $$S$$ these outcomes are

$$\{HHT, HTH, THH, HHH\} \quad (4 \text{ outcomes})$$

$$P(\text{at least } 2 \text{ heads}) = \frac{4}{8} = \frac{1}{2}$$

'At most $$2$$ heads' means $$0$$, $$1$$ or $$2$$ heads — that is, anything except $$3$$ heads. It is easiest to use the complement.

$$P(\text{at most } 2 \text{ heads}) = 1 - P(3 \text{ heads}) = 1 - \frac{1}{8} = \frac{7}{8}$$

'No head' means all three coins show tails. From $$S$$ this is the single outcome

$$\{TTT\} \quad (1 \text{ outcome})$$

$$P(\text{no head}) = \frac{1}{8}$$

'$$3$$ tails' means all three coins show tails. From $$S$$ this is the single outcome

$$\{TTT\} \quad (1 \text{ outcome})$$

$$P(3 \text{ tails}) = \frac{1}{8}$$

When three coins are tossed once, each coin can land in $$2$$ ways, so the sample space has $$2^3 = 8$$ equally likely outcomes:

$$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$$

The outcomes with exactly $$2$$ tails (and hence exactly $$1$$ head) are

$$\{HTT, THT, TTH\} \quad (3 \text{ outcomes})$$

$$P(\text{exactly two tails}) = \frac{\text{favourable outcomes}}{\text{total outcomes}} = \frac{3}{8}$$

'No tail' means all three coins show heads. From $$S$$ this is the single outcome

$$\{HHH\} \quad (1 \text{ outcome})$$

$$P(\text{no tail}) = \frac{1}{8}$$

'At most $$2$$ tails' means $$0$$, $$1$$ or $$2$$ tails — everything except $$3$$ tails. Using the complement,

$$P(\text{at most } 2 \text{ tails}) = 1 - P(3 \text{ tails}) = 1 - \frac{1}{8} = \frac{7}{8}$$

The probabilities are consistently defined only if they obey the basic rules. One such rule: since $$A \cap B$$ is contained in $$A$$,

$$P(A \cap B) \le P(A)$$

Here $$P(A \cap B) = 0.6$$ while $$P(A) = 0.5$$, so

$$P(A \cap B) = 0.6 > 0.5 = P(A)$$

This violates the rule $$P(A \cap B) \le P(A)$$. Hence the probabilities are not consistently defined.

Using the addition rule $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$, solve for $$P(A \cap B)$$:

$$P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.5 + 0.4 - 0.8 = 0.1$$

For the data to be consistent, $$P(A \cap B)$$ must satisfy $$P(A \cap B) \le P(A)$$ and $$P(A \cap B) \le P(B)$$ (and be non-negative). Here

$$0 \le P(A \cap B) = 0.1 \le 0.4 = P(B) \le 0.5 = P(A)$$

All the conditions hold, so the probabilities are consistently defined.

By the addition theorem of probability,

$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

Substituting the given values:

$$P(A \cup B) = \frac{1}{3} + \frac{1}{5} - \frac{1}{15}$$

Taking the common denominator $$15$$:

$$P(A \cup B) = \frac{5}{15} + \frac{3}{15} - \frac{1}{15} = \frac{5 + 3 - 1}{15} = \frac{7}{15}$$

Start from the addition theorem:

$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

Rearranging to make $$P(B)$$ the subject:

$$P(B) = P(A \cup B) + P(A \cap B) - P(A)$$

Substituting the given values:

$$P(B) = 0.6 + 0.25 - 0.35 = 0.5$$

Start from the addition theorem:

$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

Rearranging to make $$P(A \cap B)$$ the subject:

$$P(A \cap B) = P(A) + P(B) - P(A \cup B)$$

Substituting the given values:

$$P(A \cap B) = 0.5 + 0.35 - 0.7 = 0.15$$

We are given $$P(E) = \frac{1}{4}$$, $$P(F) = \frac{1}{2}$$ and $$P(E \text{ and } F) = P(E \cap F) = \frac{1}{8}$$.

