Chapter 9: Probability

Let $D$ denote the event that a selected person tests positive for iron deficiency.

(a) Has Iron Deficiency:

$$P(D) = \frac{70}{100} = 0.70 = \frac{7}{10}$$

(b) Does Not Have Iron Deficiency:

Using the complementary law of probability:

$$P(D') = 1 - P(D) = 1 - 0.70 = 0.30 = \frac{3}{10} \text{ (or } 30\% \text{)}$$

Mathematical Calculation:

Since events $A$ and $B$ are mutually exclusive (disjoint), they cannot happen together, which means $P(A \cap B) = 0$.

By the Addition Theorem for mutually exclusive events:

$$P(A \cup B) = P(A) + P(B) = 0.35 + 0.45 = 0.80$$

Let $S$ be the event of success with $P(S) = p$.

(a) Probability of Failure:

Failure is the exact complement of success ($F = S'$). Therefore:

$$P(F) = 1 - P(S) = 1 - p$$

(b) Probability of Neither Success nor Failure:

In a binary trial with sample space $S_{total} = \{\text{Success}, \text{Failure}\}$, the outcomes are exhaustive. The event "neither success nor failure" corresponds to the impossible event $\emptyset$.

$$P(\text{Neither}) = 0$$

Interpretation: Every trial must terminate in exactly one valid outcome defined in the sample space; hence, an impossible null outcome has zero probability.

Let $M$ be the event of studying Mathematics and $S$ be the event of studying Science.

Given total students $n(U) = 40$:

• $P(M) = \frac{n(M)}{n(U)} = \frac{25}{40}$
• $P(S) = \frac{n(S)}{n(U)} = \frac{18}{40}$
• $P(S \cap M) = \frac{n(S \cap M)}{n(U)} = \frac{10}{40}$

We require the conditional probability $P(S | M)$:

$$P(S | M) = \frac{P(S \cap M)}{P(M)} = \frac{10/40}{25/40} = \frac{10}{25} = \frac{2}{5} = 0.4$$

Hence, the probability is $\frac{2}{5}$ or $40\%$.

Sample space of a single die throw: $S = \{1, 2, 3, 4, 5, 6\} \implies n(S) = 6$.

Event $A = \text{even} = \{2, 4, 6\} \implies P(A) = \frac{3}{6} = \frac{1}{2}$.

Event $B = \text{prime} = \{2, 3, 5\} \implies P(B) = \frac{3}{6} = \frac{1}{2}$.

Intersection $A \cap B = \{2\} \implies P(A \cap B) = \frac{1}{6}$.

From the problem statement:

• Probability a driver is young: $P(Y) = 30\% = 0.30$
• Conditional probability of an accident given the driver is young: $P(A | Y) = 40\% = 0.40$

Using the Multiplication Theorem of Probability:

$$P(A \cap Y) = P(Y) \cdot P(A | Y) = 0.30 \times 0.40 = 0.12$$

Thus, the probability that an accident involves a young driver is $0.12$ (or $12\%$).

Total number of cards $= 52$.

Let $K$ be the event of drawing a King, and $H$ be the event of drawing a Heart.

• Number of Kings $= 4$
• Number of Hearts $= 13$
• Number of cards that are both King and Heart (King of Hearts) $= 1$

Number of cards that are a King or a Heart:

$$n(K \cup H) = n(K) + n(H) - n(K \cap H) = 4 + 13 - 1 = 16$$

The number of cards that are neither a King nor a Heart:

$$n(\text{Neither}) = 52 - 16 = 36$$

$$P(\text{Neither}) = \frac{36}{52} = \frac{9}{13}$$

Correct Option: (b) 9/13

Total elementary outcomes when throwing 2 dice: $n(S) = 6 \times 6 = 36$.

Possible sums range from $2$ to $12$.

1. Sum is Prime: $\{2, 3, 5, 7, 11\}$

• Sum = 2: $(1,1)$ $\to 1$ outcome
• Sum = 3: $(1,2), (2,1)$ $\to 2$ outcomes
• Sum = 5: $(1,4), (2,3), (3,2), (4,1)$ $\to 4$ outcomes
• Sum = 7: $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$ $\to 6$ outcomes
• Sum = 11: $(5,6), (6,5)$ $\to 2$ outcomes

Total favorable prime sum outcomes $= 1 + 2 + 4 + 6 + 2 = 15$.

2. Sum is Multiple of 4: $\{4, 8, 12\}$

• Sum = 4: $(1,3), (2,2), (3,1)$ $\to 3$ outcomes
• Sum = 8: $(2,6), (3,5), (4,4), (5,3), (6,2)$ $\to 5$ outcomes
• Sum = 12: $(6,6)$ $\to 1$ outcome

Total favorable multiple of 4 outcomes $= 3 + 5 + 1 = 9$.

