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