Statistics for Data Science I · Quiz 2 · Sep 2025

**1. Use the continuity condition at $x = 0$:** For the piecewise function $f(x)$ to be continuous at $x = 0$, the left-hand limit, right-hand limit, and function value at $x = 0$ must all be equal: $$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$$ This continuity condition yields $q = 8$. **2. Compute $\frac{q}{2}$:** $$\frac{q}{2} = \frac{8}{2} = 4$$ **Answer: 4**

**1. Count the total number of arrangements of the 5 books:** There are $5$ distinct books in total, which can be arranged on a shelf in: $$5! = 120\text{ ways}$$ **2. Subtract the arrangements where the 2 novels ARE together:** - Treat the $2$ novels as a single unit. Together with the $3$ textbooks, there are $4$ units, which can be arranged in $4! = 24$ ways. - Within the novel unit, the $2$ novels can be ordered in $2! = 2$ ways. - Thus, the number of arrangements with the two novels together is $4! \times 2! = 24 \times 2 = 48$. **3. Subtract from the total:** $$120 - 48 = 72$$ **Answer: Option D** A: **Incorrect:** $120 = 5!$ is the total number of unrestricted arrangements, including those where the novels are together. B: **Incorrect:** Subtracts $24$ instead of $48$ ($4! \times 2!$). C: **Incorrect:** $24 = 4!$ is only half the number of arrangements where the novels are together. D: **Correct:** $5! - 4!2! = 120 - 48 = 72$ ways.

**1. Count the number of valid letter pairs for the first two positions:** There are $26$ capital letters ($5$ vowels and $21$ consonants): - Total ways to choose $2$ distinct capital letters in order: $26 \times 25 = 650$. - Number of ways to choose $2$ distinct letters with **no vowels** (both consonants): $21 \times 20 = 420$. - Therefore, the number of distinct letter pairs with **at least one vowel** is: $$650 - 420 = 230$$ **2. Multiply by the choices for the third position:** The third position is a digit from $\{1, 2, 3, 4, 5\}$ ($5$ choices): $$\text{Total codes} = 230 \times 5 = 1150$$ **Answer: Option D** A: **Incorrect:** $950$ undercounts the valid vowel-containing pairs. B: **Incorrect:** $1300$ includes pairs where the two vowels are identical ($5 \times 5 = 25$ repeated vowels times $2$). C: **Incorrect:** Multiplies by $10$ digits instead of the $5$ digits $\{1,2,3,4,5\}$. D: **Correct:** $(26 \times 25 - 21 \times 20) \times 5 = 230 \times 5 = 1150$.

**1. Define the events for drawing 2 balls without replacement from 4 red and 2 blue balls (6 total):** - $A$: The first ball drawn is red. - $B$: The second ball drawn is blue. **2. Compute $P(A \cap B)$ and $P(B)$:** - Probability that the first is red AND the second is blue: $$P(A \cap B) = P(A)P(B \mid A) = \frac{4}{6} \times \frac{2}{5} = \frac{8}{30}$$ - Probability that the second ball is blue (by symmetry, or law of total probability): $$P(B) = P(R_1 B_2) + P(B_1 B_2) = \frac{4}{6}\cdot\frac{2}{5} + \frac{2}{6}\cdot\frac{1}{5} = \frac{8 + 2}{30} = \frac{10}{30} = \frac{1}{3}$$ **3. Compute the conditional probability $P(A \mid B)$:** $$P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{8/30}{10/30} = \frac{8}{10} = \frac{4}{5}$$ (Intuitively: given the second ball is blue, $1$ blue ball is used up, leaving $4$ red and $1$ blue ball—$5$ equally likely candidates—for the first draw, so $P(A \mid B) = \frac{4}{5}$.) **Answer: Option B** A: **Incorrect:** $2/3 = 4/6$ is the unconditional probability $P(A)$, not $P(A \mid B)$. B: **Correct:** Given the second ball is blue, 4 of the remaining 5 balls that could have been drawn first are red, so $P(A \mid B) = 4/5$. C: **Incorrect:** $1/3$ is $P(B)$, the probability that the second ball is blue. D: **Incorrect:** $2/5$ is $P(B \mid A)$, the probability that the second ball is blue given the first is red.

