Given μ = 225, σ = 55
P(X≤100)
When X = 100, Z = \(\frac { X-\mu }{ \sigma } \)
= \(\frac { 100-225 }{ 55 } \) = -2.272
∴ P(X≤100) = P(Z≤-2.272)

 = 0.5-P(0 [By symmetry]
= 0.5-0.4884 = 0.0116
∴ Probability of one customer to have a bill for Rs.100 or less is 0.0116.
∴ Out of 500 customers, the number of persons to have a bill for Rs. 100 or less is 0.0116 x 500 = 5.8 ≌ 6.

Given μ = 3500, σ = 500
P(X<3100)
When X = 3100,
Z = \(\frac { X-\mu }{ \sigma } =\frac { 3100-3500 }{ 500 } \)
= -0.8
∴ P(X<3100) = P(Z<-0.8)


= 0.5-P(0 [By symmetry]
= 0.5-0.2881= 0.2119
Hence, the probability that a baby is born with weight less than 3100g is 0.2119.

Given μ = 500, σ = 100
P(X< 585)
When X = 585, Z = \(\frac { X-\mu }{ \sigma } \)
= \(\frac { 585-500 }{ 100 } =\frac { 85 }{ 100 } \) = 0.85
P(X < 858) = P(Z< 0.85)
P(-∞  = 0.5 + 0.3023
= 0.8023

When his score is 585,
80.23% of people have scored less than Rahul
∴ Rahul will be admitted to the university.

Given average = λ = \(\frac { 3 }{ 20 } \) = 0.15
P(will not be more than one failure)
= P(X ≤ 1)
= P(X = 0) + P(X = 1)
= \(\frac { { e }^{ -\lambda }.{ \lambda }^{ 0 } }{ 0! } +\frac { { e }^{ -\lambda }.{ \lambda }^{ 1 } }{ 1! } \)
= e^-λ (1+λ) =e^-0.15 (1+0.15)
= (0.8607) (1.15)
= (0.98981 [e^-0.15 =0.8607]
∴ Probability that there will not be more than one failure = 0.98981

Let p be the probability of a patient to recover
Given q = 75% = \(\frac { 75 }{ 100 } \)  = 0.75
∴ P = 1 - q = 1 - 0.75 = 0.25
n = 6
P (4 will recover) = P(X = 4)
= 6C₄ (0.25)⁴ (0.75)²
[∵ P(x) = nC_x p^xq^n-x n = 6, x = 4
= 6C₂ (0.25)⁴ (0.75)²

= 15 (0.25)⁴ (0.75)²
P(X = 4) = 0.03295

Let p be the probability of fatal accident in a year.
Given p = \(\frac { 1 }{ 1200 } \) and n=300
∴ λ = np = 300 x \(\frac { 1 }{ 1200 } =\frac { 1 }{ 4 } \) = 0.25
Hence X follows a poisson distribution with
p(x,λ) = \(\frac { e^{ -\lambda }.{ \lambda }^{ x } }{ x! } \)
∴ P (atleast 2 fatal accidents)
= P(X≥2)
= 1-P(X<2)
= 1 - [P(X = 0) + P(X =1)]
= 1-\(\left[ \frac { { e }^{ -\lambda }.{ \lambda }^{ 0 } }{ 0! } +\frac { { e }^{ -\lambda }.{ \lambda }^{ 1 } }{ 1! } \right] \)
= 1-e^-λ (1+λ)
= 1-e^-0.25 (1+0.25)
= 1 - 0.7788 (1.25) = 1 - 0.9735
=  0.0265
Hence, the probability of atleast 2 fatal accidents in a year = 0.0265.

Let p be the probability of defective electric bulb.
p = \(\frac { 5 }{ 100 } \) and n = 120

Hence λ follows Poisson distribution with
p(x,λ) = \(\frac { e^{ -\lambda }.{ \lambda }^{ x } }{ x! } \)
∴ P (no defective bulb) = P(X = 0)
= \(\frac { { e }^{ -6 }.6^{ 0 } }{ 0! } \) = e^-6  [λ = 6 and x = 0]
P(X=0) = 0.0025   [∵ e^-6 = 0.0025]

Let X be probability of mortality rate for a certain disease
∴ p = \(\frac { 7 }{ 1000 } \) and n = 400

Hence X follows poisson distribution with p(x, λ)
= \(\frac { e^{ -\lambda }.{ \lambda }^{ x } }{ x! } \)
∴ P (just 2 deaths) = P(X = 2)
= \(\frac { e^{ -2.8 }(2.8)^{ 2 } }{ 2! } \)    [∵ p(x,λ) =\(\frac { e^{ -\lambda }.{ \lambda }^{ x } }{ x! } \), λ = 2.8, x = 2]
= \(\frac { (0.06)(2.8)^{ 2 } }{ 2 } \) = (0.03)(2.8)² = 0.2352
∴ Probability for just 2 deaths on account of this disease = 0.2352.

