MHB Multiple choice test : random variable

[mathmari]

Hey!

A multiple choice test consists of 10 questions. For every question there are five possible answers, of which exactly one is correct. A test candidate answers all questions by chance.
(a) Give a suitable random variable with value range and probability distribution in order to work on part (b) with it.
(b) Determine (with intermediate steps) the probability that
(i) exactly 4 questions were answered correctly,
(ii) more than 4 questions have been answered correctly,
(iii) all tasks have been answered correctly,
(iv) at least half of the questions were answered correctly,
(v) at least 5 and at most 8 questions have been answered correctly.For (a) :
Let $X$ be a random variable that describes the number of correct answers out of $10$, right?
For each correct answer the probability is equal to $\frac{1}{5}$ and each wrong answer has the probability $\frac{4}{5}$.

Is that correct so far? :unsure:

 

[I like Serena]

For (a) :
Let $X$ be a random variable that describes the number of correct answers out of $10$, right?
For each correct answer the probability is equal to $\frac{1}{5}$ and each wrong answer has the probability $\frac{4}{5}$.

Hey mathmari!

Yep. (Nod)

We still need to identify the probability distribution function for (a) though.

 

[mathmari]

Yep. (Nod)

So is the range of the random variable $\{0,1,2,3,4,5,6,7,8,9,10\}$ ? :unsure:

We still need to identify the probability distribution function for (a) though.

Do we have to calculate for each value of the random variable the corresponding probability? :unsure:

 

[I like Serena]

So is the range of the random variable $\{0,1,2,3,4,5,6,7,8,9,10\}$ ?

Yep. (Nod)

Do we have to calculate for each value of the random variable the corresponding probability?

It think we should just identify the name of the probability distribution and its parameters.
If we want to, we can also calculate the corresponding probabilities of the possible outcomes.

 

[mathmari]

It think we should just identify the name of the probability distribution and its parameters.

What do you mean by the name of probability distribution an its parameters? Is it $P(X=k)$ with $k\in\{0, 1, 2, \ldots , 10\}$? :unsure:

If we want to, we can also calculate the corresponding probabilities of the possible outcomes.

This is then part (b), or not? :unsure:

 

[I like Serena]

What do you mean by the name of probability distribution an its parameters? Is it $P(X=k)$ with $k\in\{0, 1, 2, \ldots , 10\}$?

You wrote:

For each correct answer the probability is equal to $\frac{1}{5}$ and each wrong answer has the probability $\frac{4}{5}$.

This describes a Bernoulli distribution with parameter $p=\frac 15$, which is for an experiment with a single yes-no question.
However, we don't have a single yes-no question, but we have 10 questions.
We are looking for a distribution for the number of successes in a sequence of $n$ independent experiments, each asking a yes–no question, and each with its own Boolean-valued outcome: success (with probability $p$) or failure (with probability $q = 1 − p$). (Sweating)

This is then part (b), or not?

More or less. We can use a table with the probabilities of each possible outcome to help us answer each of the questions in (b).
But we can also find the answers for (b) in a different fashion.

 

[mathmari]

This describes a Bernoulli distribution with parameter $p=\frac 15$, which is for an experiment with a single yes-no question.
However, we don't have a single yes-no question, but we have 10 questions.
We are looking for a distribution for the number of successes in a sequence of $n$ independent experiments, each asking a yes–no question, and each with its own Boolean-valued outcome: success (with probability $p$) or failure (with probability $q = 1 − p$). (Sweating)

So do we have a binomial distribution? :unsure:

More or less. We can use a table with the probabilities of each possible outcome to help us answer each of the questions in (b).
But we can also find the answers for (b) in a different fashion.

Do we have the following ?

(i) $p_X(4)=P(X=4)=\binom{10}{4}p^4(1-p)^{10-4}=\binom{10}{4}\left (\frac{1}{5}\right )^4\left (\frac{4}{5}\right )^{6}$
(ii) $P(X>4)=1-P(X\leq 4)=1-\sum_{i=0}^4P(X=i)=1-\sum_{i=0}^4\binom{10}{i}p^i(1-p)^{10-i}=1-\sum_{i=0}^4\binom{10}{i}\left (\frac{1}{5}\right )^i\left (\frac{4}{5}\right )^{10-i}$
(iii) $p_X(10)=P(X=10)=\binom{10}{10}p^{10}(1-p)^{10-10}=\left (\frac{1}{5}\right )^{10}$
(iv) $P(X\geq 5)=P(X>4)=1-\sum_{i=0}^4\binom{10}{i}\left (\frac{1}{5}\right )^i\left (\frac{4}{5}\right )^{10-i}$
(v) $P(5\leq X\leq 8)=P(X\leq 8)-P(X<5)=P(X\leq 8)-P(X\leq 4)=[1-P(X> 8)]-[1-P(X> 4)]=[1-P(X=9)-P(X=10)]-\left [1-1+\sum_{i=0}^4\binom{10}{i}\left (\frac{1}{5}\right )^i\left (\frac{4}{5}\right )^{10-i}\right ]=[1-\binom{10}{9}\left (\frac{1}{5}\right )^9\left (\frac{4}{5}\right )^{10-9}-\left (\frac{1}{5}\right )^{10}]-\left [1-1+\sum_{i=0}^4\binom{10}{i}\left (\frac{1}{5}\right )^i\left (\frac{4}{5}\right )^{10-i}\right ]$

:unsure:

 

[I like Serena]

Yep. All correct. (Sun)