Total outcomes = 6!
Favourable outcomes = 4! x 3!
\(\frac { 1 }{ 5 } \)

{2}

\(\frac { 5 }{ 11 } \)

\(\frac { 1 }{ 3 } \)

Let us represent the red ball by R and 3 balck balls by B₁, B₂, B₃
{(R, B₁), (R, B₂), (R, B₃), (B₁, R), (B₁, B₂), (B₁, B₃), (B₂, R),(B₂, B₁), (B₂, B₃),  (B₃, R), (B₃, B₁), (B₃, B₂), 

Total number of possible outcomes = \(^{ 52 }C_{ 4 }\)
We know that, there are 4 suits each having 13 cards.
\(\therefore\) Number of favourable outcomes
=\(^{ 13 }C_{ 1 }\times ^{ 13 }C_{ 1 }\times ^{ 13 }C_{ 1 }\times ^{ 13 }C_{ 1 }=\left( 13 \right) ^{ 4 }\)
Hence, P(getting one card from each suit) = \(\frac { \left( 13 \right) ^{ 4 } }{ ^{ 52 }C_{ 4 } } \)

We know that, if odds against of an event are n:m, then probability of occurrence of this event is \(\frac { m }{ m+n } \)
\(\therefore \) Required probability = \(\frac { 7 }{ 7+4 } =\frac { 7 }{ 11 } \)

On throwing of two dice, we have sample space
\(S=\left\{\begin{array}{l} (1,1),(1,2),(1,3),(1,4),(1,5),(1,6), \\ (2,1),(2,2),(2,3),(2,4),(2,5),(2,6) \\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6), \\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6) \\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6), \\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \end{array}\right\}\)
B: Getting on the odd number on the first die
B^'Getting an even number on the first die
=\(\left\{ (2,1),(2,2),(2,3),(2,4),(2,5),(2,6),\\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6),\\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6), \right\} \)

We have, \(P(A\cup B)\) = 0.6 and \(P(A\cap B)\) = 0.2
We know, \(P(A\cup B)\) = P(A) + P(B) -\(P(A\cap B)\)
\(\therefore \)  0.6 = P(A) + P(B) - 0.2
P(A) + P(B) = 0.6 + 0.2 = 0.8 ...(i)
Now,
\(P(\overline { A } )+P(\overline { B } )\) 
=  P(A) +1-P(B)
= 2 - (P(A) + P(B))
= 2 - 0.8
= 1.2

(i) \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\)
\(=0.54+0.69-0.35\)
\(=0.88\)
(ii) \(P(A'\cap B')=P((A\cup B)')=1-P(A\cup B)\)
\(=1-0.88=0.12\)
(iii) \(P(A\cap B')=P(A)-P(A\cap B)=0.54-0.35=0.19\)
(iv) \(P(B\cap A')=P(B)-P(A\cap B)=0.69-0.35=0.34\)

When the order of the birth of a girl or a boy is considered, the sample space is given by S = {GG, GB, BG, BB}

\(\frac { 1 }{ 2 } \)

\(\frac { 1 }{ 13 } \)

P(getting neither a doublet nor a total of 10)
=1 - P(getting a doublet or a total of 10)
Ans. \(\frac { 7 }{ 9 } \)

Here the sample space S = {I, 2, 3,4,5,6}
\(\therefore \) n(S) = 6
(i) Let A be the event of getting a prime number
A = {2, 3, 5} \(\Rightarrow \) n(A) = 3
\(Thus\ P(A)=\frac { n(A) }{ n(S) } =\frac { 3 }{ 6 } =\frac { 1 }{ 2 } \)
(ii) Let B be the event of getting a number greater than or equal to 3
B = {3, 4, 5, 6} \(\Rightarrow \) n(B) = 4
\(Thus\ P(B)=\frac { n(B) }{ n(S) } =\frac { 4 }{ 6 } =\frac { 2 }{ 3 } \)
Let C be the event of getting a number less than or equal to 1
C = {I} \(\Rightarrow \) n(C) = 1
\(Thus\ P(C)=\frac { n(C) }{ n(S) } =\frac { 1 }{ 6 } \)
(iv) Let D be the event of getting a number more than 6
\(D=\phi \Rightarrow n(D)=0\)
\(Thus\ P(D)=\frac { n(D) }{ n(S) } \frac { 0 }{ 6 } =0\)
Let E be the event of getting a number less than 6
E = {I, 2, 3, 4, 5} \(\Rightarrow \) n(E) = 5
\(Thus\ P(E)=\frac { n(E) }{ n(S) } \frac { 5 }{ 6 } \)

Taking A, B, C events from question 6 above we have
\(i)\ A\cap B=\phi \)
Thus A and B are mutually exclusive and exhaustive events.
∴ True.
\( ii)\ A\cap B\ =\phi \ and\ A\cap B=S\)
Thus A and B are mutually exclusive and exhaustive events.
∴ True.
(iii) B′={(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}=A.
∴ True.
(iv) A∩C = {(2,1)(2,2)(2,3)(4,1)}=ϕ
Thus A and C are not mutually exclusive events.
∴ False
(v) A∩B′ = A ≠ ϕ
Thus A and B are not mutually exclusive events.
(vi) Since A′ = B and B′ = A, A∩B = ϕ
B∩C = {(1,1),(1,2),(1,3),(1,4),(3,1),(3,2)} ≠ ϕ
A∩C = {(2,1)(2,2)(2,3)(4,1)} = ϕ
Thus A′, B′ and C are not mutually exclusive.
∴ False.