(i) Given that five fair coins are tossed once. Since the coins are fair coins the probability of getting an head in a single coin is
 \(p=\frac { 1 }{ 2 } \) and \(q=1-p=\frac { 1 }{ 2 } \) 
Let X denote the number of heads that appear in five coins. X is binomial random variable that takes on the values 0, 1, 2,  3, 4 and 5 and \(p=\frac { 1 }{ 2 } \) That is \(X\sim B\left( 5,\cfrac { 1 }{ 2 } \right) \) 
Therefore the binomial distribution is
\(f(x)=\left( \begin{matrix} n \\ x \end{matrix} \right) p*\left( 1-p \right) ^{ n-x }\), x = 0, 1, 2,..,n
becomes
\(f(x)=\left( \begin{matrix} 5 \\ x \end{matrix} \right) \left( \cfrac { 1 }{ 2 } \right) ^{ x }\left( \cfrac { 1 }{ 2 } \right) ^{ n-x }\), x = 0, 1, 2,..,5
That is 
\(f(x)=\left( \begin{matrix} 5 \\ x \end{matrix} \right) \left( \cfrac { 1 }{ 2 } \right) ^{ n }\), x = 0, 1, 2...,n
(ii) A fair die is rolled ten times and X denotes the number of times 4 appeared. X is binomial
random variable that takes on the values 0, 1, 2, 3,...10 , with n = 10 and \(p=\cfrac { 1 }{ 6 } \). That is \(X\sim B\left( 10,\cfrac { 1 }{ 6 } \right) \) 
Probability of getting a four in a die is \(p=\frac { 1 }{ 2 } \) and \(q=1-p=\frac { 5 }{ 6 } \)
Therefore the binomial distribution is 
\(f(x)=\left( \begin{matrix} 10 \\ x \end{matrix} \right) \left( \cfrac { 1 }{ 6 } \right) ^{ x }\left( \cfrac { 5 }{ 6 } \right) ^{ 10-x }\) x = 0, 1, 2,...,10

(i) The sample space S = {H,T}\(\times\){H,T}
That is S = {TT,TH,HT,HH}
(ii)  Let X : S ⟶R be the number of tails
Then X (TT) = 2 (2 Tails)
X (TH ) = 1 (1 Tail)
X (HT) = 1 (1 Tail)
and X (HH) = 0 (0 Tails).
Then X is a random variable that takes on the values 0, 1 and 2.
Let X (ω) denotes the number of tails, this gives
\(\\ \\ \\ \\ \\ \\ X\left( \omega \right) =\begin{cases} \begin{matrix} 2 & if\omega =TT \end{matrix} \\ \begin{matrix} 1 & if\omega =HT,TH \end{matrix} \\ \begin{matrix} 0 & if\omega =HH \end{matrix} \end{cases}\) 
The inverse images of 1 {TH, HT} . That is X^-1{1} = {TH, HT}.
(iii) Number of elements in inverse images are shown in the table.

Values of the Random Variable	0	1	2	Total
Number of elements in inverse image	1	2	1	4

Given \(f(x)=\begin{cases} \begin{matrix} kxe^{ -2x } & forx>0 \end{matrix} \\ \begin{matrix} 0 & for\quad x\le 0 \end{matrix} \end{cases}\)
Since the given function is a probability density function 
\(\int _{ -\infty }^{ \infty }{ f(x)dx } \) = 1 
\(\Rightarrow k\int _{ 0 }^{ \infty }{ { xe }^{ -2x }dx=1 } \)
\(\Rightarrow k \frac { 1! }{ \left( 2 \right) ^{ 2 } } =1\) 
[\(\int _{ 0 }^{ \infty }{ { x }^{ n }e^{ -ax } } =\frac { n! }{ { a }^{ +1 } } \), Here a = 2, n = 1]
\(\Rightarrow \frac { k }{ 4 } =1\\ \Rightarrow k=4\)

Let X be the random variable of getting apples Given 5 mangoes and 4 apples are in an urn 
= {0, 1,2,3}
The sample space consists of ⁹C₃ = 84
X = 0, X (3 mangoes) = ⁵C₃ = 10
X = 1, X (2 mangoes and 1 apples) = ⁵C₂ x ⁴C₁ = 40
X = 2, X (1 mangoes and 2 apples) = ⁵C₁ x ⁴C₂ = 30
X = 3, X (apples) = ⁴C₃ = 4

Values of random variable	0	1	2	3	Total
No of points in inverse image	10	40	30	4	84

Let X be the random variable of number of tails when three coins tossed.
S = {HHH, HHT, THH, HTH, HTT, THT, TTH,TTT}
n(S) = 8
Let X denote the number of tarits occured.
X^-1 (0) {HHH} = 1
X^-1 (1) = {HHT, THT, HTH} = 3
X^-1 (2) =, {HTT, THT, TTH} = 3
X^-1 (3) = {TTT} = 1
ஃ X takes the values 0, 1, 2, 3.

