Since a 3digit number cannot start with 0, therefore hundredth place can be filled up in 4 ways.
Now as repetition of  digits is allowed, therefore each of  ten's and unit's place can be filled in 5 ways.
Thus , the total number of possible3 -digit number
= 4 x 5 x 5 = 100
Note that, there are three digits numbers, which has the same digits viz. 222,444,666 and 888
\(\Rightarrow \)  Number of favourable outcomes = 4
Hence ,P (3digit number with same digits) = \(\frac { 4 }{ 100 } =\frac { 1 }{ 25 } \)

A={(H,H)}, B={(H,T),(T,H)} and C={(H,T),(T,H),(T,T)}
A not B=A-B={(H,H)}

{2}

{3,5}

\(\frac { 2 }{ 7 } \)

30

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), 

Let the balls in the box be represented by R₁,R_2,and B₁ ,B_2, B₃
{(H,1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, R₁), (T, R₂), (T, B₁), (T, B₂), (T, B₂), (T, B₃)}

Here total number of persons = 5 one spokesperson is selected out of 5 persons in \(^{ 5 }C_{ 1 }\)=ways
Let A be the event that person is male and Bbe the event that person is over 35 years. There are 3male and one person can be selected in \(^{ 3 }C_{ 1 }\)ways
\(\therefore \quad P(A)=\frac { ^{ 3 }C_{ 1 } }{ ^{ 5 }C_{ 1 } } =\frac { 3 }{ 5 } \)
There are 2 person who are over 35 years. So one person can be selected in \(^{ 2 }C_{ 1 }\) ways
\(\therefore \ P(B)=\frac { ^{ 2 }C_{ 1 } }{ ^{ 5 }C_{ 1 } } =\frac { 2 }{ 5 } \)
There is 1 person who is male and over 35 years.
\(\therefore P(A\cap B)=\frac { ^{ 1 }C_{ 1 } }{ ^{ 5 }C_{ 1 } } =\frac { 1 }{ 5 } \)
\(\text{Now P(either male or over 35 years)}=P(A\cup B)\)
\(=P(A)+P(B)-P(A\cap B)\)
\(=\frac { 3 }{ 5 } +\frac { 2 }{ 5 } -\frac { 1 }{ 5 } =\frac { 3+2-1 }{ 5 } =\frac { 4 }{ 5 } \)

Total number of tickets = 10,000 Number of prize bearing tickets = 10
(a) Let A be the event that one ticket is prize bearing ticket.
\(\therefore \ n(A)=^{ 10 }C_{ 1 }\)
\( \therefore \ P(A)=\frac { ^{ 10 }C_{ 1 } }{ ^{ 10000 }C_{ 1 } } =\frac { 10 }{ 10000 } =\frac { 1 }{ 1000 } \)
\(Now\ P(\overline { A } )=1-\frac { 1 }{ 1000 } =\frac { 999 }{ 1000 } \)
b) Let B be the event that two tickets are prize bearing tickets.
\(\therefore \ n(B)=^{ 10 }C_{ 2 }\)
\(\therefore \ P(B)=\frac { ^{ 10 }C_{ 2 } }{ ^{ 10000 }C_{ 2 } } =\frac { 10! }{ 2!8! } =\frac { 2!9998! }{ 10000! } \)
\(=\frac { 10\times 9 }{ 2 } \times \frac { 2 }{ 10000\times 9999 } \)
\(=\frac { 1 }{ 1111000 } \)
\(Now\ P(\overline { B } )=1-\frac { 1 }{ 1111000 } =\frac { 1110999 }{ 1111000 } \)
c) Let C be the event that ten tickets are not prize bearing tickets.
Therefore  number of non prize bearing tickets.
\(=10000-10=9990\)
\(\therefore \ P(C)=\frac { ^{ 9990 }C_{ 10 } }{ ^{ 10000 }C_{ 10 } }\)

Let E₃ be the event of getting 2 women.
Then, n(E₃) = \({ C }_{ 2 }\times ^{ 7 }{ C }_{ 2 }\) = 1 x 21 =21
Hence, required probability = \(\frac { 21 }{ 126 } =\frac { 1 }{ 6 } \)

0.87

\(\frac { 1 }{ 4 } \)

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

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
\(Thus\ P(E)=\frac { n(E) }{ n(S) } \frac { 5 }{ 6 } \)