Question

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

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(A)}& \text{Poonam flipped a fair coin five times. In the first three flips, the coins came up heads exactly twice and in the last three}& \text{(P)}& \frac{3}{4}\\ & \text{flips, the coin also came up heads exactly twice. The probability that the third flip was head, is}& & \\ \text{(B)}& \text{A box contains 10 transistors of which 2 are defective. Transistors are drawn one by one without replacement unless a}& \text{(Q)}& \frac{2}{5}\\ & \text{non-defective one is chosen. The probability that atmost 3 transistors are drawn, is}& & \\ & \text{A box contains 1 black and 1 white ball. A ball is drawn randomly and replaced in the box with an additional ball of the}& & \\ \text{(C)}& \text{same colour, then a second ball is drawn randomly from the box containing 3 balls. The probability that the first drawn}& \text{(R)}& \frac{4}{5}\\ & \text{ball was white given that at least one of the two balls drawn was white, is}& & \\ \text{(D)}& \text{Let}P\left(A\right)=0.7\text{and}P\left(B\right)=\mathrm{0.3.}\text{Let}{B}^c\text{denote the complement the event}B.\text{Then the smallest value of}P(A\cap B^c)\text{is}& \text{(S)}& \frac{2}{3}\\ & & \text{(T)}& \text{None of these}\end{array}$

Which of the following is a CORRECT combination?

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

Answer 1

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

$\left(C\right)$


$A:1$st drawn is white
$B:$ at least one of the two balls drawn is white
$P\left(A/B\right)=\frac{P(A\cap B)}{P\left(B\right)}=\frac{\frac{1}{2}\cdot \frac{2}{3}+\frac{1}{2}\cdot \frac{1}{3}}{\frac{1}{2}\cdot \frac{1}{3}+\frac{1}{2}\cdot \frac{2}{3}+\frac{1}{2}\cdot \frac{1}{3}}=\frac{\frac{2}{6}+\frac{1}{6}}{\frac{1}{6}+\frac{2}{6}+\frac{1}{6}}=\frac{3}{4}$

$\left(D\right)$

$P(A\cap B^c)=P\left(A\right)-P(A\cap B)$
$P(A\cap B^c)$ is minimum if $P(A\cap B)$ is maximum
And $P(A\cap B{)}_{max}=0.3$
$\therefore (A\cap B^c{)}_{min}=0.7-0.3=0.4=\frac{4}{10}=\frac{2}{5}$