# Probability and Statistics

 Question 1
A remote village has exactly 1000 vehicles with sequential registration numbers starting from 1000 . Out of the total vehicles, $30 \%$ are without pollution clearance certificate. Further, even- and oddnumbered vehicles are operated on even- and oddnumbered dates, respectively.
If 100 vehicles are chosen at random on an evennumbered date, the number of vehicles expected without pollution clearance certificate is.
 A 15 B 30 C 50 D 70
GATE CE 2023 SET-2   Engineering Mathematics
Question 1 Explanation:
Since $30 \%$ of the total vehicles are without pollution clearance certificate.
Out of the 100 chosen vehicle, $30 \%$ i.e. $100 \times 0.3=30$ vehicle are expected to be without pollution clearance certificate.
 Question 2
Which of the following probability distribution functions (PDFs) has the mean greater than the median? A Function 1 B Function 2 C Function 3 D Function 4
GATE CE 2023 SET-2   Engineering Mathematics
Question 2 Explanation: Option (B) is correct.

 Question 3
The expected number of trials for first occurrence of a "head" in a biased coin is known to be 4. The probability of first occurrence of a "head" in the second trial is ___ (Round off to 3 decimal places).
 A 0.125 B 0.188 C 0.254 D 0.564
GATE EE 2023   Engineering Mathematics
Question 3 Explanation:
Let probability of head $=\mathrm{P}$
Let probability of Tail $=q=P-1$
$\begin{array}{|c|c|c|c|c|} \hline Trial & 1 & 2 & 3 & ... \\ \hline Probability & [\mathrm{P} & \mathrm{qP} & [latex]\mathrm{q}^{2} \mathrm{P} & .... \\ \hline \end{array}$
$\therefore$ Expected No. of trail
\begin{aligned} & =\sum_{n=1}^{\infty} p+2 P q+3 q^{2} P+\ldots \\ & =P\left(1+2 q+3 q^{2}+\ldots\right) \\ & =P(1-q)^{-2} \\ & =\frac{P}{P^{2}}=\frac{1}{P} \end{aligned}

Given : Trial $=4$
$\frac{1}{P}=4 \Rightarrow P=\frac{1}{4}$
$\therefore \quad \mathrm{q}=\frac{3}{4}$

Now, probability of head for second trail. \begin{aligned} & =\mathrm{qP} \\ & =\frac{3}{4} \times \frac{1}{4}=\frac{3}{16}=0.1875 \end{aligned}
 Question 4
A random variable $X$, distributed normally as $N(0,1)$, undergoes the transformation $Y=h(X)$, given in the figure. The form of the probability density function of $Y$ is (In the options given below, $a, b, c$ are non-zero constants and $g(y)$ is piece-wise continuous function) A $a \delta(y-1)+b \delta(y+1)+g(y)$ B $a \delta(y+1)+b \delta(y)+c \delta(y-1)+g(y)$ C $a \delta(y+2)+b \delta(y)+c \delta(y-2)+g(y)$ D $a \delta(y+1)+b \delta(y-2)+g(y)$
GATE EC 2023   Engineering Mathematics
Question 4 Explanation:
$X=N(0,1)$ $f_X(x)=\frac{1}{\sqrt{2\pi}}e^{-x^2/2}$ \begin{aligned} Y= & -1 ; x \leq-2 \\ & 0 ;-1 \leq x \leq 1 \\ & 1 ; x \geq 2 \\ & x+1 ;-2 \leq x \leq-1 \\ & x-1 ; 1 \leq x \leq 2 \end{aligned}
$Y$ is taking discrete set of values and a continuous range of values, so it is mixed random variable.
From the given options, density function of '$Y$' will be.
$f_{Y}(y)=a \delta(y+1)+b \delta(y)+c \delta(y-1)+g(y)$
 Question 5
The probabilities of occurrences of two independent events $A$ and $B$ are $0.5$ and $0.8$, respectively. What is the probability of occurrence of at least $A$ or $B$ (rounded off to one decimal place)?
 A 0.4 B 0.2 C 0.9 D 0.7
GATE CE 2023 SET-1   Engineering Mathematics
Question 5 Explanation:
\begin{aligned} & P(A)=0.5 \\ & P(B)=0.8 \end{aligned}
Probability of occurence of atleast $A$ or $B=$ $\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
\begin{aligned} P(A \cap B) & =P(A)+P(B)-P(A \cap B) \\ & =0.5+0.8-P(A) \times P(B) \\ & =0.5+0.8-0.5 \times 0.8 \\ & =1.3-0.4 \\ & =0.9 \end{aligned}

There are 5 questions to complete.

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