Let p be the probability of driving foreign cars
∴ p = \(\frac { 18 }{ 28 } \) = 0.64 and
q = 1 - p = 1 - 0.64 = 0.36
n = 5
∴ P (atleast 3 of them drive foreign cars)
= P(X ≥3) = 1-P(X<3)
= 1-[P(X = 0) + P(X = 1) + P (X = 2)]
= 1-[⁵C₀ (0.64)⁰ (0.36)⁵ + ⁵C₁ (0.64)¹ + ⁵C₂ (0.64)² (0.36)³ (0.36)⁴ ]
= 1-[(0.36)⁵ + 5(0.64) (0.36)⁴ + \(\frac { 5\times 4 }{ 2\times 1 } \) (0.64)² (0.36)³ ]
= 1-[0.0060 + 0.0537 + 0.1911]
= 1-(0.2508) = 0.7492

We are given mean μ = 2.25 and standard deviation σ = 0.25.
Probability that weight of fish is less than 2 kg is P(X < 2.0)
When x = 20 \(Z=\frac { X-\mu }{ \sigma } =\frac { 2.0-2.25 }{ 0.25 } =P(Z<-1.0)=P(Z>1.0)\)
= 0.5 – 0.3413 = 0.1587
Therefore 15.87% of fishes weigh less than 2 kg.

Let x devotes the set of twins on a day
P(twin birth) = p = 1/80 = 0.0125 and n = 30
The value of mean λ = np = 30 × 0.0125 = 0.375
Hence, X follows poisson distribution with p(x)\(\frac { { e }^{ -\lambda }{ \lambda }^{ x } }{ x! } \)
The probability is 
P(2 or more) = 1 – [p (x = 0) + p (x = )] \(=1-\left[ \frac { { e }^{ -0.375 }{ (0.375) }^{ 0 } }{ 0! } +\frac { { e }^{ -0.375 }{ (0.375) }^{ 1 } }{ 1! } \right] \)
\(=1-{ e }^{ -0.375 }[1+0.375]\)
\(=1-(0.6873\times 1.375)\)
= 0.055

Probability p of guessing an answer correctly is p = \(\frac{1}{2}\)
⇒ q = \(\frac{1}{2}\)
Probability of guessing correctly x answers in 10 questions
\(P(X=x)=p(x)^{ n }{ C }_{ x }{ q }^{ n-x }=10Cx\left( \frac { 1 }{ 2 } \right) \left( \frac { 1 }{ 2 } \right) \)
The required probability P(X ≥ 6) = P(6) + P(7) + P(8) + P(9) + P(10)
\(={ \left( \frac { 1 }{ 2 } \right) }^{ 10 }\left[ { 10C }_{ 6 }+{ 10C }_{ 7 }+{ 10C }_{ 8 }+{ 10C }_{ 9 }+{ 10C }_{ 10 } \right] \)
\(=\left[ \frac { 1 }{ 1024 } \right] [210+120+45+10+1]\)
\(=\frac { 193 }{ 512 } \)

The parameters of Binomial distribution are n and p
For Binomial distribution Mean = np = 20
Standard deviation =\(\\ \\ \\ \sqrt { npq } =4\)   
∴ npq = 16
⇒ npq/np = 16/20 = 4/5
q = 4/5
⇒ p = 1–q = 1– (4/5) = 1/5
Since np = 20
n = \(\frac{20}{p}\)
n = 100