Question

Given two independent events A and B such that P (A) = 0.3 and P (B) = 0.6. Find
(i) P (A ∩ B)
(ii) P (A ∩ $\overline{B}$)
(iii) P ($\overline{A}$ ∩ B)
(iv) $P\left(\overline{A}\cap \overline{B}\right)$
(v) P (A ∪ B)
(vi) P (A/B)
(vii) P (B/A)

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ \text{A and B are independent events.} \\ P\left(A\right)=0.3 \\ P\left(B\right)=0.6 \\ \left(\mathrm{i}\right)P\left(A\cap B\right)=P\left(A\right)P\left(B\right) \\ =0.3\times 0.6 \\ =0.18 \\ \left(\mathrm{ii}\right)P\left(A\cap \overline{B}\right)=P\left(A\right)P\left(\overline{B}\right) \\ =P\left(A\right)\left[1-P\left(B\right)\right] \\ =0.3\times \left[1-0.6\right] \\ =0.3\times 0.4 \\ =0.12 \\ \left(\mathrm{iii}\right)P\left(\overline{A}\cap B\right)=P\left(\overline{A}\right)P\left(B\right) \\ =P\left(B\right)\left[1-P\left(A\right)\right] \\ =0.6\times 0.7 \\ =0.42 \\ \left(\mathrm{iv}\right)P\left(\overline{A}\cap \overline{B}\right)=P\left(\overline{A}\right)P\left(\overline{B}\right) \\ =\left[1-P\left(A\right)\right]\left[1-P\left(B\right)\right] \\ =0.7\times 0.4 \\ =0.28 \\ \left(\mathrm{v}\right)P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right) \\ =0.3+0.6-0.18 \\ =0.9-0.18 \\ =0.72 \\ \left(\mathrm{vi}\right)P\left(\frac{A}{B}\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)} \\ =\frac{0.18}{0.6} \\ =0.3 \\ \left(\mathrm{vii}\right)P\left(\frac{B}{A}\right)=\frac{P\left(A\cap B\right)}{P\left(A\right)} \\ =\frac{0.18}{0.3} \\ =0.6 \end{gathered}$$