Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store
seo-qna
SearchIcon
banner

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
VerifiedVerified
573k+ 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:
True/FalseMultiple choiceTotal
Easy300500800
Difficult200400600
Total5009001400

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 \[P\left( {B\left| A \right.} \right)\].