Question

A student is given a quiz with 10 true or false questions, and he answers by sheer guessing. If X is the number of questions answered correctly, what is the probability distribution of X? If the student passes the quiz by getting X max correct passers, what is the probability that the student passes the quiz?

Answer 1

We will first find the number of possibilities of answering the questions. Then we will find a number of ways to answer X questions correctly. Then we will find the probability of answering X questions correct out of 10 question by relation P(X) = $\dfrac{\text{n(X)}}{\text{n(S)}}$, where X is the event set of getting X questions correct and S is the event set of answering all the questions. Then, we will substitute values of X from 0 to 10 and tabulate the data to find the probability distribution. Then we will find the probability of passing if the student has to answer X questions correctly to pass the test.

Complete step-by-step answer:
The number of ways of answering a true or false question is 2, since the answer can either be true or false.
Thus, the number of answers to the 1st question is 2, the number of ways of answering the 2nd question is true and so on.
Therefore, the number of ways of answering 10 true or false questions is $2\times 2\times 2...$ = ${{2}^{10}}$.
Let S be the event of answering 10 true or false questions with guessing.
Thus, number of elements in S = n(S) = ${{2}^{10}}$ = 1024
It is given that X is the number of questions answered correctly.
Now, we need to choose X number of questions from 10 questions, which are correct.
Number of ways of choosing r elements from n elements is given by $^{n}{{C}_{r}}$, where $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.
Therefore, the number of ways of choosing X questions from 10 questions is given as $^{10}{{C}_{X}}$.
Let X be the event of getting X questions correct.
Thus, number of elements in event set X = n(X) = $^{10}{{C}_{X}}$
Now, as we know, probability of getting an event A true is given by P(A) = $\dfrac{\text{n(A)}}{\text{n(S)}}$.
Therefore, probability of getting X questions correct out of 10 questions, P(X) = $\dfrac{^{10}{{C}_{X}}}{1024}$.
If the student got no correct questions, then X = 0.
$\Rightarrow$ P(X = 0) = $\dfrac{^{10}{{C}_{0}}}{1024}$
$\Rightarrow$ P(X = 0) = $\dfrac{1}{1024}$
$\Rightarrow$ P(X = 0) = 0.00098
Similarly, if the students got one correct question, then X = 1.
$\Rightarrow$ P(X = 1) = $\dfrac{^{10}{{C}_{1}}}{1024}$
$\Rightarrow$ P(X = 1) = $\dfrac{10}{1024}$
$\Rightarrow$ P(X = 1) = 0.0098
In the similar fashion, we will calculate the probability for X = 2, 3, 4, 5….10 and tabulate the data.
Thus, the probability distribution table is:

X	0	1	2	3	4	5	6	7	8	9	10
P(X)	0.00098	0.0098	0.044	0.117	0.205	0.246	0.205	0.117	0.044	0.0098	0.00098

Now, if the student has to answer X questions correctly to pass the test,
He can answer X questions correctly, or X + 1 questions correctly or X + 2 questions correctly and so on until he answers 10 questions correctly.
Therefore, probability will be P(passing) = $\dfrac{^{10}{{C}_{X}}}{1024}+\dfrac{^{10}{{C}_{X+1}}}{1024}+\dfrac{^{10}{{C}_{X+2}}}{1024}...$
It can also be written as P(passing) = $\sum\limits_{X}^{10}{\dfrac{^{10}{{C}_{X}}}{1024}}$

Note: For easy calculation of $^{n}{{C}_{r}}$, students can keep in mind that $^{n}{{C}_{r}}{{=}^{n}}{{C}_{n-r}}$. Therefore $^{10}{{C}_{1}}{{=}^{10}}{{C}_{10-1}}{{=}^{10}}{{C}_{9}}$.