**1. List the outcomes for each event from the 4 equally likely tickets $\{111, 122, 212, 221\}$:** - $A_1$ ($1^{\text{st}}$ digit is $1$): $\{111, 122\} \implies P(A_1) = \frac{2}{4} = \frac{1}{2}$ - $A_2$ ($2^{\text{nd}}$ digit is $1$): $\{111, 212\} \implies P(A_2) = \frac{2}{4} = \frac{1}{2}$ - $A_3$ ($3^{\text{rd}}$ digit is $1$): $\{111, 221\} \implies P(A_3) = \frac{2}{4} = \frac{1}{2}$ Thus, **$P(A_1) = P(A_2) = P(A_3) = \frac{1}{2}$** (**Option C is correct**). **2. Check pairwise and mutual independence:** - Every pair intersects only at ticket $111$: $$P(A_1 \cap A_2) = P(A_2 \cap A_3) = P(A_1 \cap A_3) = P(\{111\}) = \frac{1}{4} = \frac{1}{2} \times \frac{1}{2}$$ Thus, **$A_1, A_2, A_3$ are pairwise independent** (**Option A is correct**). - For all three events together, $A_1 \cap A_2 \cap A_3 = \{111\}$, so: $$P(A_1 \cap A_2 \cap A_3) = \frac{1}{4} \ne P(A_1)P(A_2)P(A_3) = \frac{1}{8}$$ Hence, $P(A_1 \cap A_2 \cap A_3) = \frac{1}{4} \ne 0$ (Option D is false) and they are **not** mutually independent (Option B is false). **Answer: Options A, C** A: **Correct:** For each pair $i \ne j$, $P(A_i \cap A_j) = 1/4 = P(A_i)P(A_j)$, so the events are pairwise independent. B: **Incorrect:** $P(A_1 \cap A_2 \cap A_3) = 1/4 \ne (1/2)^3 = 1/8$, so they are not mutually independent. C: **Correct:** Each event contains 2 of the 4 tickets, so $P(A_1) = P(A_2) = P(A_3) = 1/2$. D: **Incorrect:** Ticket $111$ belongs to all three events, so $P(A_1 \cap A_2 \cap A_3) = 1/4 \ne 0$.

**1. Use the addition rule for two events $E = A \cup B$ and $C$:** For any two events $E$ and $C$: $$P(E \cup C) = P(E) + P(C) - P(E \cap C)$$ Setting $E = A \cup B$, we have $E \cup C = A \cup B \cup C$ and $E \cap C = (A \cup B) \cap C$. **2. Substitute the given probabilities:** - $P(A \cup B) = 0.7$ - $P(C) = 0.4$ - $P(A \cup B \cup C) = 0.9$ $$0.9 = 0.7 + 0.4 - P((A \cup B) \cap C) \implies P((A \cup B) \cap C) = 1.1 - 0.9 = 0.2$$ **Answer: 0.2**

**1. Define the independent events $T$ (likes tea) and $C$ (likes coffee):** - $P(T) = 0.6 \implies P(T^c) = 1 - 0.6 = 0.4$ - $P(C) = 0.5 \implies P(C^c) = 1 - 0.5 = 0.5$ **2. Compute the probability of liking Ordered/Exactly One of tea or coffee:** Since $T$ and $C$ are independent: $$P(\text{exactly one}) = P(T \cap C^c) + P(T^c \cap C) = P(T)P(C^c) + P(T^c)P(C)$$ $$= (0.6)(0.5) + (0.4)(0.5) = 0.30 + 0.20 = 0.5$$ **Answer: 0.5**

**1. Recall the relationship between permutations ${}^nP_r$ and combinations ${}^nC_r$:** $${}^nP_r = r! \cdot {}^nC_r$$ **2. Substitute the given values ${}^nC_r = 84$ and ${}^nP_r = 504$:** $$r! = \frac{{}^nP_r}{{}^nC_r} = \frac{504}{84} = 6$$ Since $3! = 6$, we get: $$r = 3$$ **Answer: 3**