Let p be the probability of mice recovering from vitamin A deficiency
Given p = 0.73 ⇒ q = 1 - 0.73 = 0.27 and n = 5
∴  P (atleast 3 mice recover) = P(X ≥ 3)
P(X≥3) = 1-P(X<3)
= 1 - [P(X = 0) + P(X = 1) + P (X = 2)]
= 1-[⁵C₀ (0.73)⁰ (0.27)⁵ + ⁵C₁ (0.73)¹ (0.27)⁴ + ⁵C₂ (0.73)² (0.27)³ ]
= 1-[(0.27)⁵ + 5(0.73)(0.27)⁴ + \(\frac { 5\times 4 }{ 2\times 1 } \)+(0.73)² (0.27)³ ]
= 1 - [0.00143 + 0.01939 + 0.10489]
= 1 - 0.1257
P(X≥3) = 0.8743

Let p be the probability of drug's side effect
∴ p = \(\frac { 3 }{ 100 } \) =.03
⇒ q = 1-p = 1-0.03 = 0.97 and n = 10
P (atleast one person will have side effect)
= P(X ≥ 1)
= 1 - P(X < 1)
= 1 - [P(X =0)]       ∵ p(x) =ⁿC_x p^xq^n-x, n=10, x=0
= 1-[¹⁰C₀ (0.03)⁰ (0.97)¹⁰ ]
= 1-(0.97)¹⁰    [∵ ¹⁰C₀ = 1 and (0.03)⁰ = 1]
= 1-0.7374
P(X≥1) = 0.2626

Given mean of the binomial distribution is 4.
⇒ np =4  ...(1)
Variance=3 ⇒ npq =3..(2)
(2) \(\div \) (1) given, \(\frac { npq }{ np } =\frac { 3 }{ 4 } \Rightarrow q=\frac { 3 }{ 4 } \)
∴ p=1-q =\(1-\frac { 3 }{ 4 } =\frac { 1 }{ 4 } \)
Substituting p=\(\frac { 1 }{ 4 } \) in (1) we get,
n\(\left( \frac { 1 }{ 4 } \right) \) =4 ⇒ n=16
∴ The binomial distribution is ⁿC_x p^xq^n-x, x=0,1,2....n
⇒ ¹⁶C_x \(\left( \frac { 1 }{ 4 } \right) ^{ x }\left( \frac { 3 }{ 4 } \right) ^{ 16-x }\) x=0,1,2,......16.
∴ P(X=15) =¹⁶C₁₅ \(\left( \frac { 1 }{ 4 } \right) ^{ 15 }\left( \frac { 3 }{ 4 } \right) ^{ 1 }\)
=¹⁶C₁ \(\frac { 1 }{ { 4 }^{ 15 } } .\frac { 3 }{ 4 } =16.\frac { (3) }{ 4^{ 16 } } =\frac { 4^{ 2 }.(3) }{ 4^{ 16 } } \)
P(X=15) =\(\frac { 3 }{ { 4 }^{ 14 } } \).

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

Let X be p probability random variable denoting the number of defective yarns.
Mean = np = 1.2, n = 6
6p = 1.2
\(p=\frac{1.2}{6}=0.2\)
q = 1-p = 1-0.2 = 0.8
P(X <2) = P(X = 0)+ P(X = 1)
P(X = x) = nC_x p^x q^n-x
P(X<2) = 6C₀(0.2)⁰(0.8)⁰+6C₁(0.2)¹ (8)⁵
= (0.32768)2
= 0.6553

(i) Binomial distribution is symmetrical if p = q = 0.5. It is skew symmetric if p≠q. It is positively skewed if p < 0.5 and it is negatively skewed if p > 0.5.
(ii) For binomial distribution, variance is less than mean.
Variance = npq = (np)q < np < mean.

We are given mean μ = 800 and standard deviation σ = 100
Probability that the procurement of milk between 800 litres to 1000 litres per day is
P(800 < X < 1000)
\(P(\frac { 800-800 }{ 100 })\) < z < \((\frac { 1000-800 }{ 100 })\)
P(0 < Z < 2) = 0.4772 (table value)
Therefore 47.75 percent of societies procure milk between 800 litres to 1000 litres per day.

