A student is taking a multiple-choice exam in which each question has four choices. Assume that the student has no knowledge of the correct answers to any of the questions. She has decided on a strategy in which she will place four balls (marked A, B, C,

Добавлена: 15.04.2026
Обновлена: 15.04.2026
Предмет: Другое
Автор: Кэмп

Условие:

A student is taking a multiple-choice exam in which each question has four choices. Assume that the student has no knowledge of the correct answers to any of the questions. She has decided on a strategy in which she will place four balls (marked A, B, C, and D) into a box. She randomly selects one ball for each question and replaces the ball in the box. The marking on the ball will determine her answer to the question. There are five multiple-choice questions on the exam. What is the probability that she will get a) five questions correct? b) at least four questions correct? c) no questions correct? d) no more than two questions correct?

Решение:

Given:

Number of questions, $n = 5$
Number of choices per question, $k = 4$
Probability of getting a question correct, $p = \frac{1}{4}$
Probability of getting a question wrong, $q = 1 - p = \frac{3}{4}$

The Binomial Probability Formula:\nThe probability of getting exactly $r$ correct answers out of $n$ questions is given by the formula:

\nP(X = r) = \binom{n}{r} p^r q^{n-r}

\nwhere $\binom{n}{r}$ is the binomial coefficient, calculated as:

\binom{n}{r} = \frac{n!}{r!(n-r)!}

a) Probability of getting five questions correct:\nTo fin...

Внутри — полный разбор, аргументация, алгоритм решения, частые ошибки и как отвечать на каверзные вопросы препода, если спросит

Попробуй решить по шагам

Попробуй один шаг и продолжи в режиме обучения или посмотри готовое решение

Какая формула используется для расчета вероятности получения ровно $r$ правильных ответов из $n$ вопросов, если каждый вопрос имеет $k$ вариантов ответа и студент выбирает ответы случайным образом?

$P(X = r) = \binom{n}{r} p^r (1-p)^{n-r}$