# Probability and Statistics Question & Answer

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**Q.1 a)50% of all automobile accidents leads to property damage of Rs 100, 40% lead to damage of Rs. 500 and 10% lead to damage of Rs 1800. If a car has 5% chance of being in an accident in a year what expected value of the property damage due to possible accident?**

**Answer :**

To calculate the expected value of the property damage due to a possible accident, we need to multiply the possible outcomes by their respective probabilities and sum them up.

Given:

- 50% of accidents lead to property damage of Rs. 100.
- 40% of accidents lead to property damage of Rs. 500.
- 10% of accidents lead to property damage of Rs. 1800.
- The car has a 5% chance of being in an accident in a year.

Let’s calculate the expected value of the property damage:

Expected Value = (Probability of Rs. 100 damage * Rs. 100) + (Probability of Rs. 500 damage * Rs. 500) + (Probability of Rs. 1800 damage * Rs. 1800)

Expected Value = (0.50 * 100) + (0.40 * 500) + (0.10 * 1800) = 50 + 200 + 180 = 430

Therefore, the expected value of the property damage due to a possible accident is Rs. 430.

**b)The probability distribution of a discrete random variable x is given by the following:
Table:
x – 0,1,2,3,4,5
p(x) – k, 5k, 10k, 10k, 5k, k
Find the value of k, mean and standard deviation of x.**

**Answer :**

To find the value of k, mean, and standard deviation of the discrete random variable x, we can use the given probability distribution.

Let’s first determine the value of k: Since the sum of probabilities in a probability distribution must equal 1, we can set up the equation:

k + 5k + 10k + 10k + 5k + k = 1

Combining like terms: 32k = 1

Dividing both sides by 32: k = 1/32

Now that we have the value of k, we can calculate the mean (expected value) of x:

Mean (μ) = Σ(x * p(x))

μ = (0 * k) + (1 * 5k) + (2 * 10k) + (3 * 10k) + (4 * 5k) + (5 * k) = 0 + 5k + 20k + 30k + 20k + 5k = 80k

Substituting the value of k: μ = 80 * (1/32) = 2.5

The mean of x is 2.5.

Next, let’s calculate the standard deviation of x:

Standard Deviation (σ) = √[Σ((x – μ)^2 * p(x))]

σ = √[((0 – 2.5)^2 * k) + ((1 – 2.5)^2 * 5k) + ((2 – 2.5)^2 * 10k) + ((3 – 2.5)^2 * 10k) + ((4 – 2.5)^2 * 5k) + ((5 – 2.5)^2 * k)]

Simplifying the equation:

σ = √[(6.25 * k) + (2.25 * 5k) + (0.25 * 10k) + (0.25 * 10k) + (2.25 * 5k) + (6.25 * k)] = √[31.25k]

Substituting the value of k:

σ = √[31.25 * (1/32)] = √[0.9765625] ≈ 0.988

The standard deviation of x is approximately 0.988.

To summarize:

- The value of k is 1/32.
- The mean of x is 2.5.
- The standard deviation of x is approximately 0.988.

**Q.2 a)Verify that the function (x) = 2x+1/25, x = 0, 1, 2, 3, 4, is a probability mass function or not? If it is a probability mass function, determine the cumulative distribution function of X and using that find the probability P (2<=X<4).**

**Answer : **

To verify if the function f(x) = (2x + 1)/25 is a probability mass function (PMF), we need to check two conditions:

- The values of f(x) are non-negative for all possible values of x.
- The sum of all values of f(x) is equal to 1.

Let’s evaluate these conditions:

- Non-negativity: f(0) = (2
*0 + 1)/25 = 1/25 > 0 f(1) = (2*1 + 1)/25 = 3/25 > 0 f(2) = (2*2 + 1)/25 = 5/25 = 1/5 > 0 f(3) = (2*3 + 1)/25 = 7/25 > 0 f(4) = (2*4 + 1)/25 = 9/25 > 0The values of f(x) are non-negative for all x = 0, 1, 2, 3, 4. - Sum of probabilities: f(0) + f(1) + f(2) + f(3) + f(4) = (1/25) + (3/25) + (1/5) + (7/25) + (9/25) = 1The sum of probabilities is equal to 1.

Since both conditions are satisfied, the function f(x) = (2x + 1)/25 is a probability mass function (PMF).

To determine the cumulative distribution function (CDF) of X, we need to calculate the cumulative probabilities up to each value of x.

