(i) In random experiment of tossing three coins, the sample  space is 
S = (HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
\(\Rightarrow \)  n(S) = 8
Let E₁ be the event of getting 3 heads. Then, outcomes favourable to E₁ is {HHH}.
Thus, n(E₁) = 1
\(\therefore \)  \(P({ E }_{ 1 })=\frac { n({ E }_{ 1 }) }{ n{ (S) } } =\frac { 1 }{ 8 } \)
\(\text { (ii) Let } C \text { be the event of the occurrence of } 2 \text { heads. Accordingly, } \mathrm{C}=\left\{\mathrm{HH}_{1} \mathrm{HTH}_{\mathrm{d}} \mathrm{THH}\right\}\)
\(\therefore P(G)=\frac{n(C)}{n(S)}=\frac{3}{8}\)
(iii) In random experiment of tossing three coins, the sample  space is 
S = (HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
\(\Rightarrow \) n(S) = 8
Let E₂ be the event of getting atleast 2 heads.
Then, outcome favourable to E₂ are {HHT, HTH, THH, and HHH}
Thus, n(E₂) = 4
\(\therefore \)   \(P({ E }_{ 2 })=\frac { n({ E }_{ 2 }) }{ n{ (S) } } =\frac { 4 }{ 8 } =\frac { 1 }{ 2 } \)
(iv) In random experiment of tossing three coins, the sample  space is 
S = (HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
\(\Rightarrow \) n(S) = 8
Let E₃ be the event of getting atmost 2 tails. Then, outcomes favourable to E₃ are {HHH, HHT, HTH, THH, TTH, THT, HTT}.
Thus, n(E₃) = 7
\(\therefore \)  \(P({ E }_{ 3 })=\frac { n({ E }_{ 3 }) }{ n{ (S) } } =\frac { 7 }{ 8 } \)
(v) \( \text { Let } \mathrm{F} \text { be the event of the accurrence of no head. }\)
\(\text { Accordingly, } \mathrm{F}=\{\mathrm{TTT}\} \)
\(\therefore \mathrm{P}(\mathrm{F})=\frac{n(\mathrm{~F})}{n(\mathrm{~S})}=\frac{1}{8}\)
(vi) \( \text { Let } \mathrm{G} \text { be the event of the accurrence of 3 tails. }\)
\(\text { Accordingly, } \mathrm{F}=\{\mathrm{TTT}\} \)
\(\therefore \mathrm{P}(\mathrm{G})=\frac{n(\mathrm{~G})}{n(\mathrm{~S})}=\frac{1}{8}\)
(vii) In random experiment of tossing three coins, the sample  space is 
S = (HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
⇒  n(S) = 8
Let E₄ be the event of getting exactly two tails, Then, outcomes favourable to E₄ are {TTH, HHT, THT}.
Thus, n(E₄) = 3
 \(\therefore \)  \(P({ E }_{ 4 })=\frac { n({ E }_{ 4 }) }{ n{ (S) } } =\frac { 3 }{ 8 } \)
(viii) In random experiment of tossing three coins, the sample  space is 
S = (HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
⇒ n(S) = 8
Let E₅ be the event of getting no tail. Then, outcomes favourable to E₅ is {HHH}.
Thus ,  n(E₅) = 1
\(\therefore \)  \(P({ E }_{ 5 })=\frac { n({ E }_{ 5 }) }{ n{ (S) } } =\frac { 1 }{ 8 } \)
(ix)  In random experiment of tossing three coins, the sample  space is 
S = (HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
⇒ n(S) = 8
\(P({ E }_{ 7 })=\frac { n({ E }_{ 8 }) }{ n{ (S) } } =\frac { 7 }{ 8 } \)

