Question

Match $\text{List I}$ with the $\text{List II}$ and select the correct answer using the code given below the lists :

$\begin{array}{cccc}& \text{List - I}& & \text{List - II}\\ & \text{There are two boxes. In the first box, there are}2\text{white and}1\text{black balls, where in the second box, there is}1\text{white and}& & \\ \text{(A)}& 5\text{black balls. A ball is drawn from the first box and kept in the second and a ball drawn from the second box was found}& \text{(P)}& \frac{1}{8}\\ & \text{to be white. The chance that a black ball was transferred from first to the second box is}& & \\ \text{(B)}& 3\text{identical unfair coins are flipped. The probability that there are two heads and one tail is}\frac{2}{9}.\text{The probability that}& \text{(Q)}& \frac{1}{4}\\ & \text{there are two tails and one head is}\frac{4}{9}.\text{The probability that the coin will flip head, is}& & \\ \text{(C)}& \text{There are}n\text{distinct points on a circle. If two of these points are joined to form a line and then another two points}& \text{(R)}& \frac{1}{3}\\ & \text{(different from the first two) are joined to form another line, the probability that these two lines intersect inside the circle, is}& & \\ \text{(D)}& \text{If three fair dice are thrown, and the sum is an odd number, the probability that all three dice show an odd number is}& \text{(S)}& \frac{1}{5}\\ & & \text{(T)}& \frac{4}{7}\end{array}$

Which of the following is a CORRECT combination?

• A. $\text{(C)}\to \text{(R)},\text{(D)}\to \text{(Q)}$
• B. $\text{(C)}\to \text{(R)},\text{(D)}\to \text{(R)}$
• C. $\text{(C)}\to \text{(S)},\text{(D)}\to \text{(Q)}$
• D. $\text{(C)}\to \text{(P)},\text{(D)}\to \text{(S)}$

Answer 1

The correct option is A $\text{(C)}\to \text{(R)},\text{(D)}\to \text{(Q)}$

$\left(C\right)$
$n\left(A\right):$ Number of favourable points inside the circle
$n\left(S\right):$ Total number of points
$n\left(S\right)=\frac{{}^n{C}_2\cdot {}^{n-2}{C}_2}{2}=\frac{1}{2}\frac{n(n-1)}{2}\cdot \frac{(n-2)(n-3)}{2}$
$n\left(A\right)={}^n{C}_4$
$\therefore$ Required probability,
$p=\frac{n\left(A\right)}{n\left(S\right)}=\frac{n(n-1)(n-2)(n-3)}{24}\cdot \frac{8}{n(n-1)(n-2)(n-3)}=\frac{1}{3}$

$\left(D\right)$
The total number of ways of throwing an odd number on all $3$ dice is $3^3=27$
Total number of ways such that sum is odd, $n\left(S\right)=\frac{\left(6\right)\left(6\right)\left(6\right)}{2}=108$
Therefore the probability that all dice are odd is $\frac{27}{108}=\frac{1}{4}$