\((i)\ \mathrm{P}(0.5<x<0.75) \)
\( =\mathrm{F}(0.75)-\mathrm{F}(0.5)=(0.75)^{2}-(0.5)^{2} \)
\( =0.3125 \)
\((ii) \ \mathrm{P}(\mathrm{X} \leq 0.5) \)
\( =\mathrm{P}(-\infty<x \leq 0.5)=\mathrm{F}(0.5)-\mathrm{F}(-\infty) \)
\( =(0.5)^{2}-0=0.25 \)
\((iii) \ \mathrm{P}(\mathrm{X}>0.75) \)
\( =\mathrm{P}(0.75 \leq x<\infty)=\mathrm{F}(\infty)-\mathrm{F}(0.75) \)
\( =1-(0.75)^{2}=0.4375 \)

S = {H, TH, TTH, TTTH, TTTTH}
 X take values 1, 2, 3, 4, 5
P(X = 1)= P(H) \( =\frac{1}{2} \)
 P(X = 2) = P(TH) \( =\frac{1}{2} \times \frac{1}{2}=\frac{1}{4} \)
 P(X = 3) = P(TTH) \( =\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}=\frac{1}{8} \)
 \( P(X=4) =P(T T T H)=\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}=\frac{1}{16} \)
\(P(X=5) =\mathrm{P}(\mathrm{TTTTH})+\mathrm{P}(\mathrm{TTTTT}) \)
\( =\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}+\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \)
\( =\frac{1}{32} \times \frac{1}{32}=\frac{1}{16} \)
The probability distribution of X is

Mean  \( =\Sigma_{i} \mathrm{P}_{i}=1\left(\frac{1}{2}\right)+2\left(\frac{1}{4}\right)+3\left(\frac{1}{8}\right)+ 4\left(\frac{1}{16}\right)+5\left(\frac{1}{16}\right) \)
\( =\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}+\frac{5}{16}=\frac{8+8+6+4+5}{16}=\frac{31}{16} \)
\( \therefore E(X)=\frac{31}{16} \)

(i) Since the random variable X is the probability distribution function, \(\Sigma p_{i}=1\)
.:.0+k+ 2k+2k + 3k+k² + 2k² + 7k² + k = 1
\( \Rightarrow 10 k^{2}+9 k-1=0 \Rightarrow 10(k-1)(k+1)=0 \)
\( \Rightarrow k=\frac{1}{10} \)
(ii) P(X < 6) = P(X = 0) + P(X-1) + P(X = 2)
+ P(X = 3) +P(X = 4) + P(X - 5)
= 0 + k + 2k + 2k + 3k+k²
\( =k^{2}+8 k=\left(\frac{1}{10}\right)^{2}+8\left(\frac{1}{10}\right) \)
\( =\frac{1}{100}+\frac{8}{10}=\frac{1+80}{100}=\frac{81}{100} \)
(iii) \( \mathrm{P}(\mathrm{X} \geq 6) =\mathrm{P}(\mathrm{X}=6)+\mathrm{P}(\mathrm{X}=7) \)
\( =2 k^{2}+7 k^{2}+k=9 k^{2}+k \)
\(=9\left(\frac{1}{10}\right)^{2}+\frac{1}{10}=\frac{9}{100}+\frac{1}{10}=\frac{19}{100} \)
(iv) \( \mathrm{P}(0<\mathrm{X}<5) =\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2) +\mathrm{P}(\mathrm{X}=3)+\mathrm{P}(\mathrm{X}=4) \)
\(= k+2 k+2 k+3 k=8 k=\frac{8}{10} \)
\( =\frac{4}{5} \)

Let X denote the number of kings.
The random variable X take values 0, 1, 2.
Let P(s) denote the probability of getting a king
 \( P(s)=\frac{4}{52} \)
\( P(F)=1-\frac{4}{52}=\frac{48}{52}\)
P(X = 0) = P(not getting a king in two draws)
= P(FF)
= P(F). P(F)
\(=\frac{48}{52} \times \frac{48}{52}=\frac{144}{169}\)
P(X = 1) = P(getting one king in two draws)
P(FS) + P(SF)
\( =\frac{48}{52} \times \frac{4}{52}+\frac{4}{52} \times \frac{48}{52} \)
\( =2 \times \frac{4}{52} \times \frac{48}{52}=\frac{24}{169} \)
 P(X = 2) = P(getting 2 kings in 2 draws)
\(=\mathrm{P}(\mathrm{SS})=\frac{4}{52} \times \frac{4}{52}=\frac{1}{169}\)
Thus, the probability distribution of X is 

