P\((A\cup B)\) = P(A) + P(B) - P(\(A\cap B\))
P\((A\cup B)\) = P(A) + P(B) - P(A)P(B) (since A and B are independent)
That is, 0.9 = 0.4 + P(B) − (0.4) P(B)
0.9 − 0.4 = (1− 0.4) P(B)
Therefore, P(B) = \(\frac{5}{6}\).

\(P(A)=\frac{3}{4}, P(B) =\frac{2}{5}, A \cup B=S \)
\(\Rightarrow P(A \cup B) =1\)
\(P(A \cap B) =P(A)+P(B)-P(A \cup B)\)
\(=\frac{3}{4}+\frac{2}{5}-1\)
\(=\frac{15+8}{20}-1=\frac{23-20}{20}=\frac{3}{20} \)
\(P(A / B) =\frac{P(A \cap B)}{P(B)} \)
\(=\frac{3 / 20}{2 / 5}=\frac{3}{8}\)

Given P(A) = 0.5, P(B) = 0.8
⇒ P(B/A) = 0.8
We kmow P(B/A) = \(\frac{P(A\cap B)}{P(A)}\)
⇒ 0.8 = \(\frac{P(A\cap B)}{0.5}\)
⇒ \(P(A\cap B)=(0.8)(0.5)=0.4\)
(i) Now P(A./B) = \(\frac{P(A\cap B)}{P(B)}=\frac{0.4}{0.8} =\frac{1}{2}=0.5\)
(ii) \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\)
= 0.5 + 0.8 - 0.4 = 1.3 - 0.4
= 0.9

Let A'and B be the availability of first and second fire engine respectively, then A and B are independent.
Then \(P(A)=P(B)=0.96 \)
\(P(\bar{A})=P(\bar{B})=1-0.96=0.04\)
(i) P(a fire engine is available when needed)
\(=P(A \cap \bar{B})+P(\bar{A} \cap B)+P(A \cap B)\)
\(=P(A) \cdot P(\bar{B})+P(\bar{A}) \cdot P(B)+P(A) \cdot P(B)\)
\(=0.96 \times 0.04+0.04 \times 0.96+0.96 \times 0.96\)
\(=0.96(0.04+0.04+0.96) \)
\(=0.96 \times 1.04 \)
\(=0.9984\)
(ii) P (Neither is available when needed)
\(=P(\bar{A} \cap \bar{B}) \)
\(=P(\bar{A}) \cdot P(\bar{B})\)
\(=0.04 \times 0.04\)
\(=0.0016\)

Let be the event of speaks the truth, be the event of speaks the truth
∴ \(\bar { A } \) is the event of X not speaking the truth and \(\bar { B } \) is the event of Y not speaking the truth.
Let C be the event that they will contradict each other.
Given that
P(A) = 0.70 ⇒ P(\(\bar { A } \)) = 1 - P(A) = 0.30
P(B) = 0.90 ⇒ P(\(\bar { B } \)) = 1 - P(B) = 0.10
C = (A speaks truth and B does not speak truth or B speaks truth and A does not speak truth)
C =\(\left[ (A\cap \bar { B } )\cup (\bar { A } \cap B) \right] \) (see figure)
since \((A\cap \bar { B } )\) and \((\bar { A } \cap B)\) are mutually exclusively,
P(C) = \((A\cap \bar { B } )+(\bar { A } \cap B)\)
= P(A)P(\(\bar { B } \)) + P(\(\bar { A } \))P(B)
( Since A, B are independent event A, \(\bar { B } \) are also independent events
= (0.70) (0.10) + (0.30) (0.90)
= 0.070 + 0.270 = 0.34
P(C) = 0.34
 

