Ex 14.1 - Question 1

Question: A die is rolled. Let $E$ be the event “die shows $4$” and $F$ be the event “die shows even number”. Are $E$ and $F$ mutually exclusive?

Solution:

When a die is rolled, the sample space $S$ is given by:

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

The event $E$ (“die shows $4$”) is:

$$E = \{4\}$$

The event $F$ (“die shows even number”) is:

$$F = \{2, 4, 6\}$$

To check if $E$ and $F$ are mutually exclusive, we find their intersection $E \cap F$:

$$E \cap F = \{4\} \cap \{2, 4, 6\} = \{4\}$$

Since $E \cap F \neq \phi$ (the intersection is not empty, as the element $4$ is common to both events), $E$ and $F$ are not mutually exclusive.

Ex 14.1 - Question 2

Question: A die is thrown. Describe the following events:
(i) $A$: a number less than $7$
(ii) $B$: a number greater than $7$
(iii) $C$: a multiple of $3$
(iv) $D$: a number less than $4$
(v) $E$: an even number greater than $4$
(vi) $F$: a number not less than $3$
Also find $A \cup B$, $A \cap B$, $B \cup C$, $E \cap F$, $D \cap E$, $A – C$, $D – E$, $E \cap F'$, $F'$

Solution:

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

Description of Events:

• (i) $A$ (a number less than $7$): $A = \{1, 2, 3, 4, 5, 6\}$
• (ii) $B$ (a number greater than $7$): $B = \phi$
• (iii) $C$ (a multiple of $3$): $C = \{3, 6\}$
• (iv) $D$ (a number less than $4$): $D = \{1, 2, 3\}$
• (v) $E$ (an even number greater than $4$): $E = \{6\}$
• (vi) $F$ (a number not less than $3$, i.e., $\ge 3$): $F = \{3, 4, 5, 6\}$

Finding Set Operations:

• $A \cup B = \{1, 2, 3, 4, 5, 6\} \cup \phi = \{1, 2, 3, 4, 5, 6\}$
• $A \cap B = \{1, 2, 3, 4, 5, 6\} \cap \phi = \phi$
• $B \cup C = \phi \cup \{3, 6\} = \{3, 6\}$
• $E \cap F = \{6\} \cap \{3, 4, 5, 6\} = \{6\}$
• $D \cap E = \{1, 2, 3\} \cap \{6\} = \phi$
• $A - C = \{1, 2, 3, 4, 5, 6\} - \{3, 6\} = \{1, 2, 4, 5\}$
• $D - E = \{1, 2, 3\} - \{6\} = \{1, 2, 3\}$
• $F' = S - F = \{1, 2, 3, 4, 5, 6\} - \{3, 4, 5, 6\} = \{1, 2\}$
• $E \cap F' = \{6\} \cap \{1, 2\} = \phi$

Ex 14.1 - Question 3

Question: An experiment involves rolling a pair of dice and recording the numbers that come up. Describe the following events:
$A$: the sum is greater than $8$, $B$: $2$ occurs on either die
$C$: the sum is at least $7$ and a multiple of $3$.
Which pairs of these events are mutually exclusive?

Solution:

When a pair of dice is rolled, the sample space $S$ contains $36$ outcomes of the form $(x, y)$ where $x, y \in \{1, 2, 3, 4, 5, 6\}$.

1. Description of Events in Set-Roster Form:

• $A$ (sum greater than $8$):
  $$A = \{(3,6), (4,5), (5,4), (6,3), (4,6), (5,5), (6,4), (5,6), (6,5), (6,6)\}$$
• $B$ ($2$ occurs on either die):
  $$B = \{(1,2), (2,2), (3,2), (4,2), (5,2), (6,2), (2,1), (2,3), (2,4), (2,5), (2,6)\}$$
• $C$ (sum is at least $7$ and a multiple of $3$, i.e., sum $\in \{9, 12\}$):
  Sum $9$: $(3,6), (4,5), (5,4), (6,3)$
  Sum $12$: $(6,6)$
  $$C = \{(3,6), (6,3), (5,4), (4,5), (6,6)\}$$

2. Checking for Mutual Exclusivity:

• $A$ and $B$: $A \cap B = \phi$ (No common outcome between sums $>8$ and outcomes containing $2$). Thus, $A$ and $B$ are mutually exclusive.
• $B$ and $C$: $B \cap C = \phi$ (No outcome containing $2$ has a sum of $9$ or $12$). Thus, $B$ and $C$ are mutually exclusive.
• $A$ and $C$: $A \cap C = \{(3,6), (4,5), (5,4), (6,3), (6,6)\} \neq \phi$. Thus, $A$ and $C$ are not mutually exclusive.

