Let p he the probability of getting a girls
∴ p = \(\frac { 1 }{ 2 } \) [∵ one favourable event and total no of events is 2]
⇒ q = 1-p = 1-\(\frac { 1 }{ 2 } \) = \(\frac { 1 }{ 2 } \) and n = 3
∴ (getting exactly 2 girls)
= P(X = 2)
= ³C₂\(\left( \frac { 1 }{ 2 } \right) ^{ 2 }\left( \frac { 1 }{ 2 } \right) ^{ 1 }\)   [∵ p(x) = ⁿC_x p^xq^n-x, n = 3 and x = 2]
= \(3\left( \frac { 1 }{ 2 } \right) ^{ 3 }=3\times \left( \frac { 1 }{ 2 } \right) ^{ 3 }=3\times \frac { 1 }{ 8 } \) = 0.375
P (getting exactly 2 girls) = 0.375.

The binomial distribution can be used under the following conditions.
(i) The number of trials en' is finite,
(ii) The trials are independent of each other.
(iii) The probability of success 'p' is constant for each trial.
(iv) In every trial there are only two possible outcomes namely success or failure.

A random variable X is said to follow binomial distribution with parameter n and p, if it assumes only non-negative value and its probability mass function is given by 
P(X = x) = P(x) = q = 1-p \(=\begin{cases} \begin{matrix} { { n }_{ C } }_{ x }P^{ x }{ q }^{ n-x },x=0,1,2,...n; \end{matrix} \\ \begin{matrix} 0, & otherwise \end{matrix} \end{cases}\)

Let X be a random variable follows binomial distribution with p = q = 1/2
P (3 heads) = \(5{ C }_{ x }{ \left( \frac { 1 }{ 2 } \right) }^{ x }{ \left( \frac { 1 }{ 2 } \right) }^{ 5-x }\)
\(={ 5C }_{ 3 }{ \left( \frac { 1 }{ 2 } \right) }^{ 3 }{ \left( \frac { 1 }{ 2 } \right) }^{ 5-3 }\)
\(=5{ C }_{ 3 }{ \left( \frac { 1 }{ 2 } \right) }^{ 5 }\)
\(=\frac { 5 }{ 16 } \)

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

Let X be a random variable follows poisson distribution with number of erythrocytes.
Hence, Mean λ = 8 erythrocytes/single square
P(exactly 5 erythrocytes are in one square) = P(X = 5)=\(\\ \frac { { e }^{ -\lambda }{ \lambda }^{ x } }{ x! } =\frac { { e }^{ -8 }{ 8 }^{ 5 } }{ 5! } \)
\(=\frac { 0.000335\times 32768 }{ 120 } \)
= 0.0916
The probability that exactly 5 erythrocytes are found in one square is 0.0916. i.e there are 9.16% chances that exactly 5 erythrocytes are found in one square.

In a throw of a pair of dice the doublets are (1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6, 6)
Probability of getting a doublet p = 6/36 = 1/6
⇒ q = 1 – p = 5/6 and also n = 4 is given
The probability of successes
 \(=\left( \begin{matrix} 4 \\ x \end{matrix} \right) { (\frac { 1 }{ 6 } ) }^{ x }\left( \frac { 5 }{ 6 } \right) ^{ 4-x }\)
Therefore the probability of 2 successes are
\(P(X=2)\left( \begin{matrix} 4 \\ 2 \end{matrix} \right) { (\frac { 1 }{ 6 } ) }^{ 2 }\left( \frac { 5 }{ 6 } \right) ^{ 4-2 }\)
\(=6\times \frac { 1 }{ 36 } \times \frac { 25 }{ 36 } \)
\(=\frac { 25 }{ 216 } \)

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)
\(={ \left( \frac { 1 }{ 2 } \right) }^{ 10 }\left[ { 10C }_{ 6 }+{ 10C }_{ 7 }+{ 10C }_{ 8 }+{ 10C }_{ 9 }+{ 10C }_{ 10 } \right] \)
\(=\left[ \frac { 1 }{ 1024 } \right] [210+120+45+10+1]\)
\(=\frac { 193 }{ 512 } \)