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saswata4s
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Can solve this quest by applying set theory
Let A= PPL who passed in 1st test =60
Let C= PPL who passed in 2nd test =50
Let B= overlap i.e. PPL who passed in both= 30
Then A+B=60
Therefore A=30
B+C= 50
Therefore C=20
A+B+C= 80
Remaining= 20 (PPL who did not pass at all)
=20/100= 1/5
Hence option A
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I used Venn Diagram to solve this question.
1st circle gives us 30 passed
2nd circle gives us 20 passed
inner part of both circles 1 and 2 give us 30 students. In total we have 80,but since total number of students is 100, we conclude that 20 students didn't pass the test. So, 20/100=1/5
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Please correct me if i am wrong in any of the following steps-
(i am not using Venn Diagram)

As per the question, it states that 30 students pass both the tests (i.e 30 out of 100 passed both examinations)

this means that rest of 70 students will be grouped as "students who didn't passed the both tests"

This itself implies the following-
students failed in 1st exam + students failed in 2nd exam + students failed in both exams= 70
=> 40 + 50 + students failed in both = 70
=> students failed in both = -20 (ie 20 students failed in both)

thus probability is 20/100 => 1/5

please correct me if i am wrong.
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saswata4s
100 students appeared for two examinations. 60 passed the first, 50 passed the second and 30 passed both. Find the probability that a student selected at random has failed in both the examinations?

(A) 1/5
(B) 1/7
(C) 5/7
(D) 5/6
(E) 6/7

a- Students who passed only 1st exam
b- Students who passed both the exams
c- Students who passed only 2nd exam
a+b = 60
b+c = 50
b=30
Therefore, a= 30; c=20
Hence, now we can add all the a+b+c students
i.e. 30+30+20=80
Therefore students who failed in both the examination = 100-80= 20
Probability= 20/100 = 1/5
Correct answer is (A)
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Failed 1st exam = (f1) = 40
Failed 2nd exam = (f2) = 50
Failed both exams = (fb) = 70

Ans: P(f1) + P(f2) - P(fb)
= 40/100 + 50/100 - 70/100
= 20/100
= 1/5
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