Values of random variable X	0	1	2	3	Tortal
Number of elements in reverse images	1	3	3	1	8

\( \mathrm{n}=10, \mathrm{p}=\frac{1}{5}, \mathrm{k}=4 \)
\( \mathrm{P}(\mathrm{X}=\mathrm{x})={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{x}} \mathrm{p}^{\mathrm{x}} \mathrm{q}^{\mathrm{n}-\mathrm{x}}, \mathrm{x}=0,1,2, \ldots, \mathrm{n} \)
\( \mathrm{p}=\frac{1}{5} \)
\( \mathrm{q}=1-\mathrm{p}=1-\frac{1}{5}=\frac{4}{5} \)
\( \mathrm{P}(\mathrm{X}=4)={ }^{10} \mathrm{C}_{4}\left(\frac{1}{5}\right)^{4}\left(\frac{4}{5}\right)^{6} \)
\( =210 \times\left(\frac{1}{5}\right)^{4}\left(\frac{4}{5}\right)^{6} \)
\( =210 \times \frac{4^{6}}{5^{10}}\)

Given n = 12
P = 0.9
(i) P(X = 10) = 12C₁₀(0.9)¹⁰(1 - 0.9)²
= 12C₁₀(0.9)¹⁰ (0.1)²
(ii) P(X 2: 11) = R(X = 11) + P(X = 12)
= 12C₁₁(0.9)¹¹ (0.1) + 12C₁₂(0.9)¹²(0.1)⁶
= 12C (0.9)¹¹ (0.1) + (0.9)¹²
= 12(0.9)¹¹ (0.1) + (0.9)¹²
= (0.9)¹¹ ((12)(0.1) + 0.9)
= (0.9)11 (1.2 + 0.9) = (0.9)11 (2.1)
(iii) P( at least 2 will not have a useful life of atleast 600 hours) = 1 - P( atleast 11 will have a useful life of atleast 600 hours)
= 1 - P(X ≥11)
= 1-2.1 (0.9)¹¹

Given probability mass function is

x	0	1	2	3	4
f(x)	\(\frac{1}{36}\)	\(\frac{2}{36}\)	\(\frac{3}{36}\)	\(\frac{2}{36}\)	\(\frac{3}{36}\)

(i) Since f(x) is a probability mass function.
\(\sum _{ i=1 }^{ 5 }{ f({ x }_{ i }) } =1\) 
⇒ k² + 2k² + 3k² + 2k + 3k = 1
⇒ 6k² + 5k = 1
⇒ 6k² + 5k - 1 = 0 
⇒ (k + 1) (6k - 1) = 0
⇒ k = -1 or  ⇒ \(k=\frac { 1 }{ 6 } \) 
⇒ \(k=\frac { 1 }{ 6 } \)
(ii) p(2 ≤ x < 5)
= p(x = 2) + p(x = 3) + p(x = 4)
= 2k² + 3k² + 2k = 5k² + 2k
= \(5\left( \frac { 1 }{ 36 } \right) +2\left( \frac { 1 }{ 6 } \right) \) 
= \(\frac { 5 }{ 36 } +\frac { 1 }{ 3 } =\frac { 5+12 }{ 36 } \) 
= \(\frac { 17 }{ 36 } \) 
(iii) p(3 < x) = p(x > 3)
= p(x = 4) + p(x = 5)
= 2k + 3k = 5k
= \(5\left( \frac { 1 }{ 6 } \right) \) 
= \(\frac { 5 }{ 6 } \)

Given

The random variable X take the values -1, 0, 1, 2, 3
For a discrete random variable X, we have
f(x) = p(X = x)
∴ f(-1) = p(X= -1) = F(-1) -F(0)
= 0.15-0 = 0.15
f(0) = p(X = 0) = F(0)-F(-1)
= 0.35-0.15 = 0.20
f(1) = p(X = 1) = F(1)-F(0)
= 0.60-0.35 = 0.25
f(2) = p(X=2) = F(2)-F(1)
= 0.85-0.60 = 0.25
f(3) = p(X = 3) = F = (3)-F(2)
= 1-0.85 = 0.15
(i) ஃThe probability mass function is

x	-1	0	1	2	3
f(x)	0.15	0.20	0.25	0.25	0.15

(ii) p(X<1)
= p(X = -1) + p(X = 0)
= 0.15 + 0.20 = 0.35
(iii) p(X ≥ 2)
= p(X = 2) + p(X = 3)
= 0.25 + 0.15
= 0.40

