Question

A bag contains some white and black balls, all combinations being equally likely. The total number of balls in the bag is $12$. Four balls are drawn at random from the bag without replacement.
$\begin{array}{cccc}& \text{List - I}& \text{List -II}& \\ \text{(I)}& \text{probability that all the four balls are black}& \text{(P)}& \frac{14}{33}\\ & & & \\ \text{(II)}& \text{If the bag contains 10 black and 2 white balls,}& \text{(Q)}& \frac{1}{5}\\ & \text{then the probability that all four balls are black}& & \\ \text{(III)}& \text{If all the 4 balls are black then the probability,}& \text{(R)}& \frac{70}{429}\\ & \text{that the bag contains 10 black balls}& & \\ \text{(IV)}& \text{Probability that two balls are black and two are}& \text{(S)}& \frac{13}{165}\\ & \text{white is}& & \\ & & \text{(T)}& \frac{1}{3}\\ & & \text{(U)}& \frac{2}{9}\end{array}$

Which of the following is the only INCORRECT combination?

• A. $\left(I\right)\to \left(Q\right)$
• B. $\left(II\right)\to \left(P\right)$
• C. $\left(IV\right)\to \left(S\right)$
• D. $\left(III\right)\to \left(R\right)$

Answer 1

The correct option is C $\left(IV\right)\to \left(S\right)$

Possible combinations:
$0$ black and $12$ white
$1$ black and $11$ white
$::::::$
$11$ black and $1$ white
$12$ black and $0$ white
So, the total number of the combination is $13$.

$\Rightarrow P\left(\text{any combination}\right)=P\left(E_i\right)=\frac{1}{13}$

$P\left(I\right)=\frac{1}{13}\left(\frac{{}^4{C}_4+{}^5{C}_4+{}^6{C}_4+{}^7{C}_4+{}^8{C}_4+{}^9{C}_4+{}^{10}{C}_4+{}^{11}{C}_4+{}^{12}{C}_4}{{}^{12}{C}_4}\right)$
$=\frac{1\times {}^{13}{C}_5}{13\times {}^{12}{C}_4}=\frac{1}{5}$

$P\left(II\right)=\frac{{}^{10}{C}_4}{{}^{12}{C}_4}=\frac{14}{33}$

$P\left(III\right)=\frac{P\left(E_i\right)\times P\left(II\right)}{P\left(I\right)}=\frac{70}{429}$

$P\left(IV\right)=\frac{2({}^2{C}_2\times {}^{10}{C}_2+{}^3{C}_2\times {}^9{C}_2+{}^4{C}_2\times {}^8{C}_2+{}^5{C}_2\times {}^7{C}_2)+{}^6{C}_2\times {}^6{C}_2}{13\times {}^{12}{C}_4}=\frac{1}{5}$