CDF(X) = P(X ≤ x)

For x = 0: CDF(0) = P(X ≤ 0) = f(0) = 1/25

For x = 1: CDF(1) = P(X ≤ 1) = f(0) + f(1) = 1/25 + 3/25 = 4/25

For x = 2: CDF(2) = P(X ≤ 2) = f(0) + f(1) + f(2) = 1/25 + 3/25 + 1/5 = 9/25

For x = 3: CDF(3) = P(X ≤ 3) = f(0) + f(1) + f(2) + f(3) = 1/25 + 3/25 + 1/5 + 7/25 = 16/25

For x = 4: CDF(4) = P(X ≤ 4) = f(0) + f(1) + f(2) + f(3) + f(4) = 1/25 + 3/25 + 1/5 + 7/25 + 9/25 = 25/25 = 1

Now, to find the probability P(2 ≤ X < 4), we can use the cumulative probabilities:

P(2 ≤ X < 4) = P(X ≤ 3) – P(X ≤ 1)

P(2 ≤ X < 4) = CDF(3) – CDF(1) = (16/25) – (4/25) = 12/25

Therefore, the probability P(2 ≤ X < 4) is 12/25.

**b)Statistics released by the National Highway Traffic Safety Administration and the National Safety Council show that on an average weekend night, 1 out of every 10 drivers on the road is drunk. If 400 drivers are randomly checked next Saturday night, what is the Probability that the number of drunk drivers will be
(a) Less than 30?
(b) More than 45?
(c) At least 35 but less than 45?**

**Answer : **

To solve this problem, we can use the binomial probability formula since we are dealing with a situation where each driver can be classified as drunk or not drunk, and the probability of being drunk is constant for all drivers.

Let’s denote the number of drunk drivers as X. We are given that the probability of a driver being drunk is 1/10, and the number of drivers randomly checked is 400.

(a) To find the probability that the number of drunk drivers is less than 30, we can calculate the cumulative probability of X from 0 to 29.

P(X < 30) = Σ (k=0 to 29) (400 choose k) * (1/10)^k * (9/10)^(400-k)

(b) To find the probability that the number of drunk drivers is more than 45, we can calculate the cumulative probability of X from 46 to 400.

P(X > 45) = Σ (k=46 to 400) (400 choose k) * (1/10)^k * (9/10)^(400-k)

(c) To find the probability that the number of drunk drivers is at least 35 but less than 45, we need to calculate the cumulative probability of X from 35 to 44.

P(35 ≤ X < 45) = Σ (k=35 to 44) (400 choose k) * (1/10)^k * (9/10)^(400-k)

Since these calculations involve a large number of terms, it would be more efficient to use statistical software or tools to obtain the precise probabilities. Alternatively, approximations such as the normal approximation to the binomial distribution can be used if the conditions for such approximations are met.

Please note that the above calculations assume that the drivers are checked independently, and the probability of a driver being drunk is the same for each driver.

**Q.3 a)The Probability that a WASE student will pass UPSC examination is 3/5 and that for an IIT graduate it is 2/5. The probability that both will pass the UPSC examination is 1/5. What is the probability that only one of them passes UPSC examination?
**

**Answer: **To find the probability that only one of them passes the UPSC examination, we need to consider the cases where either the WASE student passes and the IIT graduate fails, or vice versa.

Let’s denote the event that the WASE student passes as A and the event that the IIT graduate passes as B.

We know the following probabilities:

P(A) = 3/5 (Probability that the WASE student passes)

P(B) = 2/5 (Probability that the IIT graduate passes)

P(A ∩ B) = 1/5 (Probability that both pass)

To find the probability that only one of them passes, we need to subtract the probability that both pass from the total probability that at least one of them passes.

P(Only one passes) = P(A ∩ ~B) + P(~A ∩ B) (Here, ~B denotes the complement of event B, i.e., the event that B does not happen.)

We can calculate this as follows:

P(Only one passes) = P(A ∩ ~B) + P(~A ∩ B)

= P(A) * P(~B) + P(~A) * P(B)

= (3/5) * (1 – 2/5) + (1 – 3/5) * (2/5)

= (3/5) * (3/5) + (2/5) * (2/5)

= 9/25 + 4/25 = 13/25

Therefore, the probability that only one of them passes the UPSC examination is 13/25.

