(a) Consider the experiment of tossing a coin. If the coin shows head, toss it again; but if it shows a tail, then throw a die. Find the conditional…

Let \(\displaystyle A\) :The die shows a number greater than $\displaystyle 3$ and \(\displaystyle B\) :There is at least one tail \[\begin{aligned} & \mathrm{P}(B \cap A)=\mathrm{P}(T 4, T 5, T 6)=\frac{3}{12}=\frac{1}{4} \\ & \mathrm{P}(B)=\mathrm{P}(H T, T 1, T 2, T 3, T 4, T 5, T 6)=\frac{1}{4}+\frac{6}{12}=\frac{3}{4} \\ & \mathrm{P}(A \mid B)=\frac{\mathrm{P}(B \cap A)}{\mathrm{P}(B)}=\frac{1 / 4}{3 / 4}=\frac{1}{3} \end{aligned} \] \[\begin{aligned} & \text { (i) } \mathrm{E}(\mathrm{X})=\sum \mathrm{X} . \mathrm{P}(\mathrm{X}) \\ & \qquad=1\left(\frac{11}{30}\right)+2\left(\frac{1}{15}\right)+3\left(\frac{1}{10}\right)+2 \lambda\left(\frac{3}{10}\right)+3 \lambda\left(\frac{1}{15}\right)+4 \lambda\left(\frac{1}{10}\right) \\ & \text { Given } \sum \mathrm{X} . \mathrm{P}(\mathrm{X})=3.2 \\ & \therefore 24+36 \lambda=96 \\ & \Rightarrow \lambda=2 \\ & \text { (ii) } \mathrm{P}(\mathrm{X}>1)=1-\mathrm{P}(\mathrm{X}=1) \\ & \quad=1-\frac{11}{30}=\frac{19}{30} \end{aligned} \]