MCA NIMCET Probability Distribution Previous Year Question Paper and Solution with PDF

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MCA NIMCET Probability Distribution PYQ

Qus : 1 MCA NIMCET PYQ

Probability a blade is defective $=0.002$, packet of $10$ blades. Find packets with no defective blades in $10000$ packets.

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2011 PYQ

Solution

$P(\text{good})=0.998$ Probability all 10 good $=(0.998)^{10}\approx 0.9802$ So packets $=10000\times0.9802\approx9802$

Qus : 2 MCA NIMCET PYQ

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.

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2009 PYQ

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: $1 - 0.3024 = 0.6976$

Qus : 3 MCA NIMCET PYQ

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

Qus : 4 MCA NIMCET PYQ

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:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2011 PYQ

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} $

Qus : 5 MCA NIMCET PYQ

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

Qus : 6 MCA NIMCET PYQ

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

Qus : 7 MCA NIMCET PYQ

A random variable X has the distribution law as given below:

X	1	2	3
P(X=x)	0.3	0.4	0.3

The variance of the distribution is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2013 PYQ

Solution

Qus : 8 MCA NIMCET PYQ

India plays 4 matches. Probabilities of scoring $0,1,2$ points: $0.45,\ 0.05,\ 0.50$. Find probability of at least $7$ points.

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2010 PYQ

Solution

$P(8)=0.5^4=0.0625$

$P(7)=4 \cdot 0.05 \cdot 0.5^3 = 0.025$

Total $= 0.0625 + 0.025 = 0.0875$

Qus : 9 MCA NIMCET PYQ

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

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2010 PYQ

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, $k = \frac{21}{128}$

Qus : 10 MCA NIMCET PYQ

The mean and variance of a random variable X having binomial distribution are 4 and 2 respectively. The P(X = 1) is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2017 PYQ

Solution

np = 4

npq = 2

q = 1/2, p = 1/2, n = 8

p(X = 1) = 8C1 (1/2)(1/2)7

Qus : 11 MCA NIMCET PYQ

is the mean of distribution of x, then usual notation

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2017 PYQ

Solution

Mean Deviation from mean is Zero.

Qus : 12 MCA NIMCET PYQ

Let $X$ and $Y$ be two independent identically distributed Bernoulli random variables with common probability mass function: $P(X=1)=\frac{1}{2}$ and $P(X=0)=\frac{1}{2}$. If $Z=XY$, then the distribution of $Z$ is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2026 PYQ

Solution

Given,

$Z=XY$

Now, $Z=1$ only when both $X=1$ and $Y=1$.

Since $X$ and $Y$ are independent,

$P(Z=1)=P(X=1,Y=1)$

$P(Z=1)=P(X=1)P(Y=1)$

$P(Z=1)=\frac{1}{2}\times \frac{1}{2}$

$P(Z=1)=\frac{1}{4}$

Now,

$P(Z=0)=1-P(Z=1)$

$P(Z=0)=1-\frac{1}{4}$

$P(Z=0)=\frac{3}{4}$

Therefore, $Z$ follows Bernoulli distribution with

$P(Z=1)=\frac{1}{4}$ and $P(Z=0)=\frac{3}{4}$

Correct answer is option $4$.

Qus : 13 MCA NIMCET PYQ

For a given sample, the computed values of the variance and fourth central moment are $3$ and $63$ respectively. Then the underlying frequency distribution is classified as:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2026 PYQ

Solution

Given,

Variance $=3$

So,

$\mu_2=3$

Fourth central moment is:

$\mu_4=63$

Coefficient of kurtosis is:

$\beta_2=\frac{\mu_4}{\mu_2^2}$

Substitute the values:

$\beta_2=\frac{63}{3^2}$

$\beta_2=\frac{63}{9}$

$\beta_2=7$

For a normal or mesokurtic distribution,

$\beta_2=3$

Here,

$\beta_2=7>3$

So, the distribution is leptokurtic.Given,

$Z=XY$

Now, $Z=1$ only when both $X=1$ and $Y=1$.

Since $X$ and $Y$ are independent,

$P(Z=1)=P(X=1,Y=1)$

$P(Z=1)=P(X=1)P(Y=1)$

$P(Z=1)=\frac{1}{2}\times \frac{1}{2}$

$P(Z=1)=\frac{1}{4}$

Now,

$P(Z=0)=1-P(Z=1)$

$P(Z=0)=1-\frac{1}{4}$

$P(Z=0)=\frac{3}{4}$

Therefore, $Z$ follows Bernoulli distribution with

$P(Z=1)=\frac{1}{4}$ and $P(Z=0)=\frac{3}{4}$

Correct answer is option $4$.

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