Question

Match the following by appropiately matching the lists based on the information given in Column I and Column II.
​​​​​​$\begin{array}{cc}Column1& Column2\\ \text{(a) The probability of a bomb hitting a bridge is}& \\ \frac{1}{2}.\text{Two direct hits are needed to destroy it.}& \\ \text{The number of bombs recquired so that the}& \\ \text{probability of the bridge being destroyed is}& \\ \text{greater than}0.9\text{can be}& \text{(p)}4\\ \text{(b) A bag contains}2\text{red,}3\text{white,}5\text{black balls, a}& \\ \text{ball is drawn its color is noted and replaced}& \\ \text{The number of times, a ball can be drawn}& \\ \text{so that the probability of getting a red ball}& \\ \text{for the first time is atleast}1/2& \text{(q)}6\\ \text{(c) A drawer contains a mixture of red socks}& \\ \text{and blue socks, at most}17\text{in all. It so}& \\ \text{happens that when two socks are selected}& \\ \text{randomly without replacement, there is a}& \\ \text{probability of exactly}1/2\text{that both are red}& \\ \text{or both are blue. Then number of red}& \\ \text{socks in drawer can be}& \text{(r)}7\\ \text{(d) There are two red, two blue, two white and}& \\ \text{certain number (greater than}0\text{) of green}& \\ \text{socks in a drawer. If two socks are taken at}& \\ \text{random from the drawer without}& \\ \text{replacement, the probability that they}& \\ \text{are of same color is}\frac{1}{5}\text{, then the number of}& \\ \text{green socks are}& \text{(s)}10\end{array}$

• A. $a\to r,s;b\to p,q,r,s;c\to p,q,r,s;d\to p$
• B. $a\to r,s;b\to p,q,s;c\to p,q,r,s;d\to p$
• C. $a\to r,s;b\to p,s;c\to p,s;d\to p$
• D. $a\to r,s;b\to p,q,r,s;c\to p,q,r,s;d\to q,r,s$

Answer 1

The correct option is A $a\to r,s;b\to p,q,r,s;c\to p,q,r,s;d\to p$

a.
$P\left(\text{success}\right)=\frac{1}{2};P\left(\text{failure}\right)=\frac{1}{2}$
Suppose, '$n$' bombs are to be dropped. Let $E$ be the event that bridge is destroyed. Then,
$P\left(E\right)=1-P\left(0\text{or}1\text{success}\right)=1-((\frac{1}{2}{)}^n{+}^n{C}_1\times \frac{1}{2}(\frac{1}{2}{)}^{n-1})=1-(\frac{1}{2^n}+\frac{n}{2^n})\ge 0.9\Rightarrow \frac{1}{10}\ge \frac{n+1}{2^n}\text{or}\frac{2^n}{10(n+1)}\ge 1$
b.
The bag contains $2$ red, $3$ white and $5$ black balls. Hence, $P\left(S\right)=\frac{1}{5};P\left(F\right)=\frac{4}{5};$ Let $E$ be the event of getting a red ball.
$P\left(E\right)=P(S\text{or}FS\text{or}FFS\text{or}\cdots )\ge \frac{1}{2}$
$\therefore P\left(F^n\right)\le \frac{1}{2};(\frac{4}{5}{)}^n\le \frac{1}{2}$
The value of $n$ is $4$.
c.
Let there be $x$ red socks and $y$ blue socks and $x>y$. Then
$\frac{{}^x{C}_2+{}^y{C}_2}{{}^{x+y}{C}_2}=\frac{1}{2}$ or
$\frac{x(x-1)+y(y-1)}{(x+y)(x+y-1)}=\frac{1}{2}$
Multiplying both sides by $2(x+y)(x+y-1)$ and expanding, we get
$2x^2-2x+2y^2-2y=x^2+2xy+y^2-x-y$
Rearranging, we have
$x^2-2xy+y^2=x+y$ or
$(x-y{)}^2=x+y$ or
$|x-y|=x+y$
Now,
$x+y\le 17$
$x-y=\sqrt{17}$
As $x-y$ must be an integer, so
$\therefore x-y=4$
$x+y=16$
Adding both together and dividing by $2$ yields $x\le 10.$
d.
Let the number of green socks be $x>0$.
Let $E:$ be the event that two socks drawn are of the same colour.
$P\left(E\right)=P\left(RR\text{or}BB\text{or}WW\text{or}GG\right)=\frac{3}{{}^{6+x}{C}_2}+\frac{{}^x{C}_2}{{}^{6+x}{C}_2}=\frac{6}{(x+5)(x+6)}+\frac{x(x-1)}{(x+5)(x+6)}=\frac{1}{5}\Rightarrow 5(x^2-x+6)=x^2+11x+30$
or $4x^2-16x=0$
or $x=4$