Since \( f(x) \text { is a p.d.f. } \int_{-\infty}^{\infty} f(x) d x=1 \)
\( \int_{0}^{5} \mathrm{~A} x d x+\int_{5}^{10} \mathrm{~A}(10-x) d x=1 \)
\( \Rightarrow \mathrm{A}\left(\frac{x^{2}}{2}\right)_{0}^{5}+\mathrm{A}\left[\left(\frac{10-x}{-2}\right)\right]_{5}^{210}=1 \)
\( \Rightarrow \ \frac{\mathrm{A}}{2}(25)-\frac{\mathrm{A}}{2}(0-25)=1 \)
\( \Rightarrow \ \frac{25}{\mathrm{~A}}+\frac{25}{\mathrm{~A}}=1 \)
\(\Rightarrow \ \frac{50 \mathrm{~A}}{2}=1 \)
Mean \( =\mathrm{E}(x)=\int_{0}^{5} x \cdot f(x) d x+\int_{5}^{5} x \cdot f(x) d x \)
\(\Rightarrow \ \frac{x}{25} d x+\int_{0}^{5} x \times \frac{1}{25}(10-x) d x\)
\( =\frac{1}{25} \int_{0}^{5} x^{2} d x+\frac{1}{25} \int_{3}^{10}\left(10 x-x^{2}\right) d x \)
\( =\frac{1}{25}\left[\frac{x^{3}}{3}\right]_{0}^{5}+\frac{1}{25}\left[\frac{10 x^{2}}{2}-\frac{x^{3}}{3}\right]_{3}^{10} \)
\( =\frac{1}{75}(125)+\frac{1}{25}\left[\left(5(100)-\frac{1000}{3}\right)-\left(125-\frac{125}{3}\right)\right] \)
\( =\frac{5}{3}+\frac{1}{25}\left(\frac{500}{3}-\frac{250}{3}\right)=\frac{5}{3}+\frac{1}{25} \times \frac{250}{3} \)
\( =\frac{5}{3} \times \frac{10}{3}=\frac{15}{3} \)

(i) Since f(x) is a probability mass function
\( \sum_{x} f(x)=1 \)
a + 3a + 5a +7a = 1 
16a = 1  
\(a =\frac{1}{16}\)
(ii) \(\mathbf{P}(\mathbf{X} \leq 1)\)
\( P(X \leq 1) =P(X=0)+P(X=1) \)
 = a + 3a
= 4a
\( =\frac{4}{16}=\frac{1}{4} \)

Let X be random variable of dcnoting numbcr of blackchips in 3 draws
X = {0, 1, 2, 3}
Sample space consist of ⁵C₃ , elements = 56

Probability mass function

Let X be random variable of denoting numaer of butter cakes
X = {0, 1, 2} 
Sample space consist of \({ }^{10} \mathrm{C}_{2}=45\)

Probability mass function

(i) Since f is a p.d.f. \(\int_{0}^{1} f(x) d x=1\)
\( k \int_{0}^{1}(1-x)^{3} d x=1 \)
\( k\left[\frac{(1-x)^{4}}{-4}\right]_{0}^{1}=1\)
\( -\frac{k}{4}\left[(1-x)^{4}\right]_{0}^{1} =1 \)
\(-\frac{k}{4}(0-1) =1 \)
\(\frac{k}{4} =1\)
(ii) \(P\left(X<\frac{2}{3}\right)\)
\( \mathrm{P}\left(\mathrm{X}<\frac{2}{3}\right) =\int_{0}^{3 / 2} k(1-x)^{3} d x \)
\( =4\left[\frac{(1-x)^{4}}{-4}\right]_{0}^{1 / 3} \)
\( =-\left[(1-x)^{4}\right]_{0}^{2 / 3} \)
\( =-\left(\left(1-\frac{2}{3}\right)^{4}-1^{4}\right) \)
\( =\left(\frac{1}{81}-1\right)^{3} \)
\( =\frac{80}{81} \)

\( \mathrm{F}^{\prime}(\mathrm{X})=\mathrm{f}(\mathrm{x}) =\left\{\begin{array}{cl} 0, & x<0 \\ 2 x, & 0 \leq x \leq 1 \\ 0, & x>1 \end{array}\right.\\ \)
\(\mathrm{f}(\mathrm{x}) =\left\{\begin{array}{cl} 2 x, & 0 \leq x \leq 1 \\ 0, & \text { elsewhere } \end{array}\right. \)
(i) P(X < 0.3)
\( \mathrm{P}(\mathrm{X} \leq 0.3) =\mathrm{F}(0.3)=(0.3)^{2} =0.09 \)
(ii) P(X > 0.9)
\( =1-P(X \leq 0.9) \)
 = 1-(0.9)^{2} 
 = 1-0.81 
 = 0.19
(iii) P(0.1 < X < 0.5)
\( \mathrm{P}(0.1<\mathrm{X}<0.5) =\mathrm{F}(0.5)-\mathrm{F}(0.1) \)
\( = (0.5)^{2}-(0.1)^{2} \)
 = 0.25-0.01 
 = 0.24

Let X be random variable of denoting number of red balls
X = {0, 1, 2, 3}
n(S) = 7 \(\times\)7 \(\times\)7 = 343