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 be the height of soldiers follows normal distribution with mean 69.25 inches and standard deviation 3.13 then the soldiers over 6 feet tall (6ft × 12= 72 inches)
The standard normal variate
\(Z=\frac { X-\mu }{ \sigma } =\frac { 72-69.25 }{ 3.13 } =0.8786\)
P(X > 72) = P(Z > 0.8786)
= 0.5 – P(0 < Z < 0.88)
= 0.5 – 0.3106 = 0.1894
Number of soldiers expected to be over 6 feet tall in 6000 are 6000 × 0.1894 =1136

Let the distribution of heights be normally distributed with mean mean 68.22 and standard deviation = 3.286
\(Z=\frac { X-\mu }{ \sigma } =\frac { X-69.25 }{ 3.286 } \)
When X = 74
\(Z=\frac { X-\mu }{ \sigma } =\frac { 74-69.25 }{ 3.286 } =1.4455\)
Now P(Z > 74) = P(Z > 1.44)
= 0.5 – 0.4251
= 0.0749
Expected number of children to be over 74 cm out of 1200 children
= 1200 × 0.0749 ≈ 90 children

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
\(P(X=2)\left( \begin{matrix} 4 \\ 2 \end{matrix} \right) { (\frac { 1 }{ 6 } ) }^{ 2 }\left( \frac { 5 }{ 6 } \right) ^{ 4-2 }\)
\(=6\times \frac { 1 }{ 36 } \times \frac { 25 }{ 36 } \)
\(=\frac { 25 }{ 216 } \)

Here p = q = 1/2
(a) probability of A beating B in exactly 3 games out of 4
\(\left( \begin{matrix} 4 \\ 3 \end{matrix} \right) { (\frac { 1 }{ 2 } ) }^{ 3 }\left( \frac { 1 }{ 2 } \right) ^{ 4-3 }\)
= 1/4 = 25%
(b) probability of A beating B in exactly 5 games out of 8
\(\left( \begin{matrix} 8 \\ 5 \end{matrix} \right) { (\frac { 1 }{ 2 } ) }^{ 5 }\left( \frac { 1 }{ 2 } \right) ^{ 8-5 }\)
\(=\frac { 7 }{ 32 } \) = 21.875%
Clearly, the first event is more probable.

Probability of bus running late is denoted as p = 1-0.8 = 0.2
Probability of bus running according to the schedule is q = 0.8
Also given that n = 10
The binomial distribution is p(x) = 10C_x(0.2)^x(0.8)^10-x
(i) probability that exactly one is late P(x = 1) = 10C₁pq⁹
= 10C₁(0.2)(0.8)⁹
(ii) probability that at least one is late
= 1 – probability that none is late
= 1 – p(x = 0)
= 1– (0.8)¹⁰

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 p.m.f. Binomial distribution is
p(x) = ⁿC_xp^xq^n-x
Given that E(x) = 2
For the Binomial distribution mean is given by np = 2 ... (1)
Given that var (x) = 4/3
For Binomial distribution variance is given by npq=4/3   ...(2)
\(\frac { (2) }{ (1) }⇒\frac {npq}{np} =\frac { \frac { 4 }{ 3 } }{ 2 } =\frac { 4 }{ 6 } =\frac { 2 }{ 3 } \)
q = 2/3 and p = 1–2/3 = 1/3
Substitute is (1) we get
n = 6
Hence, P(X = 5) = 6C₅\({ \left( \frac { 1 }{ 3 } \right) }^{ 5 }{ \left( \frac { 2 }{ 3 } \right) }^{ 6-5 }\) = 0.0108

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

Probability of getting a degree p = 0.4
∴ q = 1– p
= 1 - 0.4
= 0.6
(i) P (one will be a graduate) = P(X = 1) = 5C₁ (0.4)(0.6)⁴
= 0.2592
(ii) P ( atleast one will be a graduate) = 1–P (none will be a graduate)
= 1-5C₀(P⁰)(Q)5-0
= 1-5C₀(0.4)⁰(0.6)5
= 1-0.0777
= 0.9222

Let ‘p’ be the probability that ‘A’ wins the game. Then we are given n = 5, p = 3/5, q = 1–\(\frac{3}{5}=\frac{2}{5}\)(since q = 1–p)
Hence by binomial probability law, the probability that out of the 5 games played, A wins ‘x’ games is given by
P(X = x) = p(x) = 5Cx\({ \left( \frac { 3 }{ 5 } \right) }^{ x }{ \left( \frac { 2 }{ 5 } \right) }^{ 5-x }\)
The required probability that ‘A’ wins atleast three games is given by 
P(X\(\le\)3) = P(X = 3) + P(X = 4) + P(X = 5)
\(=5c3{ \left( \frac { 3 }{ 5 } \right) }^{ 3 }{ \left( \frac { 2 }{ 5 } \right) }^{ 2 }5C4{ \left( \frac { 3 }{ 5 } \right) }^{ 4 }{ \left( \frac { 2 }{ 5 } \right) }^{ 1 }+5C5{ \left( \frac { 3 }{ 5 } \right) }^{ 5 }{ \left( \frac { 2 }{ 5 } \right) }^{ 0 }\)
= 0.6826