\(\frac { 7 }{ 8 } \)

\(\frac { 1 }{ 8 } \)

52

Let A = event that 1 red, 2 white and 1 black balls are drawn,
B= event that 1 red, 2 white and 1 black balls are drawn and
C= event that 2 red, 1 white and 1 black balls are drawn.
Here A, B and C are munually exclusive events.
Hence, required probability 
= P(A\(\cup \)B\(\cup \)C) = P(A) + P(B) + P(C)
Ans. \(\frac { 48 }{ 91 } \)

\(\frac { 4 }{ 9 } \)

\(\frac { 47 }{ 10829 } \)

Let F be the event of getting the sum 6. Then,
F={(1,5), (5,1), (2,4), (4,2),(3,3)} \(\Rightarrow n(F) = 5\)
\(P(F)=\frac { n\left( F \right) }{ n\left( S \right) } =\frac { 5 }{ 36 } \Rightarrow P(\overline { F } )=1-\frac { 5 }{ 36 } =\frac { 31 }{ 36 } \)
\(Hence,\ odds\ against\ getting\ the\ sum\ 6=\frac { p\left( \overline { F } \right) }{ P\left( F \right) } \)
Ans. 31:5

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

\(\frac { 2 }{ 3 } \)

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.