Probability mass function

Mean:
\( \mu=\mathrm{E}(\mathrm{X})= 0 \times \frac{64}{343}+1 \times \frac{144}{343} +2 \times \frac{108}{343}+3 \times \frac{27}{343} \\ \)
\(= \frac{9}{7} \)
Variance:
\( \mathrm{E}\left(\mathrm{X}^{2}\right)= 0 \times \frac{64}{343}+1^{2} \times \frac{144}{343}+2^{2} \times \frac{108}{343} +3^{2} \times \frac{27}{343}=\frac{117}{49} \)
Variance \( =E\left(X^{2}\right)-[E(X)]^{2} =\frac{117}{49}-\left(\frac{9}{7}\right)^{2} \)
\( =\frac{36}{49}\)

Since f(x) is a p.d.f \(\int_{-\infty}^{\infty} f(x) d x=1\)
\( \int_{-\infty}^{\infty} k e^{-|x|} d x=1 \)
\(k \int_{-\infty}^{\infty} e^{-|x|} d x=1 \)
\( 2 k \int_{0}^{\infty} e^{-|x|} d x=1 \)
[\(\because e^{-|x|}\) is an even function]
\( 2 k \int_{0}^{\infty} e^{-x} d x =1 \)
\(2 k\left[\frac{e^{-x}}{-1}\right]_{0}^{\infty} =1 \)
\(-2 \mathrm{k}\left(\mathrm{e}^{-\infty}-\mathrm{e}^{0}\right) =1 \mathrm{k} =\frac{1}{2} \)
Mean :
\( \mathrm{E}(\mathrm{X}) =\int_{-\infty}^{\infty} x f(x) d x \)
\(=\int_{-\infty}^{\infty} x k e^{-|x|} d x\)
\(=\frac{1}{2} \int_{-\infty}^{\infty} x e^{-|x|} d x\)
= 0
[\(\because x e^{-\mid x}\) is an odd function]
Variance :
\( E\left(X^{2}\right) =\int_{-\infty}^{\infty} x^{2} f(x) d x \)
\(=\int_{-\infty}^{\infty} x^{2} k e^{-|x|} d x \)
\(=k \int_{-\infty}^{\infty} x^{2} e^{-|x|} d x \)
\(=\frac{1}{2} \times 2 \int_{0}^{\infty} x^{2} e^{-x} d x\)
[\(\because x e^{-\mid x}\) is an even function]
\( =\int_{0}^{\infty} x^{2} e^{-x} d x \)
\(\because \int_{0}^{\infty} x^{n} e^{-\alpha x} d x=\frac{n !}{\alpha^{n+1}} \)
\(\mathrm{~V}(\mathbf{X})=E\left(X^{2}\right)-[E(X)]^{2}\)
= 2 - 0
= 2

n(S) = 36
A = {(4, 6), (5, 5), (6,4)}
n(A) = 3
\( p=\frac{3}{36}=\frac{1}{12} \)
\( q=\frac{11}{12} \)
n = 10
\(\mathrm{P}(\mathbf{X}=\mathbf{x})={ }^{\mathrm{n}} \mathrm{C}_{x} \mathrm{p}^{\mathrm{x}} \mathrm{q}^{\mathrm{n}-\mathrm{x}}={ }^{10} \mathrm{C}_{x}\left(\frac{1}{12}\right)^{x}\left(\frac{11}{12}\right)^{10-x}\)
x = 0, 1, 2, .... ,10
(i) P(X = 10)
\( \mathrm{P}(\mathrm{X}=10) ={ }^{10} \mathrm{C}_{10}\left(\frac{1}{12}\right)^{10} \times\left(\frac{11}{12}\right)^{0} \)
\( =1 \times \frac{1}{12^{10}} \times 1 \)
\( =\frac{1}{12^{10}}\)
(ii) P(X = 0)
\( \mathrm{P}(\mathrm{X}=0) ={ }^{10} \mathrm{C}_{0}\left(\frac{1}{12}\right)^{0} \times\left(\frac{11}{12}\right)^{10} \)
\( =1 \times 1 \times\left(\frac{11}{12}\right)^{10} \)
\( =\left(\frac{11}{12}\right)^{10} \)
(iii) P(X > 8)
\( \mathrm{P}(\mathrm{X}>8)= \mathrm{P}(\mathrm{X}=9)+\mathrm{P}(\mathrm{X}=10) \)
\(= { }^{10} \mathrm{C}_{9}\left(\frac{1}{12}\right)^{9} \times\left(\frac{11}{12}\right)^{1} +{ }^{10} \mathrm{C}_{10}\left(\frac{1}{12}\right)^{10} \times\left(\frac{11}{12}\right)^{0} \)
\(= 10 \times \frac{11}{12^{10}}+\frac{1}{12^{10}} \)
\(= \frac{1}{12^{10}}(110+1) \)
\(= \frac{111}{12^{10}} \)