(i) A′
A = {(2,1),(2,2)...(2,6),(4,1),(4,2),...(4,6),(6,1),(6,2),...(6,6)}
B = {(1,1),(1,2),...(1,6),(3,1),(3,2),...(3,6),(5,1),(5,2),...(5,6)}
C = {(1,1),(1,2),(2,1),(1,3),(3,1),(2,2),(2,3),(3,2)(1,4),(4,1)}
A^' = {(1,1),(1,2),...(1,6),(3,1),(3,2),...(3,6),(5,1),(5,2),...(5,6)}
(ii) not B
A = {(2,1),(2,2)...(2,6),(4,1),(4,2),...(4,6),(6,1),(6,2),...(6,6)}
B = {(1,1),(1,2),...(1,6),(3,1),(3,2),...(3,6),(5,1),(5,2),...(5,6)}
C = {(1,1),(1,2),(2,1),(1,3),(3,1),(2,2),(2,3),(3,2)(1,4),(4,1)}
B^' = {(2,1),(2,2)...(2,6),(4,1),(4,2),...(4,6),(6,1),(6,2),...(6,6)}
(iii) A or B
A = {(2,1),(2,2)...(2,6),(4,1),(4,2),...(4,6),(6,1),(6,2),...(6,6)}
B = {(1,1),(1,2),...(1,6),(3,1),(3,2),...(3,6),(5,1),(5,2),...(5,6)}
C = {(1,1),(1,2),(2,1),(1,3),(3,1),(2,2),(2,3),(3,2)(1,4),(4,1)}
\(A\cup B\)  = {(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)}
(iv) A and B
A = {(2,1),(2,2)...(2,6),(4,1),(4,2),...(4,6),(6,1),(6,2),...(6,6)}
B = {(1,1),(1,2),...(1,6),(3,1),(3,2),...(3,6),(5,1),(5,2),...(5,6)}
C = {(1,1),(1,2),(2,1),(1,3),(3,1),(2,2),(2,3),(3,2)(1,4),(4,1)}
\(A\cap B=\phi \)
(v) A but not C
A - C ={(2,4),(2,5),(2,6),(4,2),(4,3),(4,4),(4,5),(4,6),(6,1),(6,2),(6,3),(6,4)(6,5),(6,6)}
(vi) B or C
\(B\cup C\)={(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)}
(vii) B and C
\(B\cap C\) ={(1,1),(1,2),(1,3),(1,4),(3,1),(3,2)}
(viii) A ∩ B′ ∩ C′
\(A\cap { B }^{ ' }\cap { C }^{ ' }\)={(2,4),(2,5),(2,6),(4,2),(4,3),(4,4),(4,5),(4,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

The four digit numbers greater than 5000are randomly formed from the digits 0,1,3,5,7.
(i) Given, the digits are repeated.
The number formed is greater than 5000, the leftmost digit is either 7or 5. The remaining three places can be filled by any of the digits 0,1,3,5or 7.
Total four digit number formed greater than 5000 is 2×5×5×5−1=249.
If unit digit of a number is 0 or 5 then number is divisible by 5.
Total numbers that are divisible by 5is 2×5×5×2−1=99.
Assume E be the event when number is divisible by 5and digits are repeating.
The probability is, P( E )= 99 249 = 33 249
Thus, the probability of number divisible by 5when digits repeated is 33 249 .
(ii) Given, repetition of digit is not allowed. Then the thousands place can be filled with digits either 5or 7and remaining three places can be filled with any of the remaining four digits.
Total four digit number formed greater than 5000is 2×4×3×2=48.
If the digit at thousand place is 5 then unit place can be filled with 0 and tens and hundreds places can be filled with any of the remaining three digits.
The four digits number formed started with 5and divisible with 5is 3×2=6.
If the digit at thousand place is 7 then unit place can be filled with 0or 5.
The four digit numbers starting with 7and divisible by 5is 1×2×3×2=12.
Total number greater than 5000and divisible by 5is, 6+12=18
The probability of a number formed is divisible by 5 with repetition of digits is, 18 48 = 3 8
Thus, the required probability is 3 8 .

