Let "getting a queen'" be termed as success. X is a random variable denoting the number of queens. X takes values 0, 1, 2
P(X = 0) = P(FF)
\(=\frac{48}{52} \times \frac{47}{51}=\frac{188}{221}\)
P(X = 1) = 2P(SF)
\(=\frac{2 \times 4}{52} \times \frac{48}{51}=\frac{32}{221}\)
P(X = 2) = P(SS)
\(=\frac{4}{52} \times \frac{3}{51}=\frac{1}{221}\)
The probability distribution is

Let "getting a six" be termed as success.
Let X be the random variable denoting the number of sixes.
X takes values 0, 1, 2, 3
\( P^{\prime}(S) =\frac{1}{6}, P(\mid c)=\frac{5}{6} \)
\(P(X=0) =P(|F| i) \)
\(=\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \)
\(=\frac{125}{216} \)
\(P(X=1) =3 P(S F F) \)
\(=3 \times \frac{1}{6} \times \frac{5}{6} \times \frac{5}{6} \)
\(=\frac{75}{216} \)
\(P(X=2) =3 P(S S F) \)
\(=3 \times \frac{1}{6} \times \frac{1}{6} \times \frac{5}{6} \)
\(=\frac{15}{216}\)
\( \mathrm{P}(\mathrm{X}=3) =\mathrm{P}(\mathrm{SSS}) \)
\( =\frac{1}{6} \times \frac{1}{6} \times \frac{1}{6}=\frac{1}{216}\)
The probability distribution is

Let X be the random variable denoting the years the dog will live
\(\text {(i) } \mathrm{P}(\mathrm{X}>10) =1-\mathrm{P}(\mathrm{X} \leq 10) \)
\(=1-\left(1-\frac{25}{x^{2}}\right) \quad x=10 \\ =\frac{25}{x^{2}}=\frac{25}{100}=\frac{1}{4}\)
\((ii) \ \mathrm{P}(\mathrm{X}<8) =\mathrm{F}(8)=1-\frac{25}{8^{2}} \)
\(=1-\frac{25}{64}=\frac{39}{64} \)
\((iii)\ \mathrm{P}(12<\mathrm{X}<15) =\mathrm{F}(15)-\mathrm{F}(12) \)
\(=1-\frac{25}{15^{2}}-\left(1-\frac{25}{12^{2}}\right) \)
\(=\frac{25}{144}-\frac{25}{225} \)
\(=25\left(\frac{25-16}{3600}\right) \)
\(=25\left(\frac{9}{3600}\right)=\frac{1}{16}\)

f is a p.d.f.
\( \int_{-x}^{x} f(x) d x =1 \)
\(\int_{1}^{\prime \prime} \frac{A}{x} d x =1 \)
\(A[\log x]_{1}^{\prime} =1 \)
\(A\left[\log e^{3}-\log 1\right] =1 \)
\(\mathrm{A}[3 \log \mathrm{e}] =1 \)
\(\mathrm{A} =\frac{1}{3} \)
\(\mathrm{F}(x>\mathrm{e}) =\frac{1}{3} \int_{c}^{c^{\prime}} \frac{1}{x} d x=\frac{1}{3}[\log x]^{\prime} \)
\(=\frac{1}{3}\left[\log e^{3}-\log e\right] \)
\(=\frac{1}{3}(3 \log \mathrm{e}-1)=\frac{1}{3}(2)=\frac{2}{3} \)

Let x be the random varialble of gelting number of 3's.
x can take values 0, 1, 2.
getting is 3 is termed as success.
\( P(X=0) =P(F F) \)
\( =\frac{5}{6} \times \frac{5}{6} \)
\( =\frac{25}{36} \)
\(P(X=1) =2 P(S F) \)
\( =2 \times \frac{1}{6} \times \frac{5}{6} \)
\( =\frac{10}{36} \)
\(\mathrm{P}(X=2) =P(S S) \)
\( =\frac{1}{6} \times \frac{1}{6} \)
\( =\frac{1}{36} \)
Probability mass function is

Cumulative distribution function
 \( \Gamma(x) =\sum_{x_{1}=-0} P\left(X=x_{i}\right) \)
\(\Gamma(()) =P(X=0)=\frac{25}{36} \)
\(\Gamma(1) =P(X=0)+P(X=1) \)
\(=\frac{25}{36}+\frac{10}{36}=\frac{35}{36} \)
\(\Gamma(2) =P(X=0)+P(X=1)+P(X=2) \)
\(=\frac{25}{36}+\frac{10}{36}+\frac{1}{36}=\frac{36}{36}=1\)