**1. Analyze the composition of the two groups:** - **Ankita's group (3 students):** $2$ Science ($S$) and $1$ Arts ($A$). Ankita must select $2$ students. - **Rohit's group (3 students):** $1$ Science ($S$) and $2$ Arts ($A$). Rohit must select $2$ students. - Note that across both groups, there are only $2 + 1 = 3$ Science students and $1 + 2 = 3$ Arts students in total, and we need $2$ Science and $2$ Arts students on the 4-member team. **2. Count the valid cases:** - **Case 1:** Ankita selects $2$ Science students and $0$ Arts students, while Rohit selects $0$ Science students and $2$ Arts students: $$\binom{2}{2}\binom{1}{0} \times \binom{1}{0}\binom{2}{2} = 1 \times 1 = 1\text{ way}$$ - **Case 2:** Ankita selects $1$ Science and $1$ Arts student, while Rohit selects $1$ Science and $1$ Arts student: $$\binom{2}{1}\binom{1}{1} \times \binom{1}{1}\binom{2}{1} = (2 \times 1) \times (1 \times 2) = 4\text{ ways}$$ **3. Add the cases:** $$\text{Total ways} = 1 + 4 = 5$$ **Answer: 5**

**1. Use the Inclusion-Exclusion Principle to find how many students play at least one sport:** Let $C$ be the set of cricket players ($|C| = 60$) and $F$ be the set of football players ($|F| = 50$), with $|C \cap F| = 30$: $$|C \cup F| = |C| + |F| - |C \cap F| = 60 + 50 - 30 = 80$$ **2. Find the probability of playing neither sport:** Out of $100$ students, the number of students who play neither cricket nor football is $100 - 80 = 20$. Therefore, the probability is: $$P(\text{neither}) = \frac{20}{100} = 0.2$$ **Answer: 0.2**

**1. Analyze the properties of forming a 4-member committee from 11 distinct people:** - **Without replacement:** A person cannot be selected more than once on the same 4-member committee, so members are selected **without replacement** (**Option A is correct**). - **Order does not matter / Combination:** In a committee with no distinct roles, the order in which the 4 members are chosen does not matter, so **combinations** are used rather than permutations (**Option D is correct**, and Options B and C are incorrect). **Answer: Options A, D** A: **Correct:** The 4 committee members must be distinct individuals, so selection is without replacement. B: **Incorrect:** A committee is an unordered subset of members, so order does not matter. C: **Incorrect:** Permutations are used when order matters; here order does not matter. D: **Correct:** Because order does not matter when selecting a committee, combinations are used.

**1. Count the total number of 4-member committees from the 11 club members:** $$\text{Total committees} = \binom{11}{4} = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = 330$$ **2. Subtract the committees with NO cricket players (all 4 from the 5 football players):** $$\text{Committees with 0 cricket players} = \binom{5}{4} = 5$$ Therefore, the number of committees with **at least one** cricket player is: $$330 - 5 = 325$$ **Answer: Option B** A: **Incorrect:** $300$ undercounts the valid committees. B: **Correct:** $\binom{11}{4} - \binom{5}{4} = 330 - 5 = 325$. C: **Incorrect:** Subtracts $20$ instead of $\binom{5}{4} = 5$. D: **Incorrect:** $280$ is not equal to $\binom{11}{4} - \binom{5}{4}$.

**1. Set up the combination equation for selections that include two particular books:** If $2$ particular books are already included in the 4-book selection, we must choose the remaining $4 - 2 = 2$ books from the remaining $n - 2$ books: $$\binom{n - 2}{2} = 15$$ **2. Solve for $n$:** $$\frac{(n - 2)(n - 3)}{2} = 15 \implies (n - 2)(n - 3) = 30$$ Since $6 \times 5 = 30$, we have $n - 2 = 6 \implies n = 8$. **Answer: 8**

**1. Use $n = 8$ from Question 240 to compute $\binom{n}{4}$:** The total number of ways to issue any $4$ books from the collection of $n = 8$ books is: $$\binom{8}{4} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70$$ **Answer: 70**

**1. Identify the probabilities for each box:** - $P(C) = 0.5$ and $P(D) = 0.5$. - Box C has $6$ green and $3$ black pens ($9$ total), so $P(\text{Green} \mid C) = \frac{6}{9} = \frac{2}{3}$. - Box D has $3$ green and $6$ black pens ($9$ total), so $P(\text{Green} \mid D) = \frac{3}{9} = \frac{1}{3}$. **2. Apply the Law of Total Probability:** $$P(\text{Green}) = P(C)P(\text{Green} \mid C) + P(D)P(\text{Green} \mid D) = \frac{1}{2}\left(\frac{2}{3}\right) + \frac{1}{2}\left(\frac{1}{3}\right) = \frac{1}{3} + \frac{1}{6} = \frac{1}{2} = 0.5$$ **Answer: 0.5**