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

It is given that the box contains 1 red ball and 3 identical white balls. Let us denote the red ball with R and a white ball with W.
When two balls are drawn at random in succession without replacement, the sample space is given by
S = {RW, WR, WW}

There are four men and six women on the city council.
As one council member is to be selected for a committee at random, the sample space contains 10 (4 + 6) elements.
Let A be the event in which the selected council member is a woman.
Accordingly, n(A) = 6
\(\therefore \mathrm{P}(\mathrm{A})=\frac{\text { Number of outcomes favourable to } \mathrm{A}}{\text { Total number of possible outcomes }}=\frac{n(\mathrm{~A})}{n(\mathrm{~S})}=\frac{6}{10}=\frac{3}{5}\)
 

Let ₁,L₂,L₃ be three letters and E₁,E₂,E₃ be their corresponding envelopes.
clearly, 3 letters can be inserted into 3 addressed envelopes in \(^{ 3 }{ P }_{ 3 }=3!=6\)ways, which are shown as follows
\(\{ { (L }_{ 1 }{ ,E }_{ 1 }{ ),(L }_{ 2 }{ ,E }_{ 3 }{ ),(L }_{ 3 }{ ,E }_{ 2 }{ ),(L }_{ 1 }{ ,E }_{ 2 }),({ L }_{ 2 },{ E }_{ 1 }),({ L }_{ 3 }{ ,E }_{ 3 });({ L }_{ 1 }{ ,E }_{ 3 }{ ),(L }_{ 2 }{ ,E }_{ 2 }{ ),(L }_{ 3 },{ E }_{ 1 });\)
\({ (L }_{ 1 }{ ,E }_{ 1 }),({ L }_{ 2 }{ ,E }_{ 2 }{ ),(L }_{ 3 }{ ,E }_{ 3 }{ );(L }_{ 1 }{ ,E }_{ 2 }),({ L }_{ 2 },{ E }_{ 3 }{ ),(L }_{ 3 }{ ,E }_{ 1 }{ ),(L }_{ 1 }{ ,E }_{ 3 }{ ),(L }_{ 2 }{ ,E }_{ 1 }),({ L }_{ 3 }{ ,E }_{ 2 })\} \)
Note that ,there are 4 ways in which atleast one letter is in its proper envelope.
Required probability=\(\frac { 4 }{ 6 } =\frac { 2 }{ 3 } \)

Let M be the event that selected student studies
Mathematics and B be the event that selected student
studies Biology. Then, P(M) = 40%=0.4
P(B)=30%=0.3 and \(P(M\cap B)\)=10%=0.1
Now, P(he will study Mathematics or Biology)
\(=P(M\cup B)=P(M)+P(B)-P(M\cap B)=0.4+0.3-0.1=0.6\)
Hence,the required probability is 0.6.

Total number of  possible outcomes = \(^{ 52 }{ C }_{ 4 }\)
Number of favourable outcomes = \(^{ 13 }{ C }_{ 3 }\times ^{ 13 }{ C }_{ 1 }\)
\(\frac { ^{ 3 }{ C }_{ 3 }.^{ 13 }{ C }_{ 1 }\quad }{ ^{ 52 }{ C }_{ 4 } } \)