**b)A machine manufacturing bolt produces 50% defective bolt and follows the Binomial distribution. Suppose we select 200 bolts then by using Chebyshev’s inequality what least probability we can assert that there will be more than 65 but fewer than 135 defective bolts?
**

**Answer: **

Chebyshev’s inequality provides a bound on the probability that a random variable deviates from its mean by a certain amount. In this case, we want to find the least probability that there will be more than 65 but fewer than 135 defective bolts out of 200 bolts.

Let X be the number of defective bolts out of 200. The probability of a bolt being defective is 50%, which means the probability of success (defective bolt) is p = 0.5. Since X follows a binomial distribution, the mean (μ) and variance (σ^2) of X are given by:

μ = n * p

= 200 * 0.5

= 100 σ^2

= n * p * (1 – p)

= 200 * 0.5 * (1 – 0.5)

= 50

Now, we can use Chebyshev’s inequality to find the least probability that X falls within the range of 65 to 135.

Chebyshev’s inequality states that for any positive k:

P(|X – μ| < k * σ) ≥ 1 – 1/k^2

In our case, we want to find the least probability that X falls within the range of 65 to 135, which can be expressed as:

P(65 < X < 135) = P(|X – μ| < 35)

Here, k = 35/σ (standard deviation) because we want to find the probability within 35 units of the mean.

Using the values we calculated earlier:

k = 35/σ = 35/sqrt(50) ≈ 4.9497

Substituting this value into the inequality:

P(|X – μ| < 35) ≥ 1 – 1/(4.9497)^2 P(|X – 100| < 35) ≥ 1 – 1/24.499

We can simplify this expression:

P(|X – 100| < 35) ≥ 1 – 0.0408 P(|X – 100| < 35) ≥ 0.9592

Therefore, by using Chebyshev’s inequality, we can assert that there is at least a 95.92% probability that the number of defective bolts falls within the range of 65 to 135.

**Q.4 a) The Head of the village wants to install LED lamps in all the streets. He found that the life time of a certain LED lamps is normally distributed with a mean of 2400 hours and a standard deviation of 400 hours what number of LED lamps might be expected to fail in first 1200 hours, if the Head of the village ordered 2000 lamps in total?**

**Answer :**

To determine the number of LED lamps that might be expected to fail in the first 1200 hours, we need to calculate the probability of a lamp failing within that time frame. Since the lifetime of the lamps is normally distributed with a mean (μ) of 2400 hours and a standard deviation (σ) of 400 hours, we can use the standard normal distribution to find this probability.

Let X be the random variable representing the lifetime of an LED lamp. We want to find P(X ≤ 1200), which represents the probability that an LED lamp fails within 1200 hours.

First, we need to standardize the value 1200 using the z-score formula:

z = (x – μ) / σ.

= (1200 – 2400) / 400

= -1200 / 400

= -3

Now, we can find the probability using the standard normal distribution table or calculator. The probability of an LED lamp failing within 1200 hours can be written as P(X ≤ 1200) = P(Z ≤ -3).

Looking up the corresponding value in the standard normal distribution table, we find that P(Z ≤ -3) is approximately 0.00135.

To find the expected number of LED lamps that might fail in the first 1200 hours out of a total of 2000 lamps, we multiply the probability by the total number of lamps:

Expected number = Probability * Total number of lamps

= 0.00135 * 2000

≈ 2.7

Therefore, we can expect around 2 or 3 LED lamps to fail in the first 1200 hours.

**b) If 0.7% of the items produced by a machine are defective Use the Poisson approximation to determine the probability that at most 4 items will be defective in a random sample of 400.**

**Answer :**

In this case, we are using the Poisson approximation to determine the probability of at most 4 defective items in a random sample of 400.

The Poisson distribution can be used when the number of events (defective items) occurring in a fixed interval (random sample) is rare and the average rate of occurrence (0.7% or 0.007) is known.

The formula for the Poisson distribution is:

P(X = k) = (e^(-λ) * λ^k) / k!

where λ is the average rate of occurrence.

In this case, λ = 0.007 * 400 = 2.8

We want to find P(X ≤ 4), which represents the probability that at most 4 items will be defective.

Using the Poisson distribution, we can calculate this probability by summing the individual probabilities for each value from 0 to 4:

P(X ≤ 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

P(X = k) = (e^(-2.8) * 2.8^k) / k!

Now we can calculate the probability using these values:

P(X ≤ 4) = (e^(-2.8) * 2.8^0) / 0! + (e^(-2.8) * 2.8^1) / 1! + (e^(-2.8) * 2.8^2) / 2! + (e^(-2.8) * 2.8^3) / 3! + (e^(-2.8) * 2.8^4) / 4!

