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.
If 100 vehicles are chosen at random on an evennumbered date, the number of vehicles expected without pollution clearance certificate is.
15 | |
30 | |
50 | |
70 |
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.
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?
Function 1 | |
Function 2 | |
Function 3 | |
Function 4 |
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).
0.125 | |
0.188 | |
0.254 | |
0.564 |
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}
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 \delta(y-1)+b \delta(y+1)+g(y) | |
a \delta(y+1)+b \delta(y)+c \delta(y-1)+g(y) | |
a \delta(y+2)+b \delta(y)+c \delta(y-2)+g(y) | |
a \delta(y+1)+b \delta(y-2)+g(y) |
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)
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)?
0.4 | |
0.2 | |
0.9 | |
0.7 |
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}
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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