Let X be the thrown random variable denotes the total in two the thrown a die.
Sample space S

I/II	1	3	3	5	5	5
1	2	4	4	6	6	6
3	4	6	6	8	8	8
3	4	6	6	8	8	8
5	6	8	8	10	10	10
5	6	8	8	10	10	10
5	6	8	8	10	10	10

n (S) = 36
X = {2, 4, 6, 8, 10}

Values of the random variable	2	4	6	8	10	Total
No. of elements in inverse images	1	4	10	12	9	36

\(p(x=2)=\cfrac { 1 }{ 36 } \)
\(p(x=4)=\cfrac { 4 }{ 36 } \)
\(p(x=6)=\cfrac { 10 }{ 36 } \)
\(p(x=8)=\cfrac { 12 }{ 36 } \)
\(p(x=10)=\cfrac { 9 }{ 36 } \)
(i) Probability mass function is

x	2	4	6	8	10
f(x)	\(\\ \cfrac { 1 }{ 36 } \)	\(\cfrac { 4 }{ 36 } \)	\(\cfrac { 10 }{ 36 } \)	\(\cfrac { 12 }{ 36 } \)	\(\cfrac { 9 }{ 36 } \)

(ii) Cumulative distribution function .
F(x) = p(X ≤ x) = \(\sum_{x_i ≤ x }\)(X = x_i)
P(X<2) = 0 for \(\infty\) < x < 2
\(F(2)=\frac { 1 }{ 36 } \)
\(F(4)=\frac { 1 }{ 36 } +\frac { 4 }{ 36 } =\frac { 5 }{ 36 } \)
\(F(6)=\frac { 1 }{ 36 } +\frac { 4 }{ 36 } +\frac { 10 }{ 36 } =\frac { 15 }{ 36 } \)
\(F(8)=\frac { 1 }{ 36 } +\frac { 4 }{ 36 } +\frac { 10 }{ 36 } +\frac { 12 }{ 36 } =\frac { 27 }{ 36 }\)
\(F(10)=\frac { 27 }{ 36 } +\frac { 9 }{ 36 } =\frac { 36 }{ 36 } =1\)
∵ The cumulative distribution function n
\(F(x)=\left\{\begin{array}{lll} 0 & \text { for } & x<2 \\ \frac{1}{36} & \text { for } & x \leq 2 \\ \frac{5}{36} & \text { for } & x \leq 6 \\ \frac{15}{36} & \text { for } & x \leq 8 \\ 1 & \text { for } & x \leq 10 \end{array}\right.\)
(iii) p(4≤ X < 10) = p(x = 4) + p(x = 6) + p(x = 8)
= \(\frac { 4 }{ 36 } +\frac { 10 }{ 36 } +\frac { 12 }{ 36 } +\frac { 26 }{ 36 } =\frac { 13 }{ 18 } \)  
(iv) p(x ≥ 6) = p(x = 6) + p(x = 8) + p(x = 10)
= \(\frac { 10 }{ 36 } +\frac { 12 }{ 36 } +\frac { 9 }{ 36 } =\frac { 31 }{ 36 } \)
Sample space = {4 childrens}

Given \(p=\frac { 3 }{ 4 } \) 
n = 5
P(X = x) = nC_xp^x (1-p)^n-x
\(P(X=3)={ 5C }_{ 3 }\left( \frac { 3 }{ 4 } \right) ^{ 3 }\left( 1-\frac { 3 }{ 4 } \right) ^{ 2 }\) 
\(P(X=3)={ 5C }_{ 2 }\left( \frac { 3 }{ 4 } \right) ^{ 3 }\left( \frac { 1 }{ 4 } \right) ^{ 2 }\) [∵nCr = nC_n-r]

= \(\frac { 135 }{ 512 } \)