Calculating this expression will give us the probability that at most 4 items will be defective in the random sample of 400.

**Q.5 a)If the distribution function of a random variable is given by f(x)= {1/x^2, for x>1 0, for x<=1 Find the probabilities that this random variable will take on value **

**i) Less than 2 **

**ii) Between 3 and 4 **

**iii) Greater than 4**

**Answer :**

To find the probabilities for the given random variable, we need to integrate the probability density function (PDF) over the desired intervals. Since the provided function is a continuous distribution, we’ll use integration instead of summation.

The cumulative distribution function (CDF) can be obtained by integrating the PDF from negative infinity up to a given value x. Then, the probability of the random variable taking on a value within a specific interval is given by the difference in CDF values at the endpoints of the interval.

Let’s calculate the probabilities:

i) To find the probability that the random variable is less than 2:

P(X < 2) = ∫[1 to 2] f(x) dx

Since f(x) = 1/x^2 for x > 1, the integral becomes:

P(X < 2) = ∫[1 to 2] (1/x^2) dx

Evaluating the integral:

P(X < 2) = [-1/x] evaluated from 1 to 2 = [-1/2] – [-1/1] = -1/2 + 1 = 1/2

Therefore, the probability that the random variable is less than 2 is 1/2.

ii) To find the probability that the random variable is between 3 and 4:

P(3 < X < 4) = ∫[3 to 4] f(x) dx

Using the same logic as above, we have:

P(3 < X < 4) = ∫[3 to 4] (1/x^2) dx

Evaluating the integral:

P(3 < X < 4) = [-1/x] evaluated from 3 to 4 = [-1/4] – [-1/3] = -1/4 + 1/3 = 3/12 – 4/12 = -1/12

However, since the probability cannot be negative, we set it to zero:

P(3 < X < 4) = 0

Therefore, the probability that the random variable is between 3 and 4 is zero.

iii) To find the probability that the random variable is greater than 4:

P(X > 4) = ∫[4 to ∞] f(x) dx

Using the same logic as above, we have:

P(X > 4) = ∫[4 to ∞] (1/x^2) dx

Evaluating the integral:

P(X > 4) = [-1/x] evaluated from 4 to ∞ = [-1/∞] – [-1/4] = 0 + 1/4 = 1/4

Therefore, the probability that the random variable is greater than 4 is 1/4.

**b) Three professors, Prof. Anand, Prof. Bhanu Prakash and Prof. Chandra Shekhar appear in an interview for the post of Managing Director in a reputed company. Their chances of getting selected are 2/9, 4/9 and 1/3 respectively. The probabilities that they introduce a new policy in the company are 3/10, 1/2 and 4/5 respectively.
i) What is the probability that the new policy is introduced in the company?
ii). What is the probability the new policy introduced is done by Prof. Anand?**

**Answer :**

To find the probabilities related to the introduction of a new policy in the company, we can use the concept of conditional probability.

Let’s denote the events as follows: A: Prof. Anand gets selected as Managing Director. B: Prof. Bhanu Prakash gets selected as Managing Director. C: Prof. Chandra Shekhar gets selected as Managing Director. P: A new policy is introduced.

Given the probabilities:

P(A) = 2/9 P(B) = 4/9 P(C) = 1/3

P(P|A) = 3/10 P(P|B) = 1/2 P(P|C) = 4/5

i) To find the probability that the new policy is introduced in the company, we can use the law of total probability:

P(P) = P(P|A) * P(A) + P(P|B) * P(B) + P(P|C) * P(C)

P(P) = (3/10) * (2/9) + (1/2) * (4/9) + (4/5) * (1/3)

P(P) = 6/90 + 4/18 + 4/15

P(P) = 1/15 + 8/18 + 8/15

P(P) = (2 + 40 + 32)/90

P(P) = 74/90

Simplifying the fraction, we get: P(P) = 37/45

Therefore, the probability that the new policy is introduced in the company is 37/45.

ii) To find the probability that the new policy introduced is done by Prof. Anand, we can use Bayes’ theorem:

P(A|P) = (P(P|A) * P(A)) / P(P)

P(A|P) = (3/10) * (2/9) / (37/45)

P(A|P) = (3/10) * (2/9) * (45/37)

Simplifying the fraction, we get: P(A|P) = 3/37

Therefore, the probability that the new policy introduced is done by Prof. Anand is 3/37.

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