Assignment -II PART B - Guide to ANS

PART B    இனிய தமிழ் புத்தாண்டு வாழ்த்துக்கள்

1. Probability of Same Numbers on Dice:

When rolling two unbiased dice, each die has 6 possibilities (1, 2, 3, 4, 5, 6) resulting in 36 total outcomes (6 x 6). There are 6 favorable outcomes for getting the same number on both dice (1, 1), (2, 2), ..., (6, 6). Therefore, the probability is:

Favorable outcomes / Total outcomes = 6 / 36 = 1/6

2. Independent Events:

Independent events are those where the outcome of one event does not affect the outcome of the other event. In other words, knowing the result of one event doesn't change the probability of the other event happening.

Example: Flipping a coin and rolling a die are independent events. The outcome of heads or tails (coin) doesn't influence whether you roll a 1, 2, 3, etc. (die).

3. Wind Speed Data:

Let’s start by constructing a frequency table for the given wind speed data:

First, we’ll organize the data into bins (ranges). Since the data is discrete, we’ll create bins that cover the possible wind speed values. Let’s choose a bin width of 5 km/h.

Bin Range	Frequency
10 - 14	4
15 - 19	5
20 - 24	4

Now, let’s create a histogram to visualize this frequency distribution. The x-axis will represent the wind speed bins, and the y-axis will show the frequency of occurrence. Here’s the histogram:

Frequency
│
5 │ ▄
4 │ ▄ ▄
3 │ ▄ ▄
2 │ ▄ ▄
1 │ ▄ ▄
└───────────────
10 15 20 25
Wind Speed (km/h)

In the histogram:

• The first bar represents wind speeds between 10 and 14 km/h (4 occurrences).
• The second bar represents wind speeds between 15 and 19 km/h (5 occurrences).
• The third bar represents wind speeds between 20 and 24 km/h (4 occurrences).

This histogram provides a visual representation of the wind speed distribution at the construction site over the two-week period. If you have any further questions or need additional assistance, feel free to ask! 😊

4. Conditional Probability:

Here's the formula for conditional probability and an explanation with a sample:

Formula:

The conditional probability of event A occurring given that event B has already occurred is written as P(A|B). It can be calculated using the following formula:

P(A|B) = P(A∩B) / P(B)

• P(A|B): The probability of event A happening, given that event B has already happened.
• P(A∩B): The probability of both event A and event B occurring together (intersection of events).
• P(B): The probability of event B happening.

Explanation with Sample:

Imagine you have a bag containing 5 red marbles and 3 blue marbles (total 8 marbles). You draw one marble without replacement (meaning you don't put it back in after drawing).

• Event A: Drawing a red marble.
• Event B: Drawing a marble that is not green (since there are no green marbles, this includes both red and blue).

Let's say you first draw a marble and it's not green (either red or blue). Now, you want to know the probability of the next marble you draw being red.

Here's how to use the formula:

1. P(A∩B): After drawing the first marble (not green), there are only 7 marbles left (since you didn't put it back). There are still 5 red marbles remaining. So, the probability of drawing a red marble and a non-green marble (which has already happened) is P(A∩B) = 5/7.

2. P(B): We know from the beginning that there are 8 total marbles and none are green, so P(B) = the probability of drawing a non-green marble = 8/8 (all marbles are not green).

3. P(A|B): Now we can plug these values into the formula:

   P(A|B) = (5/7) / (8/8)

   Simplifying, P(A|B) = 5/7.

Therefore, given that you already drew a non-green marble (event B), the probability of the next marble being red (event A) is 5/7.

Key Points:

• The order of events can matter for conditional probability. In this example, knowing you already drew a non-green marble affects the probability of the next draw being red.
• The probability of event A can change depending on whether you know event B has already happened or not.
• The value of P(B) must be greater than zero for the formula to be valid.

Civil Engineer Example:

Imagine testing soil strength (event A) at a construction site. A civil engineer might be interested in the conditional probability of encountering weak foundation (event B) given poor soil strength. This information helps assess potential risks and design appropriate foundation solutions. Knowing weak soil strength increases the probability of encountering a weak foundation compared to a scenario with strong soil.

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Problem 5:

A fair coin is tossed 6 times. Find the probability of getting exactly 4 heads using Binomial distribution.

Solution:

The Binomial distribution is a statistical method that calculates the probability of getting a specific number of successes in a series of independent trials, where each trial has only two possible outcomes (success or failure). In this case, tossing a fair coin once has two outcomes: heads (success) and tails (failure).