Conclusion: The mutually exclusive pairs of events are $A$ and $B$, and $B$ and $C$.

Ex 14.1 - Question 4

Question: Three coins are tossed once. Let $A$ denote the event ‘three heads show”, $B$ denote the event “two heads and one tail show”, $C$ denote the event “three tails show” and $D$ denote the event ‘a head shows on the first coin”. Which events are (i) mutually exclusive? (ii) simple? (iii) Compound?

Solution: When three coins are tossed once, the sample space $S$ contains $2^3 = 8$ sample points: $$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$$ Let's write down the sample sets for each given event:

• $A$ (“three heads show”) $= \{HHH\}$
• $B$ (“two heads and one tail show”) $= \{HHT, HTH, THH\}$
• $C$ (“three tails show”) $= \{TTT\}$
• $D$ (“a head shows on the first coin”) $= \{HHH, HHT, HTH, HTT\}$

(i) Mutually exclusive events: Two events are mutually exclusive if their intersection is empty ($\phi$).

• $A \cap B = \{HHH\} \cap \{HHT, HTH, THH\} = \phi$ (Mutually exclusive)
• $A \cap C = \{HHH\} \cap \{TTT\} = \phi$ (Mutually exclusive)
• $B \cap C = \{HHT, HTH, THH\} \cap \{TTT\} = \phi$ (Mutually exclusive)
• $C \cap D = \{TTT\} \cap \{HHH, HHT, HTH, HTT\} = \phi$ (Mutually exclusive)

Thus, the mutually exclusive pairs/groups of events are: $A$ and $B$; $A$ and $C$; $B$ and $C$; $C$ and $D$.
(ii) Simple events: An event containing only a single sample point is called a simple event.

• $A = \{HHH\}$ contains 1 point (Simple)
• $C = \{TTT\}$ contains 1 point (Simple)

Thus, the simple events are $A$ and $C$.
(iii) Compound events: An event containing more than one sample point is a compound event.

• $B = \{HHT, HTH, THH\}$ contains 3 points (Compound)
• $D = \{HHH, HHT, HTH, HTT\}$ contains 4 points (Compound)

Thus, the compound events are $B$ and $D$.

Ex 14.1 - Question 5

Question: Three coins are tossed. Describe (i) Two events which are mutually exclusive. (ii) Three events which are mutually exclusive and exhaustive. (iii) Two events, which are not mutually exclusive. (iv) Two events which are mutually exclusive but not exhaustive. (v) Three events which are mutually exclusive but not exhaustive.

Solution:
Sample space $S$ for three tossed coins has $8$ outcomes: $$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$$ (i) Two events which are mutually exclusive: Let $E_1$ = "Getting at least two heads" = $\{HHH, HHT, HTH, THH\}$ and $E_2$ = "Getting at least two tails" = $\{TTT, TTH, THT, HTT\}$ (Note: Depending on definitions, other examples work too, e.g., $A$ and $C$ from Q4). Here $E_1 \cap E_2 = \phi$.
(ii) Three events which are mutually exclusive and exhaustive:

• $E_1$ = "Getting no heads" $= \{TTT\}$
• $E_2$ = "Getting exactly one head" $= \{HTT, THT, TTH\}$
• $E_3$ = "Getting at least two heads" $= \{HHH, HHT, HTH, THH\}$

These are mutually exclusive (intersection of any two is $\phi$) and exhaustive ($E_1 \cup E_2 \cup E_3 = S$).
(iii) Two events which are not mutually exclusive: Let $E_1$ = "Getting at most two tails" = $S \setminus \{TTT\}$ and $E_2$ = "Getting exactly two tails" = $\{HTT, THT, TTH\}$. Since $E_1 \cap E_2 = E_2 \neq \phi$, they are not mutually exclusive.
(iv) Two events which are mutually exclusive but not exhaustive: Let $E_1$ = "Getting exactly one head" = $\{HTT, THT, TTH\}$ and $E_2$ = "Getting exactly two heads" = $\{HHT, HTH, THH\}$. $E_1 \cap E_2 = \phi$ (mutually exclusive), but $E_1 \cup E_2 \neq S$ (not exhaustive as $\{HHH, TTT\}$ are missing).
(v) Three events which are mutually exclusive but not exhaustive: Let $E_1$ = "Getting exactly one tail" = $\{HHT, HTH, THH\}$, $E_2$ = "Getting exactly two tails" = $\{HTT, THT, TTH\}$, and $E_3$ = "Getting exactly three tails" = $\{TTT\}$. They are mutually exclusive, but their union does not form $S$ since $\{HHH\}$ is excluded.

