NIMCET Probability Distribution Previous Year Questions (PYQs) – Page 1 of 2
A Place for Latest Exam wise Questions, Videos, Previous Year Papers,
Study Stuff for MCA Examinations
Study Stuff for MCA Examinations
Showing 13 questions
Probability a blade is defective $=0.002$, packet of $10$ blades.
Find packets with no defective blades in $10000$ packets.
1
2
3
4
Solution
$P(\text{good})=0.998$
Probability all 10 good $=(0.998)^{10}\approx 0.9802$
So packets $=10000\times0.9802\approx9802$
An anti-aircraft gun fires a maximum of four shots.
Probabilities of hitting in the 1st, 2nd, 3rd, and 4th shot are
0.4, 0.3, 0.2 and 0.1 respectively.
Find the probability that the gun hits the plane.
1
2
3
4
Solution
Hit at least once = 1 − (miss all shots)
Miss probabilities:
1st: $(1 - 0.4) = 0.6$
2nd: $(1 - 0.3) = 0.7$
3rd: $(1 - 0.2) = 0.8$
4th: $(1 - 0.1) = 0.9$
Miss all:
$0.6 \cdot 0.7 \cdot 0.8 \cdot 0.9 = 0.3024$
Hit at least once:
$1 - 0.3024 = 0.6976$
A random variable $X$ has the probability distribution:
\[\begin{array}{c|ccccccccc}
x & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline
P(X=x) & a & 3a & 5a & 7a & 9a & 11a & 13a & 15a & 17a
\end{array}
\]The value of $a$ is:
1
2
3
4
Solution
$ \text{Total probability} = 1 $
$ a(1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17) = 1 $
$ 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 = 81 $
$ 81a = 1 $
$ a = \frac{1}{81} $
A random variable X has the distribution law as given below:
The variance of the distribution is
| X | 1 | 2 | 3 |
| P(X=x) | 0.3 | 0.4 | 0.3 |
1
2
3
4
Solution
India plays 4 matches. Probabilities of scoring $0,1,2$ points: $0.45,\ 0.05,\ 0.50$. Find probability of at least $7$ points.
1
2
3
4
Solution
$P(8)=0.5^4=0.0625$
$P(7)=4 \cdot 0.05 \cdot 0.5^3 = 0.025$
Total $= 0.0625 + 0.025 = 0.0875$
n a Poisson distribution if $P[X=3] = \frac{1}{4} P[X=4]$ then $P[X=5] = k P[X=7]$ where $k$ equals to
1
2
3
4
Solution
Poisson PMF:
$P[X=x] = e^{-\lambda} \frac{\lambda^x}{x!}$
Given:
$P(3) = \frac{1}{4} P(4)$
$\frac{\lambda^3/3!}{\lambda^4/4!} = \frac{1}{4}$
$\frac{4}{\lambda} = \frac{1}{4}$
$\lambda = 16$
Now find $k$:
$\frac{P(5)}{P(7)} = \frac{\lambda^5/5!}{\lambda^7/7!} = \frac{7 \cdot 6}{\lambda^2}$
Substitute $\lambda = 16$:
$\frac{42}{256} = \frac{21}{128}$
So,
$k = \frac{21}{128}$
The mean and variance of a random variable X having binomial distribution are 4 and 2 respectively. The P(X = 1) is
1
2
3
4
Solution
np = 4
npq = 2
q = 1/2, p = 1/2, n = 8
p(X = 1) = 8C1 (1/2)(1/2)7
NIMCET
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
NIMCET
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
Ask Your Question or Put Your Review.