Let P(X), P(Y), P(Z) are the probability of hitting a target, then
\(P(X)=\frac{3}{4}, P(Y)=\frac{4}{5}, P(Z)=\frac{2}{3}\)
P (the target is damaged by exactly 2 hits)
=\(P(X).P(Y).P(\overset{-}{Z})+P(X).P(\overset{-}{Y}).P(X)+P(\overset{-}{X}).P(Y).P(Z)\)
=\(\frac{3}{4}\times \frac{4}{5}\times \frac{1}{3}\times \frac{1}{5}\times \frac{2}{3}+\frac{1}{4}\times \frac{4}{5}\times \frac{2}{3}\)
=\(\frac{12}{60}+\frac{6}{60}+\frac{8}{60}\)
=\(\frac{12+6+8}{60}=\frac{26}{60}=\frac{13}{30}\)

Let A be the oil change, P(A) = 0.3
B be the filter change, P(B) = 0.4
And \(P(A \cap B)=0.15\)
\(\text { (i) } P(B / A) =\frac{P(A \cap B)}{P(A)}\)
\(=\frac{0.15}{0.3}=\frac{15}{30}=0.5\)
\(\text { (ii) } P(A / B) =\frac{P(A \cap B)}{P(B)}\)
\(=\frac{0.15}{0.4}=\frac{15}{40}=\frac{3}{8}=0.375\)

Let A, B, C be the events that the problems solved by 3 students. Then,
\(P(A)=\frac{1}{3}, P(B)=\frac{1}{4}, P(C)=\frac{1}{5}\)
(i) P (Problem is solved) \(=P(A \cup B \cup C)\)
\(=1-P(\overline{A \cup B \cup C})\)
\(=1-P(\bar{A} \cap \bar{B} \cap \bar{C})\)
\(=1-P(\bar{A}) P(\bar{B}) P(\bar{C})\)
\(=1-\frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} \)
\(=1-\frac{2}{5}=\frac{3}{5}\)
(ii) P (exactly one of them will solve)
\(=P(A \bar{B} \bar{C} \cup \bar{A} B \bar{C} \cup \bar{A} \bar{B} C)\)
\(=P(A) \cdot P(\bar{B}) \cdot P(\bar{C})+P(\bar{A}) \cdot P(B) \cdot P(\bar{C}) +P(\bar{A}) \cdot P(\bar{B}) \cdot P(C)\)

\(=\frac{1}{5}+\frac{2}{15}+\frac{1}{10}=\frac{1}{5}\left(1+\frac{2}{3}+\frac{1}{2}\right) \)
\(=\frac{6+4+3}{30}=\frac{13}{30}\)

\(P(B)=0.5, P(A \cup B)=0.7, P(A \cap B)=0.2\)
\(\text {WKT } P(A \cup B)=P(A)+P(B)-P(A \cap B)\)
\(0.7=P(A)+0.5-0.2\)
\(0.7=P(A)+0.3\)
\(\therefore P(A)=0.4 \text {. }\)
To prove that A and B are independent
\(\text { i.e, } P(A \cap B)=P(A) \cdot P(B) \text { is true. }\)
\(P(A \cap B) =0.2 \)
\(P(A) \cdot P(B) =0.4 \times 0.5=0.20\)
\(\text {From (1) and (2), } P(A \cap B)=P(A) \cdot P(B)=0.2\)
Hence A and B are independent.

Given P(A) = 0.5 P(B) = 0.3
A and B are mutually cxclusive events \(P(A \cap B)=0\)
\((i) P(A \cup B)=P(A)+P(B)=0.5+0.3=0.8\)
\((ii) P(A \cap \bar{B})=P(A)-P(A \cap B)=0.5-0=0.5\)
\(\text { (iii) } P{\bar{A} \cap B}=P(B)-P(A \cap B)=0.3-0=0.3\)

Sample space S = {1, 2, 3, 4, 5, 6}.
Let A be the event of die shows an odd number.
Let B be the event of getting 5.
Then, A = {1, 3, 5}, B = {5}, and A\(\cap \)B = {5}.
Therefore, P(A) = \(\frac{3}{6}\) and P(A\(\cap \)B) = \(\frac{1}{6}\)
P(getting 5 / die shows an odd number) = P(B / A)
= \(\frac { P(A\cap B) }{ P(A) } =\frac { \frac { 1 }{ 6 } }{ \frac { 3 }{ 6 } } \)
P(B/A) = \(\frac{1}{3}\) .