Question 1 |
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 1 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 2 |
Let the probability density function of a random variable x be given as
f(x)=ae^{-2|x|}
The value of 'a' is _________
f(x)=ae^{-2|x|}
The value of 'a' is _________
0.5 | |
1 | |
1.5 | |
2 |
Question 2 Explanation:
f(x)=\left\{\begin{matrix}
ae^{2x} &x \lt 0 \\
ae^{-2x} &x \gt 0
\end{matrix}\right.
Therefore,
\begin{aligned} \int_{-\infty }^{\infty }f(x)dx&=1\\ \int_{-\infty }^{0}ae^{2x}dx+\int_{0 }^{\infty }ae^{-2x}dx&=1\\ a\left [ \left [ \frac{e^{2x}}{2} \right ]_{-\infty }^0 +[ \left [ \frac{e^{-2x}}{-2} \right ]^{\infty }_0\right ]&=1\\ a\left [ \frac{1}{2}+\frac{1}{2} \right ]&=1\\ a&=1 \end{aligned}
Therefore,
\begin{aligned} \int_{-\infty }^{\infty }f(x)dx&=1\\ \int_{-\infty }^{0}ae^{2x}dx+\int_{0 }^{\infty }ae^{-2x}dx&=1\\ a\left [ \left [ \frac{e^{2x}}{2} \right ]_{-\infty }^0 +[ \left [ \frac{e^{-2x}}{-2} \right ]^{\infty }_0\right ]&=1\\ a\left [ \frac{1}{2}+\frac{1}{2} \right ]&=1\\ a&=1 \end{aligned}
Question 3 |
Suppose the probability that a coin toss shows "head" is p, where 0\lt
p\lt 1. The coin is tossed repeatedly until the first "head" appears. The expected number of tosses required is
p/\left ( 1-p \right ) | |
\left ( 1-p \right )/p | |
1/p | |
1/p^{2} |
Question 3 Explanation:
F(H) = p, let x = Number of tosses
\begin{aligned} &\begin{array}{|c|c|c|c|c|c|} \hline x & 1 & 2 & 3 & 4 & 5 \ldots \\ \hline P(x) & p & (1-p) p & (1-p)^{2} p & & \\ \hline \end{array}\\ &\begin{aligned} E(x) &=\sum x_{i} P(x)=p+2(1-p) p+3(p)(1-p)^{2}+\ldots \\ &=p\left[1+2(1-p)+3(1-p)^{2}+\ldots\right] \\ &=p[1-(1-p)]^{-2}=\frac{1}{p} \end{aligned} \end{aligned}
\begin{aligned} &\begin{array}{|c|c|c|c|c|c|} \hline x & 1 & 2 & 3 & 4 & 5 \ldots \\ \hline P(x) & p & (1-p) p & (1-p)^{2} p & & \\ \hline \end{array}\\ &\begin{aligned} E(x) &=\sum x_{i} P(x)=p+2(1-p) p+3(p)(1-p)^{2}+\ldots \\ &=p\left[1+2(1-p)+3(1-p)^{2}+\ldots\right] \\ &=p[1-(1-p)]^{-2}=\frac{1}{p} \end{aligned} \end{aligned}
Question 4 |
The probability of a resistor being defective is 0.02. There are 50 such resistors in a circuit. The probability of two or more defective resistors in the circuit (round off to two decimal places)is ________
0.1 | |
0.26 | |
0.65 | |
0.85 |
Question 4 Explanation:
\begin{aligned} p&=0.02\\ n&=50\\ \lambda &=np=50(0.02)=1\\ p(x\geq 2)&=1-p(x \lt 2)\\ &=1-(p(x=0)+p(x=1))\\ &=1-\left ( \frac{e^{-\lambda }\lambda ^0}{0!}+\frac{e^{-\lambda }\lambda ^1}{1!} \right )\\ &=1-e^{-\lambda }(1+\lambda )\\ &=1-e^{-1}(1+1)=0.26 \end{aligned}
Question 5 |
The mean-square of a zero-mean random process is \frac{kT}{C}, where k is Boltzmann's constant, T is the absolute temperature, and C is a capacitance. The standard deviation of the random process is
\frac{kT}{C} | |
\sqrt{\frac{kT}{C}} | |
\frac{C}{kT} | |
\frac{\sqrt{kT}}{C} |
Question 5 Explanation:
Given that,
\begin{aligned} E(x^2)&=\frac{kT}{C}\\ E(x)&=0\\ &=E(x^2)-(E(x))^2\\ Var(x)&=\frac{kT}{C}-0=\frac{kT}{C} \end{aligned}
Standard deviation =\sqrt{\frac{kT}{C}}
\begin{aligned} E(x^2)&=\frac{kT}{C}\\ E(x)&=0\\ &=E(x^2)-(E(x))^2\\ Var(x)&=\frac{kT}{C}-0=\frac{kT}{C} \end{aligned}
Standard deviation =\sqrt{\frac{kT}{C}}
There are 5 questions to complete.