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Correct answer is (c) a = b

Correct answer is (b) 1/2

Correct answer is (a) \(\frac{1}{27}\)

(b) -(19/6)

Expected amount to win = 1 - (25/6) = -(19/6)

Correct answer is (a) 1/10

Correct answer is (b) \(\frac{294}{251}\)

Correct answer is (b) \(\frac{n}{3} + \frac{1}{6}\)

\(X \sim N (18000, 10000)\)

\(\frac{X - 180000}{10000} \sim N(0,1)\)

\(P(185000< X < 200000) = P\left(\frac{185000-180000}{10000} < \frac{X - \mu}\sigma < \frac{200000 -180000}{10000}\right)\)

\(= P\left(\frac 12 < \frac{X - \mu}\sigma < 2\right)\)

\(= P\left(\frac{x - \mu}\sigma \le 2\right) - P\left(\frac{x - \mu}\sigma \le \frac12\right)\)

\(= 0.9772 - 0.6915\)

\(= 0.2857\)

The required probability be 0.2857.

Given mean np = 1.2 and n = 6 

p = 1.2/6 = 0.2, q = 1 – 0.2 = 0.8 

Let X be a binomial variable denoting the number of defects, 

(i.e,) X ~ B (6, 0.2) 

p.m.f is given by P (X = x) = ⁶C_x(0.2)^x(0.8)^{6 - x}

We want P(X < 2) = P(X = 0) + P (X = 1) 

= ⁶C₀(0.2)⁰(0.8)⁶ + ⁶C₁(0.2)¹ (0.8)⁵

= (0.8)⁶ + 6 (0.2) (0.8)⁵ 

= 0.262144 + 0.393216 

= 0.65536 

Thus if lengths of 6 metres are to be inspected, the probability of less than 2 defects is 0.65536.

There are 5 bad apples and 10 good apples.

\(\therefore\) Total number of apples = 5 + 10 = 15

Let X denotes the number of good apples in 3 drawn apples.

Probability of success = \(\frac{Numbere\,of\,good\,apples}{Total number\,of\,apples}\) = 10/15 = 2/3

\(\therefore\) P = 2/3

Probability of failure is q = 1 - p = 1 - 2/3 = 1/3

P(x = 0) = ³C₀p⁰q³ = 1 x 1 x \((1/3)^3\) = 1/27

P(x = 1) = ³C₁p¹q² = 3 x 2/3 x (1/3)² = 2/9

P(x = 2) = ³C₂p²q = 3 x (2/3)² x 1/3 = 4/9

P(x = 3) = ³C₃p³q⁰ = 1 x (2/3)³ x 1 = 8/27

Hence, probability distribution of the number of good apple

(a) e^-λ

P(X -x) = e^-λλ^x/x! Put x  = 0, P(X = 0) = e^-λλ⁰/0! = e^-λ

(b) De-Moivre 

Poisson distribution is a discrete frequency distribution that gives the probability of a number of independent events occurring in a fixed time. It is useful for characterizing events with very low probabilities of occurrence within some definite time or space.

(c) Z = (X - 9)/9

µ = 9, σ = 9 

Z = (X - 9)/9

Given p = 0.09 (success) 

q = 0.91 (failure) 

We have to find number of trials ‘n.’ 

According to the problem, 

P(X ≥ 1 ) > 1/3 

(We must have atleast one success) 

1 – P(X < 1) > 1/3 

1 – P(X = 0) > 1/3 

(or) P(X = 0) < 2/3 

Using p.m.f, we have, 

ⁿC₀(0.09) ⁰ (0.91) ⁿ < 2/3

(0.91)ⁿ < 2/3

we can use log tables to calculate (or) by trial method try for n = 1, 2,…… using calculator. 

We observe that (0.91)⁵ < 2/3 but (0.91)⁴ = 0.6857 > 2/3. Thus we need a minimum of 5 trials or more.

Let X be the Poisson variable denoting the number of accidents per day. 

Given that mean is 4 (i.e,) λ = 4. The p.m.f is given by P(X = x) = (e^-44^x) / x!