Let X denote the number of white balls drawn from the urn.
Since there are 4 white balls, X can take values 0,1,2,3,4.
P(X = 0)=p(gettmg no white balls)\(=\frac { { 6C }_{ 4 } }{ 10{ C }_{ 4 } } \)
\(\frac{1}{14}\)
P(X = 1) = P(getting one white ball and 3 red balls) \(\frac { { 4C }_{ 4 }\times { 6C }_{ 3 } }{ { 10 }C_{ 4 } } =\frac { 8 }{ 21 } \)
P(X = 2) = P(getting two white balls and 2 red balls) \(\frac { { 4C }_{ 2 }\times { 6C }_{ 2 } }{ { 10 }C_{ 4 } } =\frac { 8 }{ 21 } \)
P(X = 3) = P(getting 3 white balls and 1 red ball) \(\frac { { 4C }_{ 3 }\times { 6C }l }{ { 10 }C_{ 4 } } =\frac { 4 }{ 35 } \)
P(X = 4) = P(getting 4 white balls) =\(\frac { 4{ C }_{ 4 } }{ 10{ C }_{ 4 } } \)
=\(\frac{1}{210}\)
Thus the probability distribution of X is

Let X denote the number of aces in a pack of 52 playing cards.
Since there are 4 aces in a pack, and we are drawing 3 ace cards, X can take values 0, 1, 2, 3.
P(X = 0) = P(getting no ace card) \(\frac { 48{ C }_{ 3 } }{ 52{ C }_{ 3 } } \)
\(=\frac { 4324 }{ 5525 } \)
P(X=1) = P(getting one ace card and 2 other cards)
\(\\ =\frac { 4{ C }_{ 1 }\times 48{ C }_{ 2 } }{ 52{ C }_{ 3 } } =\frac { 1128 }{ 5525 } \)
P(X=2) = P(getting 2ace card and one other card)
\(=\frac { { 4C }_{ 2 }\times { 48C }_{ 1 } }{ 52{ C }_{ 3 } } =\frac { 72 }{ 5525 }\)
P(X=3) = P(getting 3 ace card) = \(\frac { { 4C }_{ 3 } }{ { 52C }_{ 3 } } \)
\(=\frac { 1 }{ 5525 } \)
Hence, the probability distribution function is

We have P(X=r)∝α^r
⇒P(X = r) = λα^r,r = 0,1,2,....
Since sum of all the probabilities in a probability distribution is 1.
P(X = 0)+P(X = 1)+P(P(X = 2)+... = 1
⇒ λα0+λα1+λα2+...= 1
⇒λ(1+α+α2+.....) = 1
⇒\(\lambda(\frac{1}{1-\alpha})\) = 1
⇒ λ = 1 - α
∴ P(X = r) = (1-α)α^r, r = 0,1,2,...
Hence P(X=0) = (1-α)α⁰ = (1-α)(1) = 1-α.

The probability distribution of X is

Σp_i = 1 ⇒ 0.1 + k + 2k + 2k + k = 1
⇒ 0.1 + 6k = 1 ⇒ 6k = 1-0.1 = 0.9
⇒ k = \(\frac{0.9}{6}\) = 0.15
And the probability that you study atleast 2 hours is P(X≥2)=P(X=2)+P(X=3)+P(X=4)
= 2k + 2k + k = 5k
= 5(0.15) = 0.75

Since the random variable X is the probability distribution function, Σp_i = 1
∴ a + 2a + 3a + 4a + 5a + 6a = 1
21a = 1 ⇒ a = \(\frac{1}{21}\)
Now, F(4) = P(X ≤ 4)
= P(X = 0) + P(X = 1) + P(X = 2)P(X = 3) + P(X = 4)
= a + 2a + 3a + 4a + 5a = 15a
= 15\((\frac{1}{21})=\frac{5}{7}\)
∴ F(4) = \(\frac{5}{7}\)

\(E(X)=\int _{ 0 }^{ 3 }{ x.f(x)dx=\int _{ 0 }^{ 3 }{ x\left( \frac { dx }{ 9 } \right) dx } } \)
\(=\frac { 2 }{ 9 } \int _{ 0 }^{ 3 }{ { x }^{ 2 }dx=\frac { 2 }{ 9 } .{ \left[ \frac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 3 } } \)
\(=\frac { 2 }{ 27 } ({ 3 }^{ 3 }-0)=\frac { 2 }{ 27 } (27)=2\)
\(\therefore E(3X+8)=3.E(X)+8\quad [\because E(8)=8]\)
= 3(2) + 8 = 6 + 8
E(3X + 8) = 14