By the addition theorem,

$$P(E \text{ or } F) = P(E \cup F) = P(E) + P(F) - P(E \cap F)$$

$$P(E \cup F) = \frac{1}{4} + \frac{1}{2} - \frac{1}{8}$$

Taking the common denominator $$8$$:

$$P(E \cup F) = \frac{2}{8} + \frac{4}{8} - \frac{1}{8} = \frac{5}{8}$$

'not $$E$$ and not $$F$$' is the event $$E' \cap F'$$. By De Morgan's law,

$$E' \cap F' = (E \cup F)'$$

so

$$P(\text{not } E \text{ and not } F) = P((E \cup F)') = 1 - P(E \cup F)$$

From part (i), $$P(E \cup F) = \frac{5}{8}$$. Therefore

$$P(\text{not } E \text{ and not } F) = 1 - \frac{5}{8} = \frac{3}{8}$$

We are given $$P(A) = 0.42$$, $$P(B) = 0.48$$ and $$P(A \text{ and } B) = P(A \cap B) = 0.16$$.

'not $$A$$' is the complement $$A'$$, and $$P(A) + P(A') = 1$$, so

$$P(\text{not } A) = 1 - P(A) = 1 - 0.42 = 0.58$$

'not $$B$$' is the complement $$B'$$, and $$P(B) + P(B') = 1$$, so

$$P(\text{not } B) = 1 - P(B) = 1 - 0.48 = 0.52$$

By the addition theorem,

$$P(A \text{ or } B) = P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

Substituting the given values:

$$P(A \cup B) = 0.42 + 0.48 - 0.16 = 0.74$$

There are $$60$$ students in all. Let $$A$$ be the event 'opted for NCC' and $$B$$ the event 'opted for NSS'. From the data,

$$P(A) = \frac{30}{60} = \frac{1}{2}, \qquad P(B) = \frac{32}{60} = \frac{8}{15}$$

$$P(A \cap B) = \frac{24}{60} = \frac{2}{5} \quad (\text{opted for both})$$

The probability of opting for NCC or NSS is, by the addition theorem,

$$P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{1}{2} + \frac{8}{15} - \frac{2}{5}$$

Taking the common denominator $$30$$:

$$P(A \cup B) = \frac{15}{30} + \frac{16}{30} - \frac{12}{30} = \frac{19}{30}$$

'Neither NCC nor NSS' is the event $$A' \cap B' = (A \cup B)'$$ — the complement of 'NCC or NSS'.

From part (i), $$P(A \cup B) = \frac{19}{30}$$, so

$$P(\text{neither}) = P((A \cup B)') = 1 - P(A \cup B) = 1 - \frac{19}{30} = \frac{11}{30}$$

'NSS but not NCC' is the event $$B \cap A' = B - A$$. The students who opted for NSS but not for NCC are those in NSS minus those who took both:

$$32 - 24 = 8 \text{ students}$$

So the probability is

$$P(B - A) = \frac{8}{60} = \frac{2}{15}$$

Equivalently, $$P(B - A) = P(B) - P(A \cap B) = \frac{8}{15} - \frac{2}{5} = \frac{8}{15} - \frac{6}{15} = \frac{2}{15}$$.

Veena visits the four cities $$A, B, C, D$$ in some order. The total number of possible orders is the number of arrangements of $$4$$ cities:

$$4! = 24$$

and each order is equally likely.

In any arrangement, $$A$$ comes either before $$B$$ or after $$B$$. By symmetry, exactly half of the $$24$$ arrangements have $$A$$ before $$B$$, i.e. $$12$$ of them.

$$P(A \text{ before } B) = \frac{12}{24} = \frac{1}{2}$$

The total number of equally likely orders is $$4! = 24$$.

