NIMCET 2027 Probability Distribution Previous Year Question Paper and Solution with PDF

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NIMCET 2027 Probability Distribution PYQ

Qus : 1 NIMCET 2027 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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NIMCET 2027 Previous Year PYQ NIMCET 2027 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 NIMCET 2027 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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NIMCET 2027 Previous Year PYQ NIMCET 2027 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 NIMCET 2027 PYQ

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NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2012 PYQ

Solution

Qus : 4 NIMCET 2027 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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NIMCET 2027 Previous Year PYQ NIMCET 2027 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 NIMCET 2027 PYQ

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NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2012 PYQ

Solution

Qus : 6 NIMCET 2027 PYQ

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NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2012 PYQ

Solution

Qus : 7 NIMCET 2027 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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NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2013 PYQ

Solution

Qus : 8 NIMCET 2027 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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NIMCET 2027 Previous Year PYQ NIMCET 2027 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 NIMCET 2027 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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NIMCET 2027 Previous Year PYQ NIMCET 2027 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 NIMCET 2027 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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NIMCET 2027 Previous Year PYQ NIMCET 2027 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 NIMCET 2027 PYQ

is the mean of distribution of x, then usual notation

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NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2017 PYQ

Solution

Mean Deviation from mean is Zero.

Qus : 12 NIMCET 2027 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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NIMCET 2027 Previous Year PYQ NIMCET 2027 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 NIMCET 2027 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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NIMCET 2027 Previous Year PYQ NIMCET 2027 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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