(i) P (no accident) = P(X = 0) = e = 0.0183 

For 100 days we have 100 × 0.0183 = 1.83 ~ 2 

Hence out of 100 days there will be no accident for 2 days.

(ii) P (atleast 2 accidents) = P (X ≥ 2) 

= 1 – P (X < 2) 

= 1 – [P(X = 1) + P(X = 0)] 

= 1 – [e^-4 (4) + e^-4 ] 

= 1 – (0.0183) (5) 

= 1 – 0.0915

 = 0.9085 

For 100 days we have 100 × 0.9085 ~ 91 

Hence out of 100 days there will be at least 2 accidents for 91 days.

(iii) P (atmost 3 accidents) = P (X ≤ 3) 

= P (X = 0 ) + P (X = 1 ) + P (X = 2) + P (X = 3) 

= e^-4 [ 1 + 4/1 + 16/2 + 64/6] 

= (0.0183) [23.6667] 

= 0.4331 

For 100 days we have 100 × 0.4331 ~ 43 

Hence out of 100 days, there will be atmost 3 accidents for 43 days.

Poisson distribution is a limiting case of binomial distribution under the following conditions: 

•  the number of trials ‘n’ is indefinitely large 

i.e, → ∞ 

•  the probability of success ‘p’ in each trial is very small, 

i.e, p → 0 

•   np = λ is finite. Thus p = λ/n and q = 1 – λ/n, λ > 0

The binomial distribution can be used under the following conditions:

•  The number of trials (or) observations ‘n’ is fixed (finite). 

•  Each observation is independent of each other. 

•  In every trial, there are only two possible outcomes – success or failure. 

•  The probability of success ‘p’ is the same for each outcome.

Given that X ~ B (5, 0.05) 

(i.e.) n = 5, p = 0.05, q = 0.95 

P(X ≤ 1) = P(X = 0) + P(X = 1) 

= ⁵C₀(0.05)⁰(0.95)⁵ + ⁵C₁(0.05)¹(0.95)⁴

= (0.95)⁵ + 5 (0.05) (0.95)⁴ 

= (0.95)⁴ + [0.95 + 0.25] 

= 0.9774

Let X be the binomial random variable denoting the number of traffic lights. 

Given n = 15, p =0.2, q = 0.8

(a) P(X = 2) = ¹⁵C₂ (0.2)² (0.8)¹³

= 105 (0.04) (0.8) [(0.8)⁴]³

= 0.2309

(b) P(X ≥ 1) = 1 - 9(X < 1)

= 1 - P(X = 0)

= 1 - ¹⁵C₀ (0.2)⁰ (0.8)¹⁵

= 1 - (0.8)¹⁵ = 0.9648

The Normal distribution is a limiting case of Binomial distribution under the following conditions: 

•  n, the number of trials is infinitely large, i.e. n → ∞ 

•  neither p (or q ) is very small.

1. p 

2. less than mean 

3. independent 

4. p > 1/2

5. two 

6. 2.1 

7. p = q 

8. zero 

9. p = 0.5

n  15, p = 1/7, q = 6/7

P(X = 2) = ¹⁵C₂(1/7)² (6/7)¹³ = 105((6)¹³/7¹⁵)

Chief Characteristics of the Normal Probability Curve are as follows:

•   The curve is bell-shaped and symmetrical about the line x = µ.

•   Mean, median, and mode of the distribution coincide. 

•   The total area under the normal curve is equal to unity. 

•   For a given µ and σ, there is only one normal distribution. 