Consider only the relative order of the three cities $$A$$, $$B$$, $$C$$. These three can appear among themselves in $$3! = 6$$ different relative orders, and all are equally likely. Only one of these six orders is '$$A$$, then $$B$$, then $$C$$'.

$$P(A \text{ before } B \text{ and } B \text{ before } C) = \frac{1}{3!} = \frac{1}{6}$$

(Counting directly: the favourable arrangements of all four cities are $$ABCD, ABDC, ADBC, DABC$$ — $$4$$ of them — giving $$\frac{4}{24} = \frac{1}{6}$$.)

The total number of equally likely orders is $$4! = 24$$.

If $$A$$ must be first and $$B$$ must be last, the arrangement looks like

$$A, \_\,, \_\,, B$$

The two middle positions are filled by the remaining cities $$C$$ and $$D$$, which can be arranged in $$2! = 2$$ ways ($$ACDB$$ and $$ADCB$$).

$$P(A \text{ first and } B \text{ last}) = \frac{2}{24} = \frac{1}{12}$$

The total number of equally likely orders is $$4! = 24$$.

$$A$$ first: the other three cities fill the remaining $$3$$ positions in $$3! = 6$$ ways.

$$A$$ second: again the other three cities fill the remaining $$3$$ positions in $$3! = 6$$ ways.

These two cases are mutually exclusive, so the number of favourable arrangements is $$6 + 6 = 12$$.

$$P(A \text{ first or second}) = \frac{12}{24} = \frac{1}{2}$$

The total number of equally likely orders is $$4! = 24$$.

'$$A$$ just before $$B$$' means $$A$$ and $$B$$ are next to each other with $$A$$ immediately followed by $$B$$. Treat the pair $$AB$$ as a single block. Then we are arranging three items — the block $$AB$$, city $$C$$ and city $$D$$ — which can be done in

$$3! = 6 \text{ ways}$$

(The block is fixed as $$AB$$, not $$BA$$, so there is no extra factor.)

$$P(A \text{ just before } B) = \frac{6}{24} = \frac{1}{4}$$

The five teams $$A, B, C, D, E$$ can finish the race in any order, so the total number of equally likely finishing orders is

$$5! = 120$$

We want $$A$$ first, $$B$$ second and $$C$$ third — these three positions are now fixed. The remaining two positions (fourth and fifth) are filled by $$D$$ and $$E$$, which can be done in

$$2! = 2 \text{ ways}$$

So there are $$2$$ favourable orders.

$$P(A \text{ 1st}, B \text{ 2nd}, C \text{ 3rd}) = \frac{2}{120} = \frac{1}{60}$$

The total number of equally likely finishing orders is again $$5! = 120$$.

Now $$A$$, $$B$$, $$C$$ must occupy the first three places in any order, and $$D$$, $$E$$ the last two places.

The three teams $$A, B, C$$ can be arranged among the first three places in $$3! = 6$$ ways, and the two teams $$D, E$$ among the last two places in $$2! = 2$$ ways. So the number of favourable orders is

$$3! \times 2! = 6 \times 2 = 12$$

$$P(A, B, C \text{ are the first three}) = \frac{12}{120} = \frac{1}{10}$$

The box contains $$10 + 20 + 30 = 60$$ marbles. Drawing $$5$$ marbles, the total number of equally likely selections is

$$\binom{60}{5}$$

'All blue' means all $$5$$ marbles are chosen from the $$20$$ blue marbles, which can be done in $$\binom{20}{5}$$ ways.

$$P(\text{all blue}) = \frac{\binom{20}{5}}{\binom{60}{5}}$$

Now $$\binom{20}{5} = \dfrac{20 \times 19 \times 18 \times 17 \times 16}{5!} = 15504$$ and $$\binom{60}{5} = \dfrac{60 \times 59 \times 58 \times 57 \times 56}{5!} = 5461512$$. Therefore

$$P(\text{all blue}) = \frac{15504}{5461512} = \frac{34}{11977}$$

It is easiest to use the complement. The opposite of 'at least one green' is 'no green marble at all', i.e. all $$5$$ marbles come from the $$10 + 20 = 30$$ non-green (red or blue) marbles.

