\((i) a=5, b=7, P(A)=\frac{a}{a+b}=\frac{5}{5+7}=\frac{5}{1 \cdot 2}\)
\( (ii) P(B)=\frac{2}{5}=\frac{a}{a+b}\)
\(a=2, a+b=5\)
\(b=3\)
The odds that the event B occurs is 2 to 3.

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) = \(\frac{5}{6}\).

Let S be the sample space and A be the event that the smallest block is to be painted in red.
n(S) = 6 P₃ = 6 \(\times\) 5 \(\times\) 4 = 120
n(A) = 5 \(\times\) 4 = 20
P(A) = \(\frac { n(A) }{ n(S) } =\frac { 20 }{ 120 } =\frac { 1 }{ 6 } \)

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)
\(=P(\bar{A} \cap \bar{B}) \)
\(=P(\bar{A}) \cdot P(\bar{B})\)
\(=0.04 \times 0.04\)
\(=0.0016\)

If the given matrix M is singular, then
\(\left| \begin{matrix} x & y \\ z & 1 \end{matrix} \right| \) = 0.
That is , x - yz = 0
Hence the possible ways of selecting (x, y, z) are
{(1,1,1), (2,1,2), (,2,2,1), (3,1,3), (3,3,1)} = A(say)
The number of favourable cases  n(A) = 5
The total number of cases are n(S) = 3³ = 27
The probability of the given matrix is a singular matrix is
P = \(\frac { n(A) }{ n(S) } =\frac { 5 }{ 27 } \)

Let A, B, and C denote the envelopes and 1, 2, and 3 denote the corresponding letters
The different combination of letters put into the envelopes are shown in the table Let c_i denote the outcomes of the events. Let X be the event of putting the letters into the exactly only one right envelopes Let Y be the event of putting none of the letters into the right envelope 
S = {C₁, C₂, C₃, C₄, C₅, C₆}, n(S) = 6
X = {C₂, C₃, C₆},  n(X) = 3
Y = {C₄, C₅}   n(Y) = 2
P(X) = \(\frac { 3 }{ 6 } =\frac { 1 }{ 2 } \) P(Y) = \(\frac { 2 }{ 6 } =\frac { 1 }{ 3 } \)

Let A_i be the event that the traffic light opens at i th cross, for i = 1, 2, 3, 4.
Let B_i be the event that the traffic light closes at i th cross, for i = 1, 2, 3, 4.
The traffic lights are all independent.
Therefore Ai and B_i are all independent events, for i = 1, 2, 3, 4.
Given that P(A_i) = 0.4, i = 1, 2, 3, 4
P(B_i) = 0.6, i =1, 2, 3, 4
(i) Probability of car crossing the first crossroad without stopping,
P(A₁) = 0.4
(ii) Probability of car crossing first two crossroads without stopping
\(P({ A }_{ 1 }\cap { A }_{ 2 })\)= P(A₁A₂) = (0.4)(0.4) = 0.16
(iii) Probability of car crossing all the crossroads, stopping at third cross
P\(({ A }_{ 1 }\cap { A }_{ 2 }{ B }_{ 3 }\cap { B }_{ 3 }\cap { A }_{ 4 })\) = P(A₁A₂B₃A₄) = (0.4)(0.4)(0.6)(0.4) = 0.0384
(iv) Probability of car crossing all the crossroads, stopping at exactly one of the crossroads is
P(B₁A₂A₃A₄ \(\cup \) A₁B₂A₃A₄ \(\cup \) A₁A₂B₃A₄ \(\cup \) A₂A₃A₃B₄)
= P(B₁A₂A₃A₄)+P(A₁B₂A₃A₄)+P(A₁A₂B₃A₄)+P(A₁A₂A₃B₄)
= 4(0.4)(0.4)(0.6)(0.4) = 4(0.0384) = 0.1536.

Let S be the sample space,
A be the event of getting an even number,
B be the event of getting multiple of three.
Therefore,
S = {1, 2, 3, 4, 5, 6} ⇒ n(S) =6 
A = {2, 4, 6} ⇒ n(A) = 3
B = {3, 6} ⇒ n(B) = 2
The required probabilities are
(i) P (getting an even number) = P(A) = \(\frac { n(A) }{ n(S) } =\frac { 3 }{ 6 } =\frac { 1 }{ 2 } \)
(ii) P (getting multiple of three) = P(B) = \(\frac { n(B) }{ n(S) } =\frac { 2 }{ 6 } =\frac { 1 }{ 3 } \)

Let A be the event of getting prime number and B be the event of getting multiple of 8 up to 100, then
\(n(A)=25, n(B)=12, n(S)=100\) and \(\mathrm{n}(\mathrm{A} \cap \mathrm{B})=0\)
Now, P (the integer chosen is a prime or multiple of 8)
\(=P(A \cup B) \)
\(=P(A)+P(B)-P(A \cap B) \)
\(=\frac{25}{100}+\frac{12}{100}-0\)

\(=\frac{37}{100}\)

(i) Let A be the event ofgetting one mango and one apple, Then
\(P(A)=\frac{5 C_1 \times 4 C_1}{9 C_2}=\frac{5}{9}\)

(ii) Let B and C be the events of getting both are mango and both are an apple respectively then
\(P(\text { Bor } C) =P(B)+P(C) \)
\(=\frac{5 C_2}{9 C_2}+\frac{4 C_2}{9 C_2}=\frac{5 \times 4}{9 \times 8}+\frac{4 \times 3}{9 \times 8} \)
\(=\frac{8}{18}=\frac{4}{9}\)