DEAR STUDENTS
Solving GATE Previous Year's Questions (PYQs) not only
clarifies concepts but also helps to improve flexibility, speed, accuracy, and
comprehension of the level of questions commonly presented in the GATE test,
which ultimately enables you to score well in the exam. Previous Year Questions
help a candidate practise and revise for GATE, allowing them to pass with a
high mark.
Engineering Mathematics Previous Year GATE Questions aid in the analysis of a subject's question structure and marking scheme, as well as in time management, which enhances overall GATE exam score. Candidates can easily crack GATE with a decent GATE Score if they practise PYQs on a regular basis.
Probability Topic Previous Year Questions (PYQ) Series of Engineering Math For GATE Exam
1. A fair dice is rolled twice. The probability that an odd number will follow an even number is
[EC:
GATE-2005]
(a)
1 / 2 (b) 1/ 6 (c) 1 /3 (d) 1/ 4
Solution:
Here the sample space S = 6
Therefore P(odd
number)= 3/6 =1/2
and P(even
number)= 3/6 =1/2
since events are independent, therefore, P(odd / even) = 1/2 *1/2 =1/4
2. An examination consists of two papers,
Paper 1 and Paper 2. The probability of failing in Paper 1 is 0.3 and that in
Paper 2 is 0.2. Given that a student has failed in Paper 2, the probability of
failing in Paper 1 is 0.6. The probability of a student failing in both the
papers is [EC: GATE-2007]
(a) 0.5 (b) 0.18 (c) 0.12 (d) 0.06
Solution:
Let A be the event that ‘failed
in paper 1’.
B be the event that ‘failed in paper 2’.
Given P(A) = 0.3, P(B) = 0.2.
Also P (A/B)= 0.6
probability of a student failing in both the papers = P (A/B) * P(B) = 0.6* 0.2 = 0.12
3
In a manufacturing plant, the probability of making a defective bolt is
0.1. The mean and standard deviation of defective bolts in a total of 900 bolts
are respectively [ME: GATE-2000]
(a) 90 and 9 (b) 9 and 90
Solution:
It’s a Poisson
distribution.
Here n =900,
p =0.1
mean(m)=
np =900* 0.1 =90
Standard
deviation = sqrt (npq)= 9
So correct answer is 90 and 9 respectively
4
The probability that two friends share the same birth-month is [ME:
GATE-1998]
(a)
1/6
(b)
1/12
(c)
1/44
(d) 1/24
Solution:
Let A =
the event that the birth month of first friend And B= that of second friend.
∴ = P(A)= 1,as 1st friend can born
in any month and P(B) =1/12 ,by the condition.
∴ Probability of two friends share
same birth-month
1
*1/12= 1/12
5 Two dice are thrown. What is the probability
that is the sum of the numbers on the two dice is eight?
[ME: GATE-2002]
(a)
5/36 (b) 5/18 (c) ¼ (d) 1/3 16. (a)
Here sample
space = 6 × 6 = 36 Here, there are five such points whose sum is 8. They are
(2,6), (3,5), (4,4), (5,3), (6,2).
Require probability 5/ 36
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