$$P(\text{no green}) = \frac{\binom{30}{5}}{\binom{60}{5}}$$

Here $$\binom{30}{5} = \dfrac{30 \times 29 \times 28 \times 27 \times 26}{5!} = 142506$$ and $$\binom{60}{5} = 5461512$$, so

$$P(\text{no green}) = \frac{142506}{5461512} = \frac{117}{4484}$$

Therefore

$$P(\text{at least one green}) = 1 - P(\text{no green}) = 1 - \frac{117}{4484} = \frac{4367}{4484}$$

The die has $$6$$ faces in all, and each face is equally likely to come up. The numbers on the faces are distributed as:

• number $$1$$ on $$2$$ faces,
• number $$2$$ on $$3$$ faces,
• number $$3$$ on $$1$$ face.

The number $$2$$ appears on $$3$$ of the $$6$$ faces, so

$$P(2) = \frac{3}{6} = \frac{1}{2}$$

The number $$1$$ appears on $$2$$ faces and the number $$3$$ appears on $$1$$ face, so

$$P(1) = \frac{2}{6} = \frac{1}{3}, \qquad P(3) = \frac{1}{6}$$

Getting a $$1$$ and getting a $$3$$ are mutually exclusive (the die shows only one number), so

$$P(1 \text{ or } 3) = P(1) + P(3) = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$$

The number $$3$$ appears on $$1$$ of the $$6$$ faces, so $$P(3) = \frac{1}{6}$$.

'not $$3$$' is the complementary event, so

$$P(\text{not } 3) = 1 - P(3) = 1 - \frac{1}{6} = \frac{5}{6}$$

Out of $$10000$$ tickets, $$10$$ win a prize, so $$10000 - 10 = 9990$$ tickets do not win a prize.

If you buy one ticket, it is chosen from all $$10000$$, and a 'no prize' outcome corresponds to one of the $$9990$$ non-winning tickets.

$$P(\text{no prize}) = \frac{9990}{10000} = \frac{999}{1000}$$

If you buy $$2$$ tickets, they are chosen from all $$10000$$ tickets in $$\binom{10000}{2}$$ equally likely ways.

'No prize' means both tickets are among the $$9990$$ non-winning tickets, which can happen in $$\binom{9990}{2}$$ ways.

$$P(\text{no prize}) = \frac{\binom{9990}{2}}{\binom{10000}{2}} = \frac{\dfrac{9990 \times 9989}{2}}{\dfrac{10000 \times 9999}{2}} = \frac{9990 \times 9989}{10000 \times 9999}$$

If you buy $$10$$ tickets, they are chosen from all $$10000$$ tickets in $$\binom{10000}{10}$$ equally likely ways.

'No prize' means all $$10$$ tickets are among the $$9990$$ non-winning tickets, which can happen in $$\binom{9990}{10}$$ ways.

$$P(\text{no prize}) = \frac{\binom{9990}{10}}{\binom{10000}{10}}$$

The $$100$$ students are split into a section of $$40$$ and a section of $$60$$. 'Same section' happens if you and your friend are both in the $$40$$-section, or both in the $$60$$-section.

Both in the section of 40. The number of ways to choose which $$40$$ students form that section is $$\binom{100}{40}$$. The number of those choices that include both you and your friend is $$\binom{98}{38}$$ (the remaining $$38$$ places filled from the other $$98$$ students). So

$$P(\text{both in 40-section}) = \frac{\binom{98}{38}}{\binom{100}{40}} = \frac{40 \times 39}{100 \times 99} = \frac{1560}{9900}$$

Both in the section of 60. Similarly,

$$P(\text{both in 60-section}) = \frac{\binom{98}{58}}{\binom{100}{60}} = \frac{60 \times 59}{100 \times 99} = \frac{3540}{9900}$$

These two cases are mutually exclusive, so

$$P(\text{same section}) = \frac{1560}{9900} + \frac{3540}{9900} = \frac{5100}{9900} = \frac{17}{33}$$

'Different sections' is the complement of 'same section'. Using the result of part (a),

$$P(\text{different sections}) = 1 - P(\text{same section}) = 1 - \frac{17}{33} = \frac{16}{33}$$

We are given $$P(A) = 0.54$$, $$P(B) = 0.69$$ and $$P(A \cap B) = 0.35$$.

