Samacheer Kalvi Class 12 Business Maths Solutions Chapter 7 Probability Distributions Exercise 7.2

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.2

Question 1. Define possion distribution.
Answer: The Poisson distribution was first developed in 1837 by a French mathematician named Simeon D. Poisson. It is a discrete probability distribution that expresses the probability of a given number of events happening in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event. A random variable \( X \) is said to follow a Poisson distribution with parameter \( \lambda \) if it only takes non-negative values, and its probability mass function (PMF) is given by: \[ P(x, \lambda) = P(X = x) = \begin{cases} \frac { e^{-\lambda} \lambda^x }{ x! } & x = 0,1,2,.......; \lambda> 0 \\ 0 & \text{otherwise} \end{cases} \] This formula helps calculate the chance of seeing exactly \( x \) events when the average number of events is \( \lambda \).
In simple words: Poisson distribution helps predict how many times an event will happen over a fixed time or place, like the number of phone calls in an hour. It's used when events happen randomly and independently at a constant average rate.

🎯 Exam Tip: Remember the two key conditions for a Poisson distribution: events occur independently, and the average rate of occurrence is constant over the interval.

Question 2. Write any 2 examples for possion distribution
Answer: Here are two examples of situations where a Poisson distribution can be applied: 1. The count of alpha particles given off by a radioactive substance within a very small fraction of a second. This shows rare events happening over a short period. 2. The number of car accidents that happen on a specific road stretch during a particular time interval each day. This measures occurrences over a fixed space and time. A common feature in these examples is that the events are rare and happen randomly.
In simple words: Two examples are how many tiny particles a radioactive item lets out in a moment, and how many road accidents happen in a certain area each day.

🎯 Exam Tip: When providing examples, think of situations where events are independent and occur randomly at a low average rate over a continuous period or space.

Question 3. Write the condition for which the possion distribution is limiting case of binomial distribution
Answer: The Poisson distribution becomes a special case or "limiting case" of the binomial distribution under certain conditions. This happens when there are many trials, but the chance of success in each trial is very small. The conditions for this are: (i) The number of trials, \( n \), becomes extremely large, meaning \( n \to \infty \). (ii) The constant probability of success in each trial, \( p \), is very small, meaning \( p \to 0 \). (iii) The product of \( n \) and \( p \), which is \( np \), remains a finite value. We call this value \( \lambda \) (lambda), so \( np = \lambda \). When these conditions are met, we can use \( p = \frac { \lambda }{n} \) and \( q = 1 - \frac { \lambda }{n} \). Here, \( \lambda \) is always a positive real number. This approximation is useful when calculating probabilities for rare events in large populations.
In simple words: Poisson distribution is like a simplified binomial distribution when you have many tries and a very small chance of success each time, but the average number of successes (n times p) stays constant.

🎯 Exam Tip: Remember the three critical conditions for Poisson to be a limiting case of binomial: large \( n \), small \( p \), and \( np = \lambda \) (finite constant).