•   The Points of inflection are given by x = µ ± σ

Given mean = 5 and standard deviation = 2 

(i.e,) np = 5 and √npq = 2 ⇒ npq = 4 

5q = 4 ⇒ q = , p = 1 – 4/5 = 1/5 

Again np = 5 gives = 5 ⇒ n = 25

So the p.m.f of the distribution is given by P (X = x) = (25, x)(1/5)^x(4/5)^25-x

(d) 0.340 

µ = 70, σ = 10 

P(72 < X < 84) = P((72 - 70)/10 < Z < (84 - 70)/10)

= P(0.2 < Z < 1.4) 

= P(0 < Z < 1.4) – P(0 < Z < 0.2) 

= 0.4192 – 0.0793 

= 0.3399 

= 0.340

(b) 0.394

µ = 1295, σ = 750 

P(X > 1500) 

= P(Z > (1500 - 1295)/750) 

= P (Z > 0.27) = 0.5 – P (0 < Z < 0.27) 

= 0.5 – 0.1064 

= 0.3936 ~ 0.394

(a) 0.138 

µ = 3.2, σ = 1.1 

P (X < 2) 

= P( Z < (2 - 3.2)/1.1) 

= P(Z < -1.09) = P(Z > 1.09) 

= 0.5 – P(0 < Z < 1.09) 

= 0.5 – 0.3621 

= 0.138

(d) all of the above statements are true

(a) 0.0062 

µ = 1500, σ = 200 

P(X < 1000) 

= P(Z < (1000 - 15000)/200) 

= P(Z < -2.5) 

= P (Z > 2.5) = 0.5 – P (0 < Z < 2.5) 

= 0.5 – 0.4938 

= 0.0062

(a) 6.00

n = 15, p = 0.4 ⇒ mean (np) = 6

(d) 8

Given, Var (X) = 4 

E(X²) – [E(X)]² = 4 

E(X²) – 2² = 4 ⇒ E(X²) = 4 + 4 = 8

(d) 1/3

Given P (X = 0) = 1 – P (X = 1) ⇒ P (X = 1) = 1 – P (X = 0) 

Let n = P(X = 0), 

∴ P(X = 1) = 1 - n 

E(X) = (0) (n) + (1) (1 – n) = 1 – n

E(X²) = (0)² (n) + (1)² (1 – n) = 1 – n

Now substituting E(X) = 3 Var (X), we get n = (1/3) (or) 1

∴ P(X = 0 ) = 1/3

When a coin is tossed 6 times, the number of heads can be 0, 1, 2, 3, 4, 5, 6. 

The corresponding number of tails will be 6, 5, 4, 3, 2, 1, 0. 

∴ X can take values 0 – 6, 1 – 5, 2 – 4, 3 – 3, 4 – 2, 5 – 1, 6 – 0 

i.e. -6, -4, -2, 0, 2, 4, 6. 

∴ X = {-6, -4, -2, 0, 2, 4, 6}.

Total number of playing cards = 52 

Number of Black cards = 26 

Number of Non-black (or) Red cards = 26 

Let ‘X’ be the random variable denotes the number of black cards. Since two black cards are drawn, ’X’ takes the values 0, 1, 2 

X (Non-black Cards) = X (26C₁ × 25C₁) = X (650) = 0 

X (1 Black Card) = X (26C₁ × 26C₀) = X (26) = 1 

X (2 Black Cards) = X (26C₁ × 25C₁) = X (650) = 2

Let X is the random variable denotes the number of tails when three coins are tossed simultaneously. 

Sample space S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} 

∴ ‘X’ takes the values 0,1, 2, 3 

i.e., X (HHH) = 0 ; X (HHT, HTH, THH) = 1 ; X (HTT, THT, TTH) = 2 ; X (TTT) = 3

(b) 1.1

P (X = 0) = P (TT) 

= P (T) P (T) = (0.4) (0.5) = 0.20 

P (X = 1) = P (HT) + P(TH) = (0.6) (0.5) + (0.4) (0.5) = 0.5 

P (X = 2) = P (HH) 

= P (H). P (H) = (0.6) (0.5) = 0.30

E[X] = ∑x P(x)

= (0).(0.20) + (1) (0.5) + (2) (0.30) 

= 0.5 + 0.6 = 1.1

Correct answer is B.