Since the coin is tossed four times, there can be a maximum of 4 heads or tails.
When 4 heads turns up, Re1 + Re1 + Re1 + Re1 = Rs 4  is the gain.
When 3 heads and 1 tail turn up, Re 1 + Re 1 + Re 1 – Rs 1.50 = Rs 3 – Rs 1.50 = Rs 1.50 is the gain.
When 2 heads and 2 tails turns up, Re 1 + Re 1 – Rs 1.50 – Rs 1.50 = – Re 1, i.e., Re 1 is the loss.
When 1 head and 3 tails turn up, Re 1 – Rs 1.50 – Rs 1.50 – Rs 1.50 = – Rs 3.50, i.e., Rs 3.50 is the loss.
When 4 tails turn up, – Rs 1.50 – Rs 1.50 – Rs 1.50 – Rs 1.50 = – Rs 6.00, i.e., Rs 6.00 is the loss.
There are 2⁴ = 16 elements in the sample space S, which is given by:
S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTTH, TTHH, HTHT, THTH, THHT, HTTT, THTT, TTHT, TTTH, TTTT}
∴ n(S) = 16
The person wins Rs 4.00 when 4 heads turn up, i.e., when the event {HHHH} occurs.
\(\text { Prabability (of winning Rs } 4.00)=\frac{1}{16}\)
\(\text { The person wins Rs } 1.50 \text { when } 3 \text { heads and one tail turn up, i.e., when the event }\{\mathrm{HHHT}, \mathrm{HHTH}, \mathrm{HTHH}\text { THHH\} occurs. }\)
\(\therefore \text { Prabability (of winning Rs } 1.50)=\frac{4}{16}=\frac{1}{4}\)
\(\text { The person lases Re } 1.00 \text { when } 2 \text { heads and } 2 \text { tails turn up, i.e., when the event }\{\mathrm{HHTT}, \mathrm{HTTH}, \mathrm{TTHH} \text { , }\text { HTHT, THTH, THHT\} occurs. }\)
\(\therefore \text { Probability (of losing Re } 1.00)=\frac{6}{16}=\frac{3}{8}\)
\(\text { The person lases Rs } 3.50 \text { when } 1 \text { head and } 3 \text { tails turn up, i.e., when the event }\{\text { HTTT } \text { THTT, TTHT }\text { TTTH\} OCCUrS. } \)
\(\text { Probability (of losing Rs } 3.50)=\frac{4}{16}=\frac{1}{4}\)
\(\text { The person lases Rs } 6.00 \text { when } 4 \text { tails turn up, i.e., when the event }\{T T T\} \text { accurs. }\)
\(\text { Prabability (of lasing Rs } 6.00)=\frac{1}{16}\)

(i) There are 13 letters in the word ASSASSINATION.
∴ Hence, n(S) = 13
There are 7 vowels in the given word
∴ probability (vowel) =\(\frac { 6 }{ 13 } \)
(ii) There are 13 letters in the word ASSASSINATION.
∴ Hence, n(S) = 13
There are 6 vowels in the given word
∴ probability (vowel) = \(\frac { 7 }{ 13 } \)

\(\text{i) here} P(A)=\frac { 1 }{ 3 } ,P(B)=\frac { 1 }{ 5 }\)
\(\text{and }P(A\cap B)=\frac { 1 }{ 15 } \)
we know that
\( P(A\cup B)=P(A)+P(B)-P(A\cap B)\)
\(=\frac { 1 }{ 3 } +\frac { 1 }{ 5 } -\frac { 1 }{ 15 } \)
\(=\frac { 5+3-1 }{ 15 } =\frac { 7 }{ 15 } \)
\(\text{Here } P(A)=0.35,(P(A\cap B)=0.25\)
\(\text{and } P(A\cup B)=0.6\)
\(we\quad know\quad that\)
\(P(A\cup B)=P(A)+P(B)-P(A\cap B)\)
\( \therefore \ 0.6=0.35+P(B)-0.25\)
\(\Rightarrow \ 0.6=0.1+P(B)\)
\(\Rightarrow \ P(B)=0.6-0.1=0.5\)
\(iii)\quad Here\quad P(A)=0.5,P(B)=0.35\)
\(and\quad P(A\cup B)=0.7\)
We know that
\(P(A\cup B)=P(A)+P(B)-P(A\cap B)\)
\(\therefore \quad 0.7=0.5+0.35-P(A\cap B)\)
\(\Rightarrow \ 0.7=0.85-P(A\cap B)\)
\(\Rightarrow P(A\cap B)=0.85-0.7=0.15\)

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.
\(=10000-10=9990\)
\(\therefore \ P(C)=\frac { ^{ 9990 }C_{ 10 } }{ ^{ 10000 }C_{ 10 } }\)

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.
\(\therefore P(A\cap B)=\frac { ^{ 1 }C_{ 1 } }{ ^{ 5 }C_{ 1 } } =\frac { 1 }{ 5 } \)
\(\text{Now P(either male or over 35 years)}=P(A\cup B)\)
\(=P(A)+P(B)-P(A\cap B)\)
\(=\frac { 3 }{ 5 } +\frac { 2 }{ 5 } -\frac { 1 }{ 5 } =\frac { 3+2-1 }{ 5 } =\frac { 4 }{ 5 } \)