Question

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(A)}& \text{Let}n\text{be a number chosen randomly from}& & \\ & \text{the set of first}100\text{natural numbers. Then}& & \\ & \text{the probability that the value of}(1+i{)}^n& & \\ & \text{is real, is}& \text{(P)}& 2\\ \text{(B)}& \text{If the coefficient of}{x}^{13}\text{in the expansion of}& & \\ & (1-x{)}^5(1+x+x^2+x^3{)}^4\text{is}k\text{, then the}& & \\ & \text{value of}\frac{k}{4}\text{is}& \text{(Q)}& 0.35\\ \text{(C)}& \text{In an examination of}9\text{papers, a candidate}& & \\ & \text{has to pass in more papers than (s)he fails in}& & \\ & \text{order to be successful. If the number of}& & \\ & \text{ways in which (s)he can be unsuccessful is}{2}^m,& & \\ & \text{then the value of}\frac{m}{4}\text{is}& \text{(R)}& 0.55\\ \text{(D)}& A,B,C\text{are three events such that}& & \\ & P\left(A\right)=0.6,P\left(B\right)=0.4,P\left(C\right)=0.5,& & \\ & P(A\cup B)=0.8,P(A\cap C)=0.3\text{and}& & \\ & P(A\cap B\cap C)=0.2\text{. If}P(A\cup B\cup C)\ge 0.85& & \\ & \text{and}P(B\cap C)\text{lies in the interval}[0.2,b]\text{,}& & \\ & \text{then the value of}b\text{is}& \text{(S)}& 1\\ & & \text{(T)}& 0.5\\ & & \text{(U)}& 0.25\end{array}$

Which of the following is the only CORRECT combination?

• A. $\left(A\right)\to \left(T\right)$
• B. $\left(B\right)\to \left(P\right)$
• C. $\left(C\right)\to \left(S\right)$
• D. $\left(D\right)\to \left(Q\right)$

Answer 1

The correct option is D $\left(D\right)\to \left(Q\right)$

$\left(A\right)$
We know that,
$(1+i{)}^n={\left(\sqrt{2}{e}^{i\pi /4}\right)}^n=2^{n/2}(\cos \frac{n\pi }{4}+i\sin \frac{n\pi }{4})$
For $(1+i{)}^n$ to be real,
$\sin \frac{n\pi }{4}=0\Rightarrow n=4m,m\in Z$
So, $n$ is a multiple of $4$.
In first $100$ natural numbers, there are $25$ numbers that are multiples of $4.$
The required probability is $\frac{25}{100}=0.25$
$\left(A\right)\to \left(U\right)$

$\left(B\right)$
$(1-x{)}^5(1+x+x^2+x^3{)}^4=(1-x{)}^5{\left[\right(1+x\left)\right(1+x^2\left)\right]}^4=(1-x\left)\right(1-x^2{)}^4(1+x^2{)}^4=(1-x\left)\right(1-x^4{)}^4=(1-x)\sum _{k=0}^4{}^4{C}_k(-1{)}^k(x^4{)}^k$
So, the coefficient of $x^{13}$ is,
$k=-{}^4{C}_3(-1{)}^3=4\therefore \frac{k}{4}=1$
$\left(B\right)\to \left(S\right)$

$\left(C\right)$
The candidate is unsuccessful if he fails in $5\text{or}6\text{or}7\text{or}8\text{or}9$ papers.
$\therefore$ Number of ways to be unsuccessful
$={}^9{C}_9+{}^9{C}_8+{}^9{C}_7+{}^9{C}_6+{}^9{C}_5={}^9{C}_0+{}^9{C}_1+{}^9{C}_2+{}^9{C}_3+{}^9{C}_4=\frac{1}{2}\sum _{k=0}^9{}^9{C}_k=\frac{1}{2}\times 2^9=2^8\Rightarrow m=8\therefore \frac{m}{4}=2$
$\left(C\right)\to \left(P\right)$

$\left(D\right)$
$P(A\cup B\cup C)=P\left(A\right)+P\left(B\right)+P\left(C\right)-P(A\cap B)-P(B\cap C)-P(A\cap C)+P(A\cap B\cap C)$
We know that,
$P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)\Rightarrow 0.8=0.6+0.4-P(A\cap B)\Rightarrow P(A\cap B)=0.2$

$\Rightarrow P(A\cup B\cup C)=0.6+0.4+0.5-0.2-P(B\cap C)-0.3+0.2\Rightarrow P(A\cup B\cup C)=1.2-P(B\cap C)$

Now, $0.85\le P(A\cup B\cup C)\le 1$
$\Rightarrow 0.85\le 1.2-P(B\cap C)\le 1\Rightarrow 0.2\le P(B\cap C)\le 0.35\Rightarrow P(B\cap C)\in [0.2,0.35]\therefore b=0.35$
$\left(D\right)\to \left(Q\right)$