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UPSC NDA Math Model Paper 9

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Question : 105 of 120

Marks: +1, -0

Of all the Testbook subscribers, it is known that 60% are preparing for NDA exam and 40% are preparing for Banking exams. Previous year result shows that 30% of the students who were preparing for NDA exam were able to clear the exam successfully while 20% of the students who were preparing for Banking exams were able to clear the exam successfully. If one Testbook subscriber is selected at random and he/she had cleared the exam. What is the probability he/she was a preparing for NDA exam?

11/13
9/13
7/13
8/13

Solution:

Let

E_1

be th event 'Student is preparing for NDA',

E_2

be the event 'Student is preparing for Banking exams and X be the event 'Student was able to clear theexam'

Given:

P(E_1) = 0.6

and

P(E_2) = 0.4

P(X \mid E_1) =

Probability 'Student was able to clear the exam given that he/she was preparing for NDA exam' = 0.3

P(X \mid E_2) =

Probability 'Student was able to clear the exam given that he/she was preparing for Banking exams' = 0.2

Here, we have to find the value of

P(E_1 \mid X).

As we know that according to bayes' theorem:

P(E_i \mid A) = \frac{P(E_i) \times P(A \mid E_i)}{\sum\limits_{i=1}^{n} P(E_i) \times P(A \mid E_i)}, \quad i = 1, 2, \ldots, n

\Rightarrow P(E_1 \mid X) = \frac{P(E_1) \cdot P(X \mid E_1)}{[P(E_1) \cdot P(X \mid E_1) + P(E_2) \cdot P(X \mid E_2)]}

\Rightarrow P(E_1 \mid X) = \frac{0.6 \times 0.3}{[0.6 \times 0.3 + 0.4 \times 0.2]}

= \frac{9}{13}

={9}/{{13}}

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