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Nia is 1 of 24 students in a class. Every month, Nia's teacher randomly selects 4 students from their class to act as class president, vice president, secretary, and treasurer. No one student can hold two positions. In a given month, what is the probability that Nia is chosen as president? Choose 1 answer: (A) (1)/(_(24)P_(4)) (B) (1)/(_(24)C_(4)) (C) (_(23)P_(3))/(_(24)P_(4)) (D) ((_(24)P_(1))*(_(24)P_(3)))/(_(24)P_(4))

Nia is 11 of 2424 students in a class. Every month, Nia's teacher randomly selects 44 students from their class to act as class president, vice president, secretary, and treasurer. No one student can hold two positions.\newlineIn a given month, what is the probability that Nia is chosen as president?\newlineChoose 11 answer:\newline(A) 124P4 \frac{1}{{ }_{24} \mathrm{P}_{4}} \newline(B) 124C4 \frac{1}{{ }_{24} \mathrm{C}_{4}} \newline(C) 23P324P4 \frac{{ }_{23} \mathrm{P}_{3}}{{ }_{24} \mathrm{P}_{4}} \newline(D) (24P1)⋅(24P3)24P4 \frac{\left({ }_{24} \mathbf{P}_{1}\right) \cdot\left({ }_{24} \mathrm{P}_{3}\right)}{{ }_{24} \mathrm{P}_{4}}

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Q. Nia is 11 of 2424 students in a class. Every month, Nia's teacher randomly selects 44 students from their class to act as class president, vice president, secretary, and treasurer. No one student can hold two positions.\newlineIn a given month, what is the probability that Nia is chosen as president?\newlineChoose 11 answer:\newline(A) 124P4 \frac{1}{{ }_{24} \mathrm{P}_{4}} \newline(B) 124C4 \frac{1}{{ }_{24} \mathrm{C}_{4}} \newline(C) 23P324P4 \frac{{ }_{23} \mathrm{P}_{3}}{{ }_{24} \mathrm{P}_{4}} \newline(D) (24P1)⋅(24P3)24P4 \frac{\left({ }_{24} \mathbf{P}_{1}\right) \cdot\left({ }_{24} \mathrm{P}_{3}\right)}{{ }_{24} \mathrm{P}_{4}}
  1. Understand the problem: Understand the problem.\newlineWe need to calculate the probability that Nia is chosen as president out of 2424 students, with 44 students being chosen for different positions.
  2. Determine total combinations: Determine the total number of ways to choose 44 students out of 2424 without regard to order.\newlineThis is a combination problem because the order in which the students are chosen does not matter. We use the combination formula:\newline(244)=24!4!×(24−4)!\binom{24}{4} = \frac{24!}{4! \times (24 - 4)!}
  3. Calculate combinations: Calculate the total number of combinations.\newline24C4=24!4!×20!_{24}C_{4} = \frac{24!}{4! \times 20!}\newline24C4=(24×23×22×21)(4×3×2×1)_{24}C_{4} = \frac{(24 \times 23 \times 22 \times 21)}{(4 \times 3 \times 2 \times 1)}\newline24C4=10626_{24}C_{4} = 10626
  4. Determine probability: Determine the probability that Nia is chosen as president.\newlineSince there is only one president, and Nia has an equal chance of being chosen as any other student, the probability is simply 11 out of the total number of combinations.\newlineP(Nia is president)=1(244)P(\text{Nia is president}) = \frac{1}{\binom{24}{4}}
  5. Substitute into formula: Substitute the total number of combinations into the probability formula.\newlineP(Nia is president)=110626P(\text{Nia is president}) = \frac{1}{10626}
  6. Choose correct answer: Choose the correct answer from the given options.\newlineThe correct answer is (B) because it represents the probability of choosing Nia as president out of the total number of combinations of 44 students from 2424.

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