Question

stat 213 assignment 1: problem 7 (1 point) suppose i give you a list of 22 problems to study, from which i will randomly pick 10 questions for an exam. for whatever reason, you prepare for this exam by completing and understanding how to solve 12 questions of the 22, so there are 10 questions you do not know how to solve. not sure how to perform some of the tasks? click to see similar problem part (a) compute the probability that you completely solve 5 of the 10 questions appearing on the midterm exam (use four decimals) part (b) to pass the midterm, you must correctly solve at least half of the 10 questions. compute the probability that you will pass the exam. (use four decimals)

Explanation:

Step1: Recall hyper - geometric probability formula

The hyper - geometric probability formula is $P(X = k)=\frac{\binom{M}{k}\binom{N - M}{n - k}}{\binom{N}{n}}$, where $N$ is the total number of items, $M$ is the number of "success" items in the population, $n$ is the sample size, and $k$ is the number of "success" items in the sample. Here, $N = 22$, $M=12$ (the number of questions you can solve), $n = 10$ (the number of questions on the exam).

Step2: Solve part (a)

For part (a), we want $k = 5$.
$P(X = 5)=\frac{\binom{12}{5}\binom{22 - 12}{10 - 5}}{\binom{22}{10}}$
First, calculate the binomial coefficients:
$\binom{12}{5}=\frac{12!}{5!(12 - 5)!}=\frac{12!}{5!7!}=\frac{12\times11\times10\times9\times8}{5\times4\times3\times2\times1}=792$
$\binom{10}{5}=\frac{10!}{5!(10 - 5)!}=\frac{10!}{5!5!}=\frac{10\times9\times8\times7\times6}{5\times4\times3\times2\times1}=252$
$\binom{22}{10}=\frac{22!}{10!(22 - 10)!}=\frac{22!}{10!12!}=646646$
$P(X = 5)=\frac{792\times252}{646646}\approx0.3081$

Step3: Solve part (b)

To pass the exam, we need $k\geq5$. So $P(X\geq5)=P(X = 5)+P(X = 6)+P(X = 7)+P(X = 8)+P(X = 9)+P(X = 10)$
$P(X = k)=\frac{\binom{12}{k}\binom{10}{10 - k}}{\binom{22}{10}}$
$P(X = 6)=\frac{\binom{12}{6}\binom{10}{4}}{\binom{22}{10}}$, $\binom{12}{6}=\frac{12!}{6!(12 - 6)!}=924$, $\binom{10}{4}=\frac{10!}{4!(10 - 4)!}=210$
$P(X = 6)=\frac{924\times210}{646646}\approx0.3029$
$P(X = 7)=\frac{\binom{12}{7}\binom{10}{3}}{\binom{22}{10}}$, $\binom{12}{7}=\frac{12!}{7!(12 - 7)!}=792$, $\binom{10}{3}=\frac{10!}{3!(10 - 3)!}=120$
$P(X = 7)=\frac{792\times120}{646646}\approx0.1470$
$P(X = 8)=\frac{\binom{12}{8}\binom{10}{2}}{\binom{22}{10}}$, $\binom{12}{8}=\frac{12!}{8!(12 - 8)!}=495$, $\binom{10}{2}=\frac{10!}{2!(10 - 2)!}=45$
$P(X = 8)=\frac{495\times45}{646646}\approx0.0345$
$P(X = 9)=\frac{\binom{12}{9}\binom{10}{1}}{\binom{22}{10}}$, $\binom{12}{9}=\frac{12!}{9!(12 - 9)!}=220$, $\binom{10}{1}=\frac{10!}{1!(10 - 1)!}=10$
$P(X = 9)=\frac{220\times10}{646646}\approx0.0034$
$P(X = 10)=\frac{\binom{12}{10}\binom{10}{0}}{\binom{22}{10}}$, $\binom{12}{10}=\frac{12!}{10!(12 - 10)!}=66$, $\binom{10}{0}=1$
$P(X = 10)=\frac{66\times1}{646646}\approx0.0001$
$P(X\geq5)=0.3081 + 0.3029+0.1470+0.0345+0.0034+0.0001=0.7960$

Answer:

Part (a): $0.3081$
Part (b): $0.7960$