Ex 14.1 - Question 6

Question: Two dice are thrown. The events $A$, $B$ and $C$ are as follows: $A$: getting an even number on the first die. $B$: getting an odd number on the first die. $C$: getting the sum of the numbers on the dice $\le 5$. Describe the events (i) $A'$ (ii) not $B$ (iii) $A$ or $B$ (iv) $A$ and $B$ (v) $A$ but not $C$ (vi) $B$ or $C$ (vii) $B$ and $C$ (viii) $A \cap B' \cap C'$

Solution:
Sample space $S$ consists of $6 \times 6 = 36$ outcomes. The sets for $A$, $B$, and $C$ are:

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

(i) $A'$ (complement of $A$): Outcomes where the first number is not even (i.e., odd): $$A' = B = \{(1,1), (1,2), \dots, (5,6)\}$$
(ii) not $B$ ($B'$): Outcomes where the first number is not odd (i.e., even): $$B' = A = \{(2,1), (2,2), \dots, (6,6)\}$$
(iii) $A$ or $B$ ($A \cup B$): $$A \cup B = S$$
(iv) $A$ and $B$ ($A \cap B$): $$A \cap B = \phi$$
(v) $A$ but not $C$ ($A - C$): Elements in $A$ that are not in $C$: $$A - C = \{(2,4), (2,5), (2,6), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$$
(vi) $B$ or $C$ ($B \cup C$): $$B \cup C = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)\}$$
(vii) $B$ and $C$ ($B \cap C$): $$B \cap C = \{(1,1), (1,2), (1,3), (1,4), (3,1), (3,2)\}$$
(viii) $A \cap B' \cap C'$: Since $B' = A$, this is $A \cap A \cap C' = A \cap C' = A - C$: $$A \cap B' \cap C' = \{(2,4), (2,5), (2,6), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$$

Ex 14.1 - Question 7

Question: Refer to question 6 above, state true or false: (give reason for your answer) (i) $A$ and $B$ are mutually exclusive (ii) $A$ and $B$ are mutually exclusive and exhaustive (iii) $A = B'$ (iv) $A$ and $C$ are mutually exclusive (v) $A$ and $B'$ are mutually exclusive. (vi) $A'$, $B'$, $C$ are mutually exclusive and exhaustive.

Solution:
(i) True. $A$ and $B$ have no common outcomes ($A \cap B = \phi$).
(ii) True. $A \cap B = \phi$ (mutually exclusive) and $A \cup B = S$ (exhaustive).
(iii) True. $B'$ represents getting an even number on the first die, which is precisely the definition of event $A$. Thus $A = B'$.
(iv) False. $A$ and $C$ share common elements like $(2,1), (2,2), (2,3), (4,1)$, so $A \cap C \neq \phi$.
(v) False. Since $B' = A$, $A$ and $B'$ is $A \cap A = A \neq \phi$.
(vi) False. Note that $A' = B$ and $B' = A$. The union of $A', B', C$ is not the entire sample space $S$ (for instance, $(6,6)$ is missing from $B$, $A$, and $C$). Thus they are not exhaustive.

Ex 14.2 - Question 1

Question: Which of the following cannot be valid assignment of probabilities for outcomes of sample Space $S = \{\omega_1, \omega_2, \omega_3, \omega_4, \omega_5, \omega_6, \omega_7\}$

Assignment	$\omega_1$	$\omega_2$	$\omega_3$	$\omega_4$	$\omega_5$	$\omega_6$	$\omega_7$
(a)	$0.1$	$0.01$	$0.05$	$0.03$	$0.01$	$0.2$	$0.6$
(b)	$\frac{1}{7}$	$\frac{1}{7}$	$\frac{1}{7}$	$\frac{1}{7}$	$\frac{1}{7}$	$\frac{1}{7}$	$\frac{1}{7}$
(c)	$0.1$	$0.2$	$0.3$	$0.4$	$0.5$	$0.6$	$0.7$
(d)	$-0.1$	$0.2$	$0.3$	$0.4$	$-0.2$	$0.1$	$0.3$
(e)	$\frac{1}{14}$	$\frac{2}{14}$	$\frac{3}{14}$	$\frac{4}{14}$	$\frac{5}{14}$	$\frac{6}{14}$	$\frac{15}{14}$

