$\begin{array}{1 1} (i) 0.18, (ii) 0.12, (iii) 0.72, (iv) 0.28 \\ (i) 0.09, (ii) 0.18, (iii) 0.36, (iv) 0.28 \\(i) 0.12, (ii) 0.18, (iii) 0.28, (iv) 0.72 \\ (i) 0.18, (ii) 0.16, (iii) 0.28, (iv) 0.68 \end{array} $

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- If A and B are independent events, \(P(A\cap\;B)=P(A)\;P(B)\)
- P (not A and not B) = \(P(\bar{A}\cap\;\bar{B})\) = \(P(\overline{A \cup B})\) = 1 - P (A $\cup$ B)
- P (A $\cup$ B) = P(A) + P(B) - P(A $\cap$ B)
- P (\(\bar{A})\) = 1 - P(A)
- If A and B are independent events, $\bar{A}$ and $\bar{B}$ are also independent.

Given P(A) = 0.3, P (B) = 0.6.

(i): Finding P (A and B):

If A and B are independent events, \(P(A\cap\;B)=P(A)\;P(B)\)

$\Rightarrow$ \(P(A\cap\;B)=P(A)\;P(B)\) = 0.3 $\times$ 0.6 = 0.18

(ii): Finding P (A and not B):

If A and B are independent events, $\bar{A}$ and $\bar{B}$ are also independent.

$\Rightarrow P (A \cap \bar{B}) = P (A) \times P (\bar{B}) = P (A) \times (1 - P (B)) = 0.3 \times (1 - 0.6) = 0.3 \times 0.4 = 0.12$

(iii): Finding P (A or B):

P (A $\cup$ B) = P(A) + P(B) - P(A $\cap$ B)

$\Rightarrow P (A \cup B) = 0.3 + 0.6 - 0.18 = 0.72.$

(iv): Finding P (neither A nor B):

P (neither A nor B) = P ($\bar{A} \cap \bar{B}$) = 1 - P (A $\cup$ B) = 1 - 0.72 = 0.28.

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