Let p be the probability of getting heads
q = 1-p
 \(p=\frac { 1 }{ 2 } \)
\(Mean=np=100\times \frac { 1 }{ 2 } =50\) 
\(Variance=npq=100\times \frac { 1 }{ 2 } \times \frac { 1 }{ 2 } =25\) 
(ii)Let p be the probability of getting 4 when a die is thrown
n = 240 
\(p=\frac { 1 }{ 6 } \) [ஃ 4 appears only one]
\(\therefore Mean=np =240\times \frac { 1 }{ 6 } =40\)
\(Variance=npq= 40\times \frac { 5 }{ 6 } \)  
\(Variance=\frac { 100 }{ 3 } \)

 

\(\mathrm{n}=9, \mathrm{p}=\frac{1}{2}, \mathrm{k}=7 \)
\( \mathrm{P}(X=x)={ }^{n} C_{x} p^{x} q^{n-x}, x=0,1,2, \ldots, n \)
\(p =\frac{1}{2} \)
\(q =1-p=\frac{1}{2} \)
\(P(X=7) ={ }^{9} C_{7}\left(\frac{1}{2}\right)^{7}\left(\frac{1}{2}\right)^{2} \)
\( =\frac{9 \times 8}{2} \times \frac{1}{2^{9}} \)
\( =36 \times \frac{1}{512}=\frac{9}{128} \)

Given n = 6, \(p=\frac { 1 }{ 3 } \), k = 3
\(P(X=k)=\left( \begin{matrix} n \\ k \end{matrix} \right) { p }^{ k }\left( 1-p \right) ^{ n-k },\) 
n = 0,1,2, ... n
\(\therefore P(X=k)=\left( \begin{matrix} n \\ k \end{matrix} \right) { p }^{ k }(1-p)^{ n-k }\) 
n = 0,1,2, ... n
\(P(X=3)=\left( \begin{matrix} 6 \\ 3 \end{matrix} \right) \left( \cfrac { 1 }{ 3 } \right) ^{ 3 }\left( 1-p \right) ^{ 6-3 }\) 
= \(\left( \begin{matrix} 6 \\ 3 \end{matrix} \right) \left( \cfrac { 1 }{ 3 } \right) ^{ 3 }\left( \cfrac { 2 }{ 3 } \right) ^{ 2 }\) 

\(P(X=3)=\frac { 160 }{ 729 } \) 

Given
\(f(x)=\cfrac { 4-x }{ 6 } \) 
\(f(x)=\cfrac { 4-1 }{ 6 } =\cfrac { 3 }{ 6 } =\cfrac { 1 }{ 2 } \) 
\(f(2)=\cfrac { 4-2 }{ 6 } =\cfrac { 2 }{ 6 } =\cfrac { 1 }{ 3 } \) 
\(f(3)=\cfrac { 4-3 }{ 6 } =\cfrac { 1 }{ 6 } \) 
ஃ The probability mass function is

x	1	2	3
f(x)	\(\cfrac { 1 }{ 2 } \)	\(\cfrac { 1 }{ 3 } \)	\(\cfrac { 1 }{ 6 } \)

Mean \(E(X)=\Sigma xf(x)\) 
= \(1\left( \cfrac { 1 }{ 2 } \right) +2\left( \cfrac { 1 }{ 3 } \right) +3\left( \cfrac { 1 }{ 6 } \right) \) 
= \(\cfrac { 1 }{ 2 } +\cfrac { 2 }{ 3 } +\cfrac { 3 }{ 6 } =\cfrac { 3+4+6 }{ 6 } \) 
= \(\cfrac { 10 }{ 6 } =\cfrac { 5 }{ 3 } =1.67\) 
\(E({ x }^{ 2 })=\Sigma { x }^{ 2 }f(x)\) 
= \({ 1 }^{ 2 }\left( \cfrac { 1 }{ 2 } \right) +{ 2 }^{ 2 }\left( \cfrac { 1 }{ 3 } \right) +3^{ 2 }\left( \cfrac { 1 }{ 6 } \right) \) 
= \(\cfrac { 1 }{ 2 } +\cfrac { 4 }{ 9 } +\cfrac { 9 }{ 6 } =\cfrac { 3+8+9 }{ 6 } \) 
= \(\cfrac { 20 }{ 6 } =\cfrac { 10 }{ 3 } =3.33\)
var(X) E(X²) - [E(X)]²
= \(\cfrac { 10 }{ 3 } -\left( \cfrac { 5 }{ 3 } \right) ^{ 2 }=\cfrac { 10 }{ 3 } -\cfrac { 25 }{ 9 } \) 
= \(\cfrac { 30-25 }{ 9 } =\cfrac { 5 }{ 9 } =0.54\)