An instructor has a question bank consisting of 300 easy True/False questions, 200 difficult True/False questions, 500 easy multiple-choice questions, and 400 difficult multiple-choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple-choice question?
Answer
628.8k+ views
Hint: In this question, first of all, draw a tabular form to write and understand the given data simply. Then calculate the probability of selecting an easy multiple-choice question and use conditional probability to get the required probability. So, use this concept to reach the solution to the given problem.
Let us define the following events as:
\[E\]: gets an easy question
\[M\]: gets a multiple-choice question
\[D\]: gets a difficult question
\[T\]: gets a True/False question
The questions in the question bank can be tabulated as follows:
So, the total number of questions = 1400
Total number of multiple-choice questions = 900
Therefore, probability of selecting an easy multiple-choice question is given by
\[P\left( {E \cap M} \right) = \dfrac{{500}}{{1400}} = \dfrac{5}{{14}}\]
And probability of selecting a multiple-choice question is given by
\[P\left( M \right) = \dfrac{{900}}{{1400}} = \dfrac{9}{{14}}\]
The probability that a randomly selected question will be an easy question, given that it is a multiple-choice question is given by
\[P\left( {E\left| M \right.} \right) = \dfrac{{P\left( {E \cap M} \right)}}{{P\left( M \right)}} = \dfrac{{\dfrac{5}{{14}}}}{{\dfrac{9}{{14}}}} = \dfrac{5}{9}\]
Thus, the required probability is \[\dfrac{5}{9}\].
Note: The probability of an event is always lying between 0 and 1 i.e., \[0 \leqslant P\left( E \right) \leqslant 1\]. We know that the probability of an event \[E\] is given by \[P\left( E \right) = \dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of outcomes}}}}\]. The condition probability of an event B is the probability that the event will occur given the knowledge that an event A has already occurred. This probability is written as \[P\left( {B\left| A \right.} \right)\].
Let us define the following events as:
\[E\]: gets an easy question
\[M\]: gets a multiple-choice question
\[D\]: gets a difficult question
\[T\]: gets a True/False question
The questions in the question bank can be tabulated as follows:
| True/False | Multiple choice | Total | |
| Easy | 300 | 500 | 800 |
| Difficult | 200 | 400 | 600 |
| Total | 500 | 900 | 1400 |
So, the total number of questions = 1400
Total number of multiple-choice questions = 900
Therefore, probability of selecting an easy multiple-choice question is given by
\[P\left( {E \cap M} \right) = \dfrac{{500}}{{1400}} = \dfrac{5}{{14}}\]
And probability of selecting a multiple-choice question is given by
\[P\left( M \right) = \dfrac{{900}}{{1400}} = \dfrac{9}{{14}}\]
The probability that a randomly selected question will be an easy question, given that it is a multiple-choice question is given by
\[P\left( {E\left| M \right.} \right) = \dfrac{{P\left( {E \cap M} \right)}}{{P\left( M \right)}} = \dfrac{{\dfrac{5}{{14}}}}{{\dfrac{9}{{14}}}} = \dfrac{5}{9}\]
Thus, the required probability is \[\dfrac{5}{9}\].
Note: The probability of an event is always lying between 0 and 1 i.e., \[0 \leqslant P\left( E \right) \leqslant 1\]. We know that the probability of an event \[E\] is given by \[P\left( E \right) = \dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of outcomes}}}}\]. The condition probability of an event B is the probability that the event will occur given the knowledge that an event A has already occurred. This probability is written as \[P\left( {B\left| A \right.} \right)\].
Recently Updated Pages
Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

What is cell theory Who formulated it class 11 biology CBSE

Phyllotaxy is the arrangement of ALeaflets BLeaves class 11 biology CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

The symbiotic association of fungi and algae is called class 11 biology CBSE

Cell theory was formulated by A Schleiden and Schwann class 11 biology CBSE