Solution: A valid probability assignment must satisfy two conditions for each outcome $\omega_i$: 1. $0 \le P(\omega_i) \le 1$ for all $i$. 2. The sum of probabilities of all elementary outcomes must be equal to 1, i.e., $\sum P(\omega_i) = 1$. Let's test each assignment:

• (a) Probabilities: $0.1, 0.01, 0.05, 0.03, 0.01, 0.2, 0.6$. Each is between $0$ and $1$.
  Sum = $0.1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6 = 1.0$.
  Status: Valid.
• (b) Probabilities: $\frac{1}{7}$ each. Each is between $0$ and $1$.
  Sum = $7 \times \frac{1}{7} = 1$.
  Status: Valid.
• (c) Probabilities: $0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7$. Each is between $0$ and $1$.
  Sum = $0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 + 0.7 = 2.8 \neq 1$.
  Status: Not a valid assignment.
• (d) Probabilities include $-0.1$ and $-0.2$, which are less than $0$.
  Status: Not a valid assignment.
• (e) Probabilities include $\frac{15}{14}$, which is greater than $1$. Also, sum = $\frac{1+2+3+4+5+6+15}{14} = \frac{36}{14} \neq 1$.
  Status: Not a valid assignment.

Therefore, assignments (c), (d), and (e) are not valid.

Ex 14.2 - Question 2

Question: A coin is tossed twice, what is the probability that atleast one tail occurs?

Solution: When a coin is tossed twice, the sample space $S$ is: $$S = \{HH, HT, TH, TT\}$$ Total number of elementary outcomes $= 4$. Let $E$ be the event of getting "at least one tail". The outcomes favorable to event $E$ are: $$E = \{HT, TH, TT\}$$ Number of favorable outcomes $= 3$. Thus, the probability of getting at least one tail is: $$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3}{4}$$

Ex 14.2 - Question 3

Question: A die is thrown, find the probability of following events: (i) A prime number will appear, (ii) A number greater than or equal to $3$ will appear, (iii) A number less than or equal to one will appear, (iv) A number more than $6$ will appear, (v) A number less than $6$ will appear.

Solution: When a die is thrown, the sample space is: $$S = \{1, 2, 3, 4, 5, 6\}$$ Total number of possible outcomes $= 6$.

• (i) A prime number will appear:
  Prime numbers in $S$ are $2, 3, 5$.
  Favorable outcomes $= \{2, 3, 5\}$ (3 outcomes).
  $$P(\text{prime}) = \frac{3}{6} = \frac{1}{2}$$
• (ii) A number greater than or equal to $3$ will appear:
  Numbers $\ge 3$ are $3, 4, 5, 6$.
  Favorable outcomes $= \{3, 4, 5, 6\}$ (4 outcomes).
  $$P(\ge 3) = \frac{4}{6} = \frac{2}{3}$$
• (iii) A number less than or equal to one will appear:
  Number $\le 1$ is $1$.
  Favorable outcomes $= \{1\}$ (1 outcome).
  $$P(\le 1) = \frac{1}{6}$$
• (iv) A number more than $6$ will appear:
  Numbers $> 6$ do not exist in $S$.
  Favorable outcomes $= \emptyset$ (0 outcomes).
  $$P(> 6) = \frac{0}{6} = 0$$
• (v) A number less than $6$ will appear:
  Numbers $< 6$ are $1, 2, 3, 4, 5$.
  Favorable outcomes $= \{1, 2, 3, 4, 5\}$ (5 outcomes).
  $$P(< 6) = \frac{5}{6}$$

Ex 14.2 - Question 4

Question: A card is selected from a pack of $52$ cards. (a) How many points are there in the sample space? (b) Calculate the probability that the card is an ace of spades. (c) Calculate the probability that the card is (i) an ace (ii) black card.

