\(P(\overline { A } \cup \overline { B } )\)=\(\left( \overline { A\cap B } \right) \)(By de Morgan's law)
1-P(A∩B)=1-0.24
=0.76​

P\(\left( \overline { A } \cap \overline { B } \right) \)=\(P(\overline { A\cup B } )\) (By de morgan's law)
=1-P(A∪B)
=1-0.71=0.29

\(P(A / \bar{B}) =\frac{P(A \cap \bar{B})}{P(\bar{B})}\)
\(=\frac{P(A)-P(A \cap B)}{1-P(B)} \)
\(=\frac{0.6-0.2}{1-0.5}=\frac{0.4}{0.5}=\frac{4}{5}\)

Let S be the sample space and A be the event of getting at least two heads.
Therefore, the event Ā denotes, getting at most one head.
n(S) = 2⁹ = 512, n(Ā) = 9C₀ + 9C₁= 1 + 9 = 10
\(P(\bar{A})=\frac{10}{512}=\frac{5}{256} \)
\( P(A)=1-P(\bar{A})=1-\frac{5}{256}=\frac{251}{256}\)

The event of getting 7 is an impossible event. Therefore, P(getting 7) = 0

\(P(\bar{A} / B) =\frac{P(\bar{A} \cap B)}{P(B)} \)
\(=\frac{P(B)-P(A \cap B)}{P(B)} \)
\(=\frac{0.5-0.2}{0.5}=\frac{0.3}{0.5}=\frac{3}{5}\)

(iii) P(queen card or 9)=\(\frac{4}{52}+\frac{4}{52}=\frac{8}{52}=\frac{2}{13}\)

S = {Pack of  52 cards}
\(\therefore\) n(S) = 52
P(ace card or a king card) = P(ace card) + P(king card)
= \(\frac{4}{52}+\frac{4}{52}=\frac{8}{52}=\frac{2}{13}\)

The sample space is S = {HH, HT, TH, TT}
⇒ n(S) = 4
(i) Let A be the event of getting one head and one tail, then
A = {HT, TH}
n(A) = 2
\(\therefore\)  \(P(A)=\frac{n(B)}{n(S)}=\frac{4}{4}=1\)
(ii) Let A be the event of getting atmose two tails, then
\(\therefore\)  B = {HH,HT,TH,TT}
\(\therefore\)  \(P(B)=\frac{n(B)}{n(S)}=\frac{4}{4}=1\)

P\( \left( A\cap \overline { B } \right) \) = P(A) - P(A∩B)
= 0.52 - 0.24 = 0.28
P\( \left( A\cap \overline { B } \right) \) = 0.28