We can use the Binomial theorem to calculate the probability of getting exactly 4 heads in 6 coin tosses. Here's how:

Steps to solve using Binomial Distribution:

1. Define the number of coin tosses (n) and number of heads (k):

   • n = 6 (number of coin tosses)
   • k = 4 (number of heads we're interested in)

2. Probability of getting a head (p) in a fair coin toss:

   • p = 0.5 (assuming a fair coin)

3. Apply the Binomial probability formula:

   • P(k heads) = (n! / (k! * (n-k)!)) * p^k * (1-p)^(n-k)
     • n! (factorial of n) = n * (n-1) * (n-2) * ... * 1
     • k! (factorial of k) = k * (k-1) * (k-2) * ... * 1

4. Calculate the probability:

   • P(4 heads) = (6! / (4! * 2!)) * 0.5^4 * (1-0.5)^(6-4) = 15 * 0.0625 * 0.25 = 0.2344 (approximately)

Answer:

The probability of getting exactly 4 heads in 6 coin tosses using the Binomial distribution is approximately 0.2344.

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Problem 6:

A car hire firm has two cars which it hires out day by day. The no. of demands for a car on each day is distributed as Poisson distribution with mean 1.5. Calculate the proportion of days on which there is no demand.

Solution:

1. Poisson Distribution and Mean: We are given that the number of demands for a car follows a Poisson distribution with a mean of 1.5 cars per day (λ = 1.5).

2. Probability of No Demand: In a Poisson distribution, the probability of getting exactly k events (demands for a car in this case) is given by:

P(X = k) = (e^-λ * λ^k) / k!

where:

• P(X = k) is the probability of k events.
• e is the mathematical constant (approximately 2.71828).
• λ (lambda) is the mean of the distribution (1.5 in this case).
• k is the number of events (0 in this case for no demand).
• ! (factorial) is the product of all positive integers less than or equal to k.

We want the probability of having no demand (k = 0). So, calculate:

P(X = 0) = (e^-1.5 * 1.5^0) / 0!

Since 0! = 1, the equation simplifies to:

P(X = 0) = e^-1.5

3. Proportion of Days with No Demand: This probability (P(X = 0)) represents the proportion of days on which there is no demand for a car.

Therefore, the proportion of days with no demand is approximately e^-1.5, which is roughly 0.223 (around 22.3%).

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Problem 7:

A non-destructive testing method is used to evaluate concrete cores drilled from a building foundation. The test result can either be a "pass" or "fail." If the probability of a single core passing the test is 0.8, what is the probability of encountering exactly 2 failures in a random sample of 5 cores tested? 

Solution:

Let's define the events:

• F: A core fails the test.
• P: A core passes the test.

We are given that the probability of a single core passing the test (event P) is 0.8. We want to find the probability of encountering exactly 2 failures (2 Fs) in a random sample of 5 cores (n = 5). This can be represented as P(FFPFF).

The binomial theorem tells us that the probability of getting k successes (failures in this case) out of n trials, with a probability of success (failure) p in each trial, is:

P(k successes) = nCr * p^k * (1-p)^(n-k)

where:

• nCr = n choose k, which is the number of combinations of k successes in n trials (can be calculated using factorials).

In this case:

• k = 2 (number of failures)
• n = 5 (number of cores tested)
• p = 0.2 (probability of failure, as success is 0.8)

We need to calculate nCr (5 choose 2), which is 10.

Now we can plug the values into the formula:

P(2 failures) = 10 * (0.2)^2 * (1 - 0.2)^(5-2) = 10 * 0.04 * 0.64 = 0.0512

Therefore, the probability of encountering exactly 2 failures in a random sample of 5 cores tested is approximately 0.0512 or 5.12% approximately.

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Problem 8:

A city gets rain on average 2 days a week. What is the chance of exactly 3 rainy days in a week? (Use Poisson distribution

Solution:

Certainly, the Poisson distribution is well-suited to model this situation where the number of events (rainy days) in a fixed interval (a week) follows an independent and identically distributed pattern.

Here's how to solve the problem using the Poisson distribution:

1: Identify the parameters

• The average number of rainy days per week (lambda) is given as 2.

2: Define the event of interest

• We need to find the probability of exactly 3 rainy days (k = 3) in a week.

3: Apply the Poisson probability formula

The probability of k rainy days in a week is given by the Poisson probability mass function (PMF):

P(k rainy days) = (e^-lambda * lambda^k) / k!

where:

• e = base of the natural logarithm (approximately 2.718)
• lambda = average number of rainy days (2 in this case)
• k = number of rainy days we're interested in (3)
• k! = factorial of k (k * (k-1) * (k-2) ... * 1)

4: Calculate the probability

Plug the values into the formula:

P(3 rainy days) = (e^-2 * 2^3) / 3!

= (2.718^-2 * 8) / 6

= (0.1353 * 8) / 6

= 0.1804 (approximately)

Answer: The chance of exactly 3 rainy days in a week is approximately 0.1804 or 18.04%.

--------------------------------------------------------------------------------------------------------------------------------------------------------------------*Class Disclaimer: Verify Answers using calculators.