Solution:

• (a) A standard pack contains $52$ cards, so each card represents a sample point. $$\text{Number of points in the sample space} = 52$$
• (b) Probability that the card is an ace of spades:
  There is only $1$ ace of spades in a deck of $52$ cards.
  $$P(\text{ace of spades}) = \frac{1}{52}$$
• (c) (i) Probability that the card is an ace:
  There are $4$ aces in a deck (one in each suit: spades, hearts, diamonds, clubs).
  $$P(\text{ace}) = \frac{4}{52} = \frac{1}{13}$$
• (c) (ii) Probability that the card is a black card:
  There are $26$ black cards in a deck ($13$ spades and $13$ clubs).
  $$P(\text{black card}) = \frac{26}{52} = \frac{1}{2}$$

Ex 14.2 - Question 5

Question: A fair coin with $1$ marked on one face and $6$ on the other and a fair die are both tossed. find the probability that the sum of numbers that turn up is (i) $3$ (ii) $12$

Solution:

When a fair coin with faces $1$ and $6$ is tossed, the possible outcomes are $\{1, 6\}$.

When a fair die is tossed, the possible outcomes are $\{1, 2, 3, 4, 5, 6\}$.

The total number of possible outcomes in the sample space $S$ is: $$n(S) = 2 \times 6 = 12$$ Specifically, the sample space is: $$S = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$$

(i) Sum of numbers is $3$:

Let $E_1$ be the event that the sum of numbers is $3$. The possible outcomes are: $$E_1 = \{(1, 2)\} \implies n(E_1) = 1$$ Therefore, the probability is: $$P(E_1) = \frac{n(E_1)}{n(S)} = \frac{1}{12}$$

(ii) Sum of numbers is $12$:

Let $E_2$ be the event that the sum of numbers is $12$. The possible outcomes are: $$E_2 = \{(6, 6)\} \implies n(E_2) = 1$$ Therefore, the probability is: $$P(E_2) = \frac{n(E_2)}{n(S)} = \frac{1}{12}$$

Ex 14.2 - Question 6

Question: There are four men and six women on the city council. If one council member is selected for a committee at random, how likely is it that it is a woman?

Solution:

Number of men on the city council $= 4$.

Number of women on the city council $= 6$.

Total number of council members $= 4 + 6 = 10$.

Total number of possible outcomes for selecting one member, $n(S) = 10$.

Let $W$ be the event that the selected council member is a woman. The number of favorable outcomes is $n(W) = 6$.

Therefore, the probability of selecting a woman is: $$P(W) = \frac{n(W)}{n(S)} = \frac{6}{10} = \frac{3}{5}$$

Ex 14.2 - Question 7

Question: A fair coin is tossed four times, and a person wins $\text{Re } 1$ for each head and loses $\text{Rs } 1.50$ for each tail that turns up. From the sample space calculate how many different amounts of money you can have after four tosses and the probability of having each of these amounts.

Solution:

When a coin is tossed $4$ times, the total number of outcomes in the sample space is $2^4 = 16$.

Let $H$ represent a Head (gain $\text{Re } 1.00$) and $T$ represent a Tail (loss $\text{Rs } 1.50$).

The possible outcomes and their corresponding net amounts are:

• HHHH (4 Heads, 0 Tails): Gain = $4 \times 1 = \text{Rs } 4.00$ (Probability = $\frac{1}{16}$)
• HHHT, HHTH, HTHH, THHH (3 Heads, 1 Tail): Gain = $3 \times 1 - 1 \times 1.50 = \text{Rs } 1.50$ (Probability = $\frac{4}{16} = \frac{1}{4}$)
• HHTT, HTTH, THHT, TTHH, HTHT, THTH (2 Heads, 2 Tails): Gain = $2 \times 1 - 2 \times 1.50 = 2 - 3 = -\text{Rs } 1.00$ (Loss of $\text{Re } 1.00$) (Probability = $\frac{6}{16} = \frac{3}{8}$)
• HTTT, THTT, TTHT, TTTH (1 Head, 3 Tails): Gain = $1 \times 1 - 3 \times 1.50 = 1 - 4.50 = -\text{Rs } 3.50$ (Loss of $\text{Rs } 3.50$) (Probability = $\frac{4}{16} = \frac{1}{4}$)
• TTTT (0 Heads, 4 Tails): Gain = $0 \times 1 - 4 \times 1.50 = -\text{Rs } 6.00$ (Loss of $\text{Rs } 6.00$) (Probability = $\frac{1}{16}$)

Thus, there are $5$ different amounts of money you can have after four tosses: $\text{Rs } 4.00$ gain, $\text{Rs } 1.50$ gain, $\text{Re } 1.00$ loss, $\text{Rs } 3.50$ loss, and $\text{Rs } 6.00$ loss.