Assignment II PART C

 

Solved Problems and Explained Theory

1. Probability of White Ball:

Two balls are drawn from the first urn (10 white, 3 black) and placed in the second urn (3 white, 5 black). Then, one ball is drawn from the second urn. We need the probability it's white.

Solution:

There are two cases to consider:

Case 1: Both balls drawn from the first urn are white (10C2 ways).

• Probability: (10C2) / (13C2) = 45 / 78

Case 2: One white and one black ball are drawn from the first urn (10C1 * 3C1 ways).

• Probability of white in second draw (considering the white ball was placed first): (3 + 1) / (8) = 4 / 8
• Probability of white in second draw (considering the black ball was placed first): (3) / (8)

Total Probability (considering both cases):

P(white ball) = (45/78) * 1 + (10/78) * (4/8) + (10/78) * (3/8) = 225/780

2. Bayes' Theorem and White Balls:

a) Bayes' Theorem:

Bayes' theorem is a statistical method to calculate the posterior probability of an event (hypothesis) based on prior knowledge (prior probability) and related evidence (likelihood). It helps update probabilities when new information becomes available.

b) Probability of All White Balls:

Information:

• Bag contains 5 balls (unknown number of white balls).
• 2 balls are drawn at random, and both are white.

Let X be the event of a white ball in the bag.

Prior Probability (difficult to know without additional information):

• We can assume equal probabilities for all possibilities (5 white, 4 white-1 black, etc.). Let P(X) = 1/5 (assuming all possibilities are equally likely).

Likelihood (probability of drawing 2 white balls given X is true):

• If all balls are white, the probability of drawing 2 white balls is 1 (certainty). So, P(data | X) = 1 (data = drawing 2 white balls).

Posterior Probability (what we need to find):

• P(X | data) = (P(data | X) * P(X)) / P(data)

Calculating P(data): This involves considering all possibilities (all white, 4 white-1 black, etc.) and their probabilities of drawing 2 white balls. However, it can be computationally expensive.

In practice, without strong prior information, a definitive answer might be challenging. We can only say that the evidence (drawing 2 white balls) strengthens the case for all white balls, but the exact probability depends on the unknown prior probabilities.

3. Bernoulli's Theorem and Defective Screws:

We previously discussed Bernoulli's Theorem and the concept of repeated independent trials with only two possible outcomes (success and failure). Now, let's apply it to find the probability of defective screws produced by an automatic machine.

Information:

• 10% of screws produced are defective (p = 0.1).
• We need to find the probability for various scenarios with 20 screws selected at random.

Bernoulli's theorem ensures independence, meaning the outcome of selecting one defective screw doesn't influence the selection of another.

Solutions:

i) Exactly 2 Defective:

This is a classic binomial distribution problem. Here's the formula:

P(k) = (nCk) * p^k * (1-p)^(n-k)

where:

• P(k): Probability of k successes (defective screws)
• n: Total number of trials (screws) = 20
• k: Number of successes (defective screws) = 2
• p: Probability of success (defective screw) = 0.1

Plug in the values:

P(k = 2) = (20C2) * (0.1)² * (0.9)¹⁸ ≈ 0.286

Therefore, the probability of getting exactly 2 defective screws in 20 selections is approximately 0.286 or 28.6%.

ii) Utmost 3 Defective:

This scenario considers all possibilities from 0 to 3 defectives. You can calculate each probability using the binomial formula and add them up:

P(0 to 3 defectives) = P(k = 0) + P(k = 1) + P(k = 2) + P(k = 3)

• Use software or tables like online binomial calculators to find each individual probability.
• Add the obtained probabilities to get the final answer.

iii) At Least 2 Defectives:

This is the complement of having 0 or 1 defective screw. We can use the following approach:

1. Calculate P(0 defectives) and P(1 defective) using the binomial formula.
2. P(at least 2 defectives) = 1 - (P(0 defectives) + P(1 defective))

iv) Between 1 and 3 Defectives (Inclusive):

This combines scenarios with 1, 2, and 3 defectives:

P(1 to 3 defectives) = P(k = 1) + P(k = 2) + P(k = 3)

Use the binomial formula for each k value and add the resulting probabilities.

Remember:

• Binomial calculators or software can simplify these calculations.
• Round the final probabilities to an appropriate number of decimal places.

4. Poisson Distribution and Phone Calls:

Properties of Poisson Distribution:

• The number of events (e.g., phone calls) in a fixed interval (e.g., minute) is independent.

• The average rate (λ) of events occurring in the interval is constant.

• The probability of exactly k events occurring is given by:

  P(k) = (e^-λ * λ^k) / k!

