n = 3
probability of getting an even number
\( =\frac{3}{6}=\frac{1}{2} \)
\( q=1-p=1-\frac{1}{2}=\frac{1}{2}\)
variance = \(\mathrm{npq}=3\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)=\frac{3}{4}\)

n = 20; p = Probability of getting a number greater than \(4=\frac{2}{6}=\frac{1}{3}\)
\( q=1-p=1-\frac{1}{3}=\frac{2}{3} \)
\(\text {Mean }=n p=20\left(\frac{1}{3}\right)=\frac{20}{3} \)
\(\text {Variance npq }=20\left(\frac{1}{3}\right)\left(\frac{2}{3}\right)=\frac{40}{9}\)

p = Probability of getting a six = \(\frac{1}{6}\)
\(q=1-\frac{1}{6}=\frac{5}{6}\)
Obtaining 3^rd six in the sixth throw of the die means that is the first five throws there are 2 sixes and the 3rd six is obtained in the sixth throw.
Required probability = P(Getting 2 sixes in first 5 throws) P (Getting six in 6^th throws)
\(=5 C_{2}\left(\frac{1}{6}\right)^{2}\left(\frac{5}{6}\right)^{3}\left(\frac{1}{6}\right)=\frac{625}{23328}\)

Let X be the random variable denoting the number of successes
\( \mathrm{n} =5, \mathrm{p}=\frac{3}{6}=\frac{1}{2}, \mathrm{q}=\frac{1}{2} \)
\(\mathrm{P}(\mathrm{X} =\mathrm{x})=n C_{x} p^{x} q^{n-x} \)
\(\mathrm{P}(x \leq 3) =1-\mathrm{P}(\mathrm{X}>3) \)
\( =1-[\mathrm{P}(\mathrm{X}=4)+\mathrm{P}(\mathrm{X}=5)] \)
\( =1-\left[5 C_{4}\left(\frac{1}{2}\right)^{5}+5 C_{5}\left(\frac{1}{2}\right)^{5}\right] \)
\( =1-\left[\frac{5}{32}+\frac{1}{32}\right]=1-\frac{6}{32}=\frac{13}{16} \)

Let X be the random variable denoting the number of hurdles knocked down by the players. Then X follows binomial distribution with 
\(n=10, p=1-\frac{5}{6}=\frac{1}{6}, q=\frac{5}{6}\)
\( \mathrm{P}(\mathrm{X}=\mathrm{x}) =\mathrm{nC}_{\mathrm{x}} \mathrm{p}^{\mathrm{x}} \mathrm{q}^{\mathrm{n}-\mathrm{x}} \)
\(=10 C_{x}\left(\frac{1}{6}\right)^{x}\left(\frac{5}{6}\right)^{10-x} \)
\(\mathrm{P}(\mathrm{X}<2) =\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1) \)
\( =\left(\frac{5}{6}\right)^{10}+10 C_{1}\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)^{9} \)
\( =\left(\frac{5}{6}\right)^{9}\left[\frac{5}{6}+\frac{10}{6}\right]=\frac{5^{10}}{2 \times 6^{9}}\)

Let p be the probability of happening of an event.
Given p = 0.002 = \(\frac { 2 }{ 1000 } \)
Also n = 1000
∴ Mean = np = \(1000\times \frac { 2 }{ 1000 } \) = 2
Hence, X follows Poisson distribution with
P(x,λ) = \(\frac { e^{ -\lambda }.{ \lambda }^{ 0 } }{ x! } \)
∴ P( event happens exactly twice)
= P(X = 2)
= \(\frac { e^{ -\lambda }.{ \lambda }^{ 2 } }{ 2! } =\frac { e^{ -2 }.{ (2 }^{ 2 }) }{ 2 } \)
= e^-2(2) = 2(0.1353) = 0.2706
∴ P(X = 2) = 0.2706

Let p be the probability of winning the prize
Given p = \(\\ \frac { 1 }{ 100 } \) and n = 50
∴ Mean = np = \(\frac { 1 }{ 100 } \times 50=\frac { 1 }{ 2 } \)
∴ λ = 0.5
Hence, X follows Poisson distribution with
P(x,λ) = \(\frac { e^{ -\lambda }.{ \lambda }^{ 0 } }{ x! } \)
∴ P(winning the prize at least once) = P(X ≥ 1)
= 1-P(X < 1)
= 1-P(X = 0)
= 1-\(\frac { e^{ -\lambda }.{ \lambda }^{ 0 } }{ x! } \) = 1-e^-0.5
= 1-0.6066
∴ P(X≥1) = 0.3934