Since prime numbers and multiples of 4 share no common sum values in $\{2,\dots,12\}$, the sets are mutually exclusive.

$$P(\text{Prime or Mult of 4}) = \frac{15 + 9}{36} = \frac{24}{36} = \frac{2}{3}$$

Correct Option: (b) 2/3

Total balls in the bag $= 5 + 4 + 3 = 12$.

Blue balls $= 4$.

Probability that the 1st ball drawn is blue: $P(B_1) = \frac{4}{12} = \frac{1}{3}$.

Since drawing is without replacement, $11$ balls remain with $3$ blue balls:

Probability that the 2nd ball drawn is blue: $P(B_2 | B_1) = \frac{3}{11}$.

$$P(B_1 \cap B_2) = \frac{4}{12} \times \frac{3}{11} = \frac{1}{3} \times \frac{3}{11} = \frac{1}{11}$$

Correct Option: (a) 1/11

For mutually exclusive events, the intersection is the empty set $\emptyset$, so $P(A \cap B) = 0$.

Applying the Addition Rule:

$$P(A \cup B) = P(A) + P(B) = 0.3 + 0.4 = 0.7$$

Correct Option: (b) 0.7

By definition of conditional probability:

$$P(B | A) = \frac{P(A \cap B)}{P(A)}$$

Rearranging to find the joint probability:

$$P(A \cap B) = P(A) \cdot P(B | A) = \frac{2}{5} \times \frac{1}{2} = \frac{1}{5}$$

Correct Option: (c) 1/5

A standard deck has $52$ cards with $4$ aces.

Probability that 1st card is an ace: $P(A_1) = \frac{4}{52} = \frac{1}{13}$.

Probability that 2nd card is an ace given 1st was an ace: $P(A_2 | A_1) = \frac{3}{51} = \frac{1}{17}$.

$$P(A_1 \cap A_2) = \frac{1}{13} \times \frac{1}{17} = \frac{1}{221}$$

Correct Option: (a) 1/221

• (a) Drawing without replacement changes the deck state $\implies$ Dependent.
• (b) Successive coin tosses have independent physical outcomes $\implies$ Independent.
• (c) Successive throws of a fair die are completely independent $\implies$ Independent.

Therefore, both (b) and (c) are independent event pairs.

Correct Option: (d) Both B and C

Total outcomes when tossing 3 coins: $2^3 = 8$.

The only outcome with no heads (all tails) is $(T, T, T) \implies 1$ outcome.

Using the complement rule:

$$P(\text{at least one head}) = 1 - P(\text{no heads}) = 1 - \frac{1}{8} = \frac{7}{8}$$

Correct Option: (c) 7/8

Total students $n(U) = 40$.

Let $C$ be cricket and $F$ be football:

• $n(C) = 28$
• $n(F) = 16$
• $n(C \cap F) = 10$

Number of students who play at least one game:

$$n(C \cup F) = n(C) + n(F) - n(C \cap F) = 28 + 16 - 10 = 34$$

$$P(C \cup F) = \frac{34}{40} = \frac{17}{20}$$

Correct Option: (a) 17/20

Total black cards in a deck $= 26$ (13 spades and 13 clubs).

Given the card is black, the reduced sample space contains $26$ cards.

Number of favorable spade cards $= 13$.

$$P(\text{Spade} | \text{Black}) = \frac{P(\text{Spade} \cap \text{Black})}{P(\text{Black})} = \frac{13/52}{26/52} = \frac{13}{26} = \frac{1}{2}$$

Correct Option: (c) 1/2

By mathematical definition, the conditional probability of event $A$ given that event $B$ has occurred (with $P(B) > 0$) is:

$$P(A | B) = \frac{P(A \cap B)}{P(B)}$$

Correct Option: (c) P(A ∩ B) / P(B)

By the fundamental multiplicative characterization, two events $A$ and $B$ are defined to be statistically independent if and only if:

$$P(A \cap B) = P(A) \cdot P(B)$$

Correct Option: (c) A and B are independent

Sample space: $S = \{1, 2, 3, 4, 5, 6\}$.

Event $A = \{2, 4, 6\}$

Event $B = \{3, 6\} \implies P(B) = \frac{2}{6}$

Intersection $A \cap B = \{6\} \implies P(A \cap B) = \frac{1}{6}$

$$P(A | B) = \frac{P(A \cap B)}{P(B)} = \frac{1/6}{2/6} = \frac{1}{2}$$

Correct Option: (c) 1/2

Let $M$ and $S$ denote liking Mathematics and Science respectively.

• $P(M) = 70\%$
• $P(S) = 40\%$
• $P(M \cap S) = 25\%$

The probability of liking only Mathematics is given by:

$$P(M \cap S') = P(M) - P(M \cap S) = 70\% - 25\% = 45\%$$

Correct Option: (b) 45%

Probability of success in a single trial: $p = 0.35$.

Probability of failure in a single trial: $q = 1 - p = 1 - 0.35 = 0.65$.