E = (X : X is a prime number)=(2, 3, 5, 7)

P(E) = P(X = 2) + (X = 3) + (X = 5) + (X = 7)

P(E) = 0.23 + 0.12 + 0.20 + 0.07 = 0.62

F=(X : X < 4) = (1, 2, 3)

P(F) = P(X = 1) + (X = 2) + (X = 3)

P(F)= 0.15 + 0.23 + 0.12 = 0.5

E∩F = (X is a prime number as well as < 4) = (2, 3)

P(E∩F) = P(X = 2) + P(X = 3)

⇒ 0.23 + 0.12 = 0.35

P(E∪F) = P(E) + P(F) - P(E∩F)

⇒ 0.62 + 0.50 - 0.35

= 0.77

The key point to solve the problem:

If a probability distribution is given then as per its definition, Sum of probabilities associated with each value of a random variable of given distribution is equal to 1

i.e. ∑(p_i) = 1

∴ P(X = -2) + P(X = -1) + P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 1

0.1 + k + 0.2 + 2k + 0.3 + k = 1

0.6 + 4k = 1

4k = 0.4

K = 0.1

Value of k = 0.1

The key point to solve the problem:

A given distribution of probabilities of a random variable X is said to be probability distribution if the sum of probabilities associated with each random variable is equal to 1

i.e. ∑(p_i) = 1

Given distribution is :

Clearly,

Sum of probabilities = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

= 0.3 + 0.2 + 0.4 + 0.1

= 1

∴ The given distribution is a probability distribution.

The key point to solve the problem:

A given distribution of probabilities of a random variable X is said to be probability distribution if the sum of probabilities associated with each random variable is equal to 1

i.e. ∑(p_i) = 1

Given distribution is :

 Clearly,

Sum of probabilities = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

= 0.1 + 0.5 + 0.2 + 0.1 + 0.1

= 1

∴ The given distribution is a probability distribution.

The key point to solve the problem:

A given distribution of probabilities of a random variable X is said to be probability distribution if the sum of probabilities associated with each random variable is equal to 1

i.e. ∑(p_i) = 1

Given distribution is :

Clearly,

Sum of probabilities = P(X = -1) + P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

= 0.3 + 0.2 + 0.4 + 0.1 + 0.05

= 1.05 ≠ 1

∴ The given distribution is not a probability distribution.

The key point to solve the problem:

A given distribution of probabilities of a random variable X is said to be probability distribution if the sum of probabilities associated with each random variable is equal to 1

i.e. ∑(p_i) = 1

Given distribution is :

Clearly,

Sum of probabilities = P(X = 0) + P(X = 1) + P(X = 2)

= 0.6 + 0.4 + 0.2

= 1.2 ≠ 1

∴ The given distribution is not a probability distribution.

(i) P(X is positive) 

= P(X = 1) + P(X = 2) + P(X = 3) 

= 0.25 + 0.15 + 0.1 

= 0.50 

(ii) P(X is non-negative) 

= P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) 

= 0.20 + 0.25 + 0.15 + 0.1 

= 0.70 

(iii) P(X is odd) 

= P(X = -3) + P(X = -1) + P(X = 1) + P(X = 3)

= 0.05 + 0.15 + 0.25 + 0.1 

= 0.55 

(iv) P(X is even) 

= P(X = -2) + P(X = 0) + P(X = 2) 

= 0.10 + 0.20 + 0.15 

= 0.45.

Correct answer is (d) 1/10

Correct answer is (b) -0.35

(i) Let X = number of unemployed graduates in a town. 

Since the population of the town is 1 lakh, X takes the finite values. 

∴ random variable X is discrete. 

Range = {0, 1, 2, …, 99999, 100000}. 

(ii) Let X = amount of syrup prescribed by a physician. 

Then X takes uncountable infinite values. 

∴ random variable X is continuous. 

(iii) Let X = gain of weight in a week 

Then X takes uncountable infinite values 

∴ random variable X is continuous. 

(iv) Let X = number of female rats selected on a specific day. 

Since the total number of rats is 20 which includes 12 males and 8 females, X takes the finite values.

∴ random variable X is discrete. Range = {0, 1, 2, 3, 4, 5}

(v) Let X = speed of .the car in km/hr. 

Then X takes uncountable infinite values 

∴ random variable X is continuous.