Phone Call Probabilities:

i) 4 or Fewer Calls:

λ = 2.5 calls/minute (average)

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

Calculate each probability using the Poisson formula and add them up. You can use software or tables for Poisson probabilities.

ii) More Than 6 Calls:

This is the complement of having 6 or fewer calls:

P(more than 6 calls) = 1 - P(k ≤ 6)

Calculate P(k ≤ 6) using the Poisson formula and subtract from 1.

5. Median and Cricket Scores:

a. Finding the Median:

1. Order the Data: Write the numbers in ascending order: 38, 42, 57, 58, 61, 65, 66, 72

2. Identify the Middle Element(s):

   • Since we have 8 numbers (even number), the median is the mean of the two middle numbers.
   • The middle two numbers are: 58 and 61.

3. Calculate the Median: Median = (58 + 61) / 2 = 119 / 2 = 59.5

Therefore, the median is 59.5.

b. Cricketers' Analysis:

1. Calculate Average Scores:

   • Add up A's scores and divide by 10 (number of games).
   • Add up B's scores and divide by 10.

   Average for A (Avg_A) = (40 + 25 + 19 + 80 + 38 + 8 + 67 + 121 + 66 + 76) / 10 = 541 / 10 = 54.1

   Average for B (Avg_B) = (28 + 70 + 31 + 0 + 14 + 111 + 66 + 31 + 25 + 4) / 10 = 379 / 10 = 37.9

2. Compare Averages:

   • Avg_A (54.1) is higher than Avg_B (37.9).

3. Calculate Standard Deviations:

   • You'll need a statistical calculator or software to find the standard deviations (SD) for A's and B's scores.
   • Standard deviation measures how spread out the data points are from the average.

4. Compare Standard Deviations (Lower SD indicates more consistency):

   • If SD_A (A's standard deviation) is lower than SD_B (B's standard deviation), A is more consistent.
   • If SD_B is lower, then B is more consistent.

Based on the calculated averages, Cricketer A is the better runner. You'll need standard deviations to determine who is more consistent.

6. Variance and Standard Deviation:

a) Variance:

i) Data: 4, 5, 2, 8, 7

Steps:

1. Calculate the mean (average): (4 + 5 + 2 + 8 + 7) / 5 = 5.2
2. Find the squared deviations from the mean for each value.
3. Calculate the variance: (average of squared deviations from the mean)

Variance = [(0.04) + (0.04) + (10.24) + (7.84) + (3.24)] / 5 ≈ 4.28

ii) Data: 6, 7, 10, 12, 13, 4, 8, 12

Follow the same steps as above to calculate the mean, squared deviations, and variance.

b) Standard Deviation:

Test Scores: [22, 99, 102, 33, 57]

1. Calculate the variance (as shown in part a).
2. Standard deviation (SD) is the square root of the variance:

SD = √(variance)

7. Binomial Distribution and Customer Arrivals:

a) Coin Toss (Binomial Distribution):

i) Exactly 2 Heads:

n = 5 trials (coin tosses) p = 0.5 (probability of heads) k = 2 successes (heads)

P(k = 2) = (5C2) * (0.5)² * (0.5)³ = 0.3125

ii) At Least 4 Heads:

This is the complement of having 0, 1, 2, or 3 heads:

P(at least 4 heads) = 1 - (P(k = 0) + P(k = 1) + P(k = 2) + P(k = 3))

Calculate each probability using the binomial formula and subtract from 1.

b) Customer Arrivals (Poisson Distribution):

Mean Arrival Rate (λ):

2 customers per minute (average)

Probability of 5 Customers:

P(k = 5) = (e^-λ * λ^k) / k!

= (e^-2 * 2^5) / 5! ≈ 0.082 (Verify Answer with Calculator)

8. Binomial vs. Poisson Distribution:

Binomial Distribution:

• Deals with a fixed number of independent trials (n).
• Each trial has only two possible outcomes (success/failure).
• The probability of success (p) remains constant throughout the trials.
• Used for situations like coin tosses, dice rolls, or passing/failing an exam on a single attempt.

Poisson Distribution:

• Models the number of events (e.g., phone calls, car accidents) occurring in a fixed interval (e.g., hour, day).
• Events are independent (occurrence of one doesn't affect another).
• The average rate (λ) of events occurring in the interval is constant.
• Used for situations like customer arrivals, machine breakdowns, or insurance claims within a specific time frame.

Examples:

Binomial:

• Tossing a coin 5 times and getting exactly 2 heads.
• Rolling a die 3 times and getting at least one 6.

Poisson:

• The number of customers entering a store in 10 minutes.
• The number of accidents occurring on a highway in a day.
• The number of website visits within an hour.

Key Differences:

Feature	Binomial Distribution	Poisson Distribution
Trial Type	Fixed number of trials	Events occur over a fixed interval
Outcomes	Only two (success/fail)	Can have more than two outcomes
Probability (p)	Constant throughout	Average rate (λ) of events
Applications	Repeated trials with two outcomes	Events occurring in a fixed interval