When 2 coins are tossed, sample space S={HH, HT, TH, TT} ⇒ n(s) = 4
∴P(X = 5) = p(getting 2 heads) =\(\frac{1}{4}\)
P(X = 2) = p(getting 1 head) = \(\frac{2}{4}=\frac{1}{2}\)
P(X = 1) = p(getting no head) = \(\frac{1}{4}\)
Hence the probability distribution function is 

\(\therefore E(x)=\sum { { x }_{ i }{ p }_{ i }=1(\frac { 1 }{ 4 } ) } +2(\frac { 1 }{ 2 } )+5(\frac { 1 }{ 4 } )\)
\(=\frac { 1 }{ 4 } +1+\frac { 5 }{ 4 } =\frac { 1+4+5 }{ 4 } \)
\(=\frac { 10 }{ 4 } =2.50\)
Hence the expected money to win is Rs. 2.50

Given f(x) = 4k(x-1)³,1 ≤ x ≤ 3
Since f(x) is a p.d.f., \(\int _{ 1 }^{ 3 }{ f(x)dx=1 } \)
\(\Rightarrow \int _{ 1 }^{ 3 }{ 4k({ x-1) }^{ 3 }dx=1\Rightarrow { \left[ \frac { 4k(x-1{ ) }^{ 4 } }{ 4 } \right] }_{ 1 }^{ 3 }=1 } \)
\(\Rightarrow k({ 2 }^{ 4 }-{ 0 }^{ 4 })=1\Rightarrow k(16)=1\)
\(\Rightarrow k=\frac { 1 }{ 16 } \)
\(p(-2\le x\le 2)=\int _{ -2 }^{ 2 }{ f(x)dx } \)
\(=\int _{ 1 }^{ 2 }{ 4k({ x-1) }^{ 3 }dx } \)
\(=\frac { 4 }{ 16 } \int _{ 1 }^{ 2 }{ { (x-1) }^{ 3 }dx } \)
\([\therefore k=\frac { 1 }{ 16 } ]\)
\(=\frac { 1 }{ 4 } { \left[ \frac { { (x-1) }^{ 4 } }{ 4 } \right] }_{ 1 }^{ 2 }\)
\(=\frac { 1 }{ 16 } ({ 1 }^{ 4 }-0)=\frac { 1 }{ 16 } \)

\(E(X)=\sum { xp(x)=-1(\frac { 1 }{ 3 } )+0(\frac { 1 }{ 6 } )+1(\frac { 1 }{ 6 } )+2\left( \frac { 1 }{ 3 } \right) } \)
\(=\frac { -1 }{ 3 } +\frac { 1 }{ 6 } +\frac { 2 }{ 3 } =\frac { -2+1+4 }{ 6 } \)
\(=\frac { 3 }{ 6 } =\frac { 1 }{ 2 } \)
\(E({ X }^{ 3 })={ \sum { x } }^{ 2 }p(x)\)
\(=1(\frac { 1 }{ 3 } )+0(\frac { 1 }{ 6 } )+1(\frac { 1 }{ 6 } )+4(\frac { 1 }{ 3 } )\)
\(=\frac { 1 }{ 3 } +\frac { 1 }{ 6 } +\frac { 4 }{ 3 } =\frac { 2+1+8 }{ 6 } =\frac { 11 }{ 6 } \)
\(\therefore E{ (2X+3) }^{ 2 }=E(4{ X }^{ 2 }+12X+9)\)
\(=4\left( \frac { 11 }{ 6 } \right) +12\left( \frac { 1 }{ 2 } \right) +9\)
\(=\frac { 22 }{ 3 } +6+9=\frac { 22 }{ 3 } +15\)
\(=\frac { 22+45 }{ 3 } =\frac { 67 }{ 3 } \)
\(\therefore E(2X+3{ ) }^{ 2 }=\frac { 67 }{ 3 } \)

\(E(X)=\sum { xp(x)=-2(\frac { 1 }{ 4 } ) } +2(\frac { 1 }{ 4 } )+5(\frac { 1 }{ 2 } )\)
\(=\frac { -2 }{ 4 } +\frac { 2 }{ 4 } +\frac { 5 }{ 2 } =\frac { 5 }{ 2 } \)
∴ 4E(X²)-V(2X)=4E(X²)-4.V(X)
=4E(X²)-4[E(X²)-E(X)²]
=4E(X²)-4E(X²)+4[E(X)]²
\(=4{ \left( \frac { 5 }{ 2 } \right) }^{ 2 }[\because E(X)=\frac { 5 }{ 2 } ]\)
\(=4\left( \frac { 25 }{ 4 } \right) =25\)