For two independent trials, no successes means both trials end in failure ($F_1 \cap F_2$):

$$P(\text{No success}) = q \times q = 0.65 \times 0.65 = 0.4225$$

Correct Option: (b) 0.4225

Total elementary outcomes when throwing 3 dice: $n(S) = 6^3 = 216$.

Let $E$ be the event of getting a sum $\ge 6$. It is easier to evaluate the complement event $E'$ (sum $< 6$, i.e. sum $= 3, 4, 5$).

• Sum = 3: $(1,1,1)$ $\to 1$ triplet
• Sum = 4: Permutations of $(1,1,2) = \frac{3!}{2!} = 3$ triplets: $(1,1,2), (1,2,1), (2,1,1)$
• Sum = 5: Permutations of $(1,1,3) \to 3$ triplets, and permutations of $(1,2,2) \to 3$ triplets $\implies 3 + 3 = 6$ triplets

Total outcomes with sum $< 6$: $n(E') = 1 + 3 + 6 = 10$.

$$P(E') = \frac{10}{216}$$

Using the complement theorem:

$$P(E) = 1 - P(E') = 1 - \frac{10}{216} = \frac{206}{216} = \frac{103}{108} \approx 0.9537$$

Answer: $\frac{103}{108}$

Total sample space $S = \{1, 2, 3, \dots, 30\} \implies n(S) = 30$.

Prime numbers between $1$ and $30$ are:

$$\{2, 3, 5, 7, 11, 13, 17, 19, 23, 29\}$$

Number of prime numbers $n(P) = 10$.

$$P(\text{Prime}) = \frac{n(P)}{n(S)} = \frac{10}{30} = \frac{1}{3}$$

Answer: $\frac{1}{3}$

Total group $= 2 + 5 = 7$ children. Selecting $2$ children has $\binom{7}{2} = \frac{7 \times 6}{2} = 21$ total combinations.

Since $A, B, C$ are mutually exclusive and exhaustive events:

$$P(A) + P(B) + P(C) = 1$$

Express all probabilities in terms of $P(A)$:

• $P(B) = \frac{3}{2}P(A)$
• $P(C) = \frac{1}{2}P(B) = \frac{1}{2}\left(\frac{3}{2}P(A)\right) = \frac{3}{4}P(A)$

Substituting into the exhaustive equation:

$$P(A) + \frac{3}{2}P(A) + \frac{3}{4}P(A) = 1$$

$$P(A)\left(1 + \frac{3}{2} + \frac{3}{4}\right) = 1$$

$$P(A)\left(\frac{4 + 6 + 3}{4}\right) = 1 \implies P(A) \cdot \frac{13}{4} = 1$$

$$P(A) = \frac{4}{13}$$

Verification: $P(B) = \frac{6}{13}, P(C) = \frac{3}{13} \implies \frac{4}{13} + \frac{6}{13} + \frac{3}{13} = 1$.

Answer: $P(A) = \frac{4}{13}$

Total possible outcomes of drawing $2$ cards from $52$: $n(S) = \binom{52}{2} = \frac{52 \times 51}{2} = 1326$.

Let $B =$ event that both are black cards ($26$ black cards in total).

$$n(B) = \binom{26}{2} = \frac{26 \times 25}{2} = 325$$

Let $K =$ event that both are kings ($4$ kings in total).

$$n(K) = \binom{4}{2} = \frac{4 \times 3}{2} = 6$$

Event $B \cap K =$ both cards are black kings ($2$ black kings: King of Spades, King of Clubs).

$$n(B \cap K) = \binom{2}{2} = 1$$

By the Addition Theorem of Probability:

$$P(B \cup K) = P(B) + P(K) - P(B \cap K) = \frac{325 + 6 - 1}{1326} = \frac{330}{1326}$$

Dividing numerator and denominator by $6$:

$$P(B \cup K) = \frac{55}{221} \approx 0.2489$$

Answer: $\frac{55}{221}$

Let $T$ be the event of having teeth cleaned, and $C$ be the event of having a cavity filled.

• $P(T) = 0.44$
• $P(C) = 0.24$
• $P(T \cup C) = 0.60$

Using the General Addition Theorem of Probability:

$$P(T \cup C) = P(T) + P(C) - P(T \cap C)$$

Rearranging for the intersection $P(T \cap C)$:

$$P(T \cap C) = P(T) + P(C) - P(T \cup C) = 0.44 + 0.24 - 0.60 = 0.68 - 0.60 = 0.08$$

Answer: $0.08$ (or $8\%$)

Let $D$ denote the event that a manufactured bulb is defective, and $F$ denote the event that a bulb fails during testing.

• Probability of a defective bulb: $P(D) = 5\% = 0.05$
• Conditional probability of failing given defective: $P(F | D) = 0.9$

By the Multiplication Theorem of Probability:

$$P(D \cap F) = P(D) \cdot P(F | D) = 0.05 \times 0.9 = 0.045$$

Answer: $0.045$ (or $4.5\%$)