The probability function for the normal distribution is
f(x) = \(\frac { 1 }{ \sigma \sqrt { 2\pi } } e^{ -\frac { 1 }{ 2 } \left( \frac { x-\mu }{ \sigma } \right) ^{ 2 } }\), -∞ Given f(x) = \(C{ e }^{ -\left( \frac { x-100 }{ 50 } \right) ^{ 2 } }\)
= \(C{ e }^{ -\frac { 1 }{ 2 } \left( \frac { x-100 }{ 25 } \right) ^{ 2 } }\)
= \(C{ e }^{ -\frac { 1 }{ 2 } \left( \frac { x-100 }{ 5 } \right) ^{ 2 } }\) ...(2)
Comparing (1) and (2), μ =100, σ = 5 and
C = \(\frac { 1 }{ \sigma \sqrt { 2\pi } } =\frac { 1 }{ 5\sqrt { 2\pi } } \)
∴ C = \(\frac { 1 }{ 5\sqrt { 2\pi } } \).

Given n = 10
Probability of selecting a defective pen = p
= \(\frac { 10 }{ 50 } =\frac { 1 }{ 5 } \)
q = 1-p = \(1-\frac { 1 }{ 5 } =\frac { 4 }{ 5 } \)
∴ P(X = x) = ⁿC_x p^xq^n-x
P (at least one pen is defective)
= P(X≥1) = 1-P(X<1)
= 1-P(X = 0)
= 1-10C₀ \(\left( \frac { 1 }{ 5 } \right) ^{ 0 }\left( \frac { 4 }{ 5 } \right) ^{ 10 }\)
= 1-\(\frac { { 4 }^{ 10 } }{ { 5 }^{ 10 } } \)

Given mean μ = 60 and S.D. σ = 5

To find P(X > 60)
When X = 60, Z =\(\frac { X-\mu }{ \sigma } =\frac { 60-60 }{ 5 } \) = 0
∴ P(X > 60) = P(Z > 0) = P (0 < Z < ∞)
= 0.5
∴ 50% of students scored more than 60 marks

Given n = 10, P(H) = \(\frac { 1 }{ 2 } \) ⇒ p =\(\frac { 1 }{ 2 } \)
q=1-p = \(1-\frac { 1 }{ 2 } =\frac { 1 }{ 2 } \)
P(X = x) = nC_x p^xq^n-x
∴ P(X = 5) = 10C₅ p⁵q⁵
= \(\frac { 10\times 9\times 8\times 7\times 6 }{ 5\times 4\times 3\times 2\times 1 } \left( \frac { 1 }{ 2 } \right) ^{ 5 }\left( \frac { 1 }{ 2 } \right) ^{ 10 }\)
= \(\frac { 6\times 7\times 6 }{ 2^{ 10 } } \)
= \(\frac { 2\times 3\times 7\times 2\times 3 }{ 2^{ 10 } } =\frac { 63 }{ { 2 }^{ 8 } } \)
= \(\frac { 63 }{ 256 } \).

Since X is a binomial variate X ~ B (5, p)
n = 5 and P(X = x) = nC_x p^x q^n-x
Given P[X = 2] = P[X = 3]
⇒ 5C₂ p²q³ = 5C₃ p³q²

⇒ q=p
we know p+q = 1 ⇒ p+p =1
⇒ 2p = 1⇒ p =\(\frac { 1 }{ 2 } =\frac { 1 }{ 2 } \).

Given mean 20 ⇒ np = 20
S.D = 4 ⇒ \(\sqrt { npq } \) = 4
∴ \(\frac { npq }{ np } =\frac { 16 }{ 20 } \Rightarrow q=\frac { 4 }{ 5 } \)
P = 1-q =\(1-\frac { 4 }{ 5 } =\frac { 1 }{ 5 } \)
Substitutingp and q in npq = 16, we get
\(n\times \frac { 1 }{ 5 } \times \frac { 4 }{ 5 } \) =16
n = \(\frac { 16\times 5\times 5 }{ 4 } \)  = 100
∴ Number of events = 100

Given n = 9 and mean = 6 ⇒ np = 6
9p = 6 ⇒ \(\frac { 6 }{ 9 } =\frac { 2 }{ 3 } \)
∴ q=1-p = \(1-\frac { 2 }{ 3 } =\frac { 1 }{ 3 } \)
Variance = npq = \(9\times \frac { 2 }{ 3 } \times \frac { 1 }{ 3 } \) = 2

Let λ be the parameter
Given 3. P(X = 2) = P(X = 4)
⇒ 3. \(\frac { { e }^{ -\lambda }.{ \lambda }^{ 2 } }{ 2! } =\frac { e^{ -\lambda }.{ \lambda }^{ 4 } }{ 4! } \)
\(\frac { 3{ \lambda }^{ 2 } }{ 2 } =\frac { { \lambda }^{ 4 } }{ 4\times 3\times 2 } \)
⇒ 36λ² = λ⁴
λ⁴-36λ² = 0
⇒ λ²(λ²-36) = 0
⇒ λ² = 0 or λ² = 36
⇒ λ = 6 since λ > 0