By the addition theorem,

$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

$$P(A \cup B) = 0.54 + 0.69 - 0.35 = 0.88$$

By De Morgan's law, $$A' \cap B' = (A \cup B)'$$, so this is the complement of $$A \cup B$$.

Using $$P(A \cup B) = 0.88$$ from part (i),

$$P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B) = 1 - 0.88 = 0.12$$

$$A \cap B'$$ is the set of outcomes in $$A$$ but not in $$B$$. The event $$A$$ splits into two disjoint parts — the part inside $$B$$ and the part outside $$B$$:

$$P(A) = P(A \cap B) + P(A \cap B')$$

Therefore

$$P(A \cap B') = P(A) - P(A \cap B) = 0.54 - 0.35 = 0.19$$

$$B \cap A'$$ is the set of outcomes in $$B$$ but not in $$A$$. Splitting $$B$$ into the part inside $$A$$ and the part outside $$A$$:

$$P(B) = P(A \cap B) + P(B \cap A')$$

Therefore

$$P(B \cap A') = P(B) - P(A \cap B) = 0.69 - 0.35 = 0.34$$

We form $$4$$-digit numbers greater than $$5000$$ using the digits $$0, 1, 3, 5, 7$$, with repetition allowed.

Total numbers (the sample space). For a number to be greater than $$5000$$, the thousands digit must be $$5$$ or $$7$$ (the only available digits that are $$\ge 5$$). The thousands digit has $$2$$ choices and each of the other $$3$$ digits has $$5$$ choices:

$$2 \times 5 \times 5 \times 5 = 250$$

But this count includes the number $$5000$$ itself, and $$5000$$ is not greater than $$5000$$. Removing it:

$$\text{total} = 250 - 1 = 249$$

Favourable numbers (divisible by 5). A number is divisible by $$5$$ when its units digit is $$0$$ or $$5$$. Here the thousands digit has $$2$$ choices, the two middle digits have $$5$$ choices each, and the units digit has $$2$$ choices:

$$2 \times 5 \times 5 \times 2 = 100$$

This again includes $$5000$$ (units digit $$0$$, divisible by $$5$$), which must be removed:

$$\text{favourable} = 100 - 1 = 99$$

$$P(\text{divisible by } 5) = \frac{99}{249} = \frac{33}{83}$$

We form $$4$$-digit numbers greater than $$5000$$ using the digits $$0, 1, 3, 5, 7$$, with no digit repeated.

Total numbers (the sample space). The thousands digit must be $$5$$ or $$7$$ ($$2$$ choices). The remaining $$3$$ places are filled by $$3$$ distinct digits chosen in order from the $$4$$ digits left, giving $$4 \times 3 \times 2 = 24$$ ways:

$$\text{total} = 2 \times 24 = 48$$

(No such number can equal $$5000$$, since $$5000$$ would need three $$0$$s; with the thousands digit $$5$$ the smallest number is $$5013 > 5000$$.)

Favourable numbers (divisible by 5). The units digit must be $$0$$ or $$5$$. Consider the two cases.

Case 1: units digit $$= 0$$. The thousands digit is $$5$$ or $$7$$ ($$2$$ choices). The two middle digits are filled from the remaining $$3$$ digits: $$3 \times 2 = 6$$ ways. This gives $$2 \times 6 = 12$$ numbers.

Case 2: units digit $$= 5$$. The thousands digit must be $$\ge 5$$, but $$5$$ is already used, so it can only be $$7$$ ($$1$$ choice). The two middle digits are filled from the remaining $$3$$ digits ($$0, 1, 3$$): $$3 \times 2 = 6$$ ways. This gives $$1 \times 6 = 6$$ numbers.

$$\text{favourable} = 12 + 6 = 18$$

$$P(\text{divisible by } 5) = \frac{18}{48} = \frac{3}{8}$$