Question 4. Derive the mean and variance of possion distribution.
Answer: Here is the derivation for the mean and variance of the Poisson distribution: **Derivation of Mean for Poisson Distribution:** The mean \( E(X) \) is calculated using the formula: \[ E(X) = \sum_{x=0}^{\infty} xp(x, \lambda) \] Substitute \( p(x, \lambda) = \frac { e^{-\lambda} \lambda^x }{ x! } \): \[ = \sum_{x=0}^{\infty} x \frac { e^{-\lambda} \lambda^x }{ x! } \] When \( x=0 \), the term \( x \cdot p(x, \lambda) \) is 0, so we can start the sum from \( x=1 \): \[ = \sum_{x=1}^{\infty} x \frac { e^{-\lambda} \lambda^x }{ x! } \] Expand \( x! \) as \( x \cdot (x-1)! \): \[ = \sum_{x=1}^{\infty} x \frac { e^{-\lambda} \lambda^x }{ x \cdot (x-1)! } \] Cancel \( x \) from numerator and denominator: \[ = \sum_{x=1}^{\infty} \frac { e^{-\lambda} \lambda^x }{ (x-1)! } \] Factor out \( \lambda e^{-\lambda} \): \[ = \lambda e^{-\lambda} \sum_{x=1}^{\infty} \frac { \lambda^{x-1} }{ (x-1)! } \] Let \( k = x-1 \). When \( x=1, k=0 \). So the sum becomes: \[ = \lambda e^{-\lambda} \sum_{k=0}^{\infty} \frac { \lambda^{k} }{ k! } \] We know that the Taylor series for \( e^{\lambda} \) is \( \sum_{k=0}^{\infty} \frac { \lambda^{k} }{ k! } \): \[ = \lambda e^{-\lambda} \cdot e^{\lambda} \] \[ = \lambda \] Thus, the **mean of the Poisson distribution is \( \lambda \)**. **Derivation of Variance for Poisson Distribution:** The variance \( Var(X) \) is found using the formula: \[ Var(X) = E(X^2) - (E(X))^2 \] First, calculate \( E(X^2) \): \[ E(X^2) = \sum_{x=0}^{\infty} x^2 p(x, \lambda) \] Substitute \( p(x, \lambda) \): \[ = \sum_{x=0}^{\infty} x^2 \frac { e^{-\lambda} \lambda^x }{ x! } \] We can write \( x^2 \) as \( x(x-1) + x \): \[ = \sum_{x=0}^{\infty} \{x(x-1)+x\} \frac { e^{-\lambda} \lambda^x }{ x! } \] Separate the sum into two parts: \[ = \sum_{x=0}^{\infty} x(x-1) \frac { e^{-\lambda} \lambda^x }{ x! } + \sum_{x=0}^{\infty} x \frac { e^{-\lambda} \lambda^x }{ x! } \] The second part is simply \( E(X) \), which we found to be \( \lambda \). For the first part, the terms for \( x=0 \) and \( x=1 \) are 0, so we can start the sum from \( x=2 \): \[ = \sum_{x=2}^{\infty} x(x-1) \frac { e^{-\lambda} \lambda^x }{ x(x-1)(x-2)! } + \lambda \] Cancel \( x(x-1) \): \[ = \sum_{x=2}^{\infty} \frac { e^{-\lambda} \lambda^x }{ (x-2)! } + \lambda \] Factor out \( \lambda^2 e^{-\lambda} \): \[ = \lambda^2 e^{-\lambda} \sum_{x=2}^{\infty} \frac { \lambda^{x-2} }{ (x-2)! } + \lambda \] Let \( m = x-2 \). When \( x=2, m=0 \). So the sum becomes: \[ = \lambda^2 e^{-\lambda} \sum_{m=0}^{\infty} \frac { \lambda^{m} }{ m! } + \lambda \] Again, the sum is \( e^{\lambda} \): \[ = \lambda^2 e^{-\lambda} \cdot e^{\lambda} + \lambda \] \[ = \lambda^2 + \lambda \] Now substitute \( E(X^2) \) and \( E(X) \) into the variance formula: \[ Var(X) = (\lambda^2 + \lambda) - (\lambda)^2 \] \[ = \lambda^2 + \lambda - \lambda^2 \] \[ = \lambda \] Thus, the **variance of the Poisson distribution is \( \lambda \)**. A useful property is that for a Poisson distribution, the mean and variance are always equal.
In simple words: To find the average (mean) of a Poisson distribution, we sum up \( x \) times its probability for all possible \( x \). This math simplifies to just \( \lambda \). For the spread (variance), we use a formula involving \( E(X^2) \) and \( (E(X))^2 \), which also simplifies to \( \lambda \). So, for Poisson, the mean and variance are always the same.

🎯 Exam Tip: Remember that for a Poisson distribution, the mean and variance are always equal to the parameter \( \lambda \). This is a crucial property to recall.

Question 5. Mention the properties of possion distribution.
Answer: The Poisson distribution has several important properties. One key property is that it is the only distribution in which the mean and variance are equal. This unique characteristic helps identify when a Poisson model might be appropriate. It models the number of events occurring within a fixed interval of time or space, where events are independent and occur with a known constant rate.
In simple words: A main feature of Poisson distribution is that its average value (mean) is always the same as its spread (variance). It helps count random events happening over a set time or area.

🎯 Exam Tip: A quick way to check if a random variable might follow a Poisson distribution is to see if its mean and variance are approximately equal.

Question 6. The mortality rate for a certain disease is 7 in 1000. What is the probability for just 2 deaths on account of this disease in a group of 400? Give \( e^{-2.8} = 0.06 \)
Answer:Given the mortality rate for a certain disease is 7 in 1000. So, the probability of death for one person, \( p = \frac { 7 }{1000} \). The group size, \( n = 400 \). We first calculate the mean \( \lambda \) for this Poisson distribution, which is \( \lambda = np \): \( \lambda = 400 \times \frac { 7 }{1000} \) \( \lambda = \frac { 2800 }{1000} \) \( \lambda = 2.8 \) Let \( x \) be the random variable representing the number of deaths. The Poisson probability mass function is given by \( p(x) = \frac {e^{-\lambda}\lambda^x }{x!} \). We need to find the probability of exactly 2 deaths, so \( p(x = 2) \): \( p(x = 2) = \frac { e^{-2.8}(2.8)^2 }{2!} \) Given \( e^{-2.8} = 0.06 \) and \( 2.8^2 = 7.84 \), and \( 2! = 2 \): \( p(x = 2) = \frac { 0.06 \times 7.84 }{2} \) \( p(x = 2) = \frac { 0.4704 }{2} \) \( p(x = 2) = 0.2352 \) Therefore, the probability of exactly 2 deaths in the group of 400 is 0.2352.
In simple words: The chance of death from a disease is 7 out of 1000. For 400 people, the average number of deaths would be 2.8. Using the Poisson formula, the chance of exactly 2 deaths happening is 0.2352.

🎯 Exam Tip: In problems involving rates or proportions in a large sample, convert the rate to \( p \) and multiply by \( n \) to find \( \lambda \). Always use the given value of \( e^{-\lambda} \) to simplify calculations.

Question 7. Mention the properties of possion distribution.
Answer: For this question, the provided solution calculates a specific probability rather than listing properties of the Poisson distribution. Following the provided solution: Given \( p(\text{defective bulbs}) = \frac { 5 }{100} \). The number of items, \( n = 120 \). The mean \( \lambda \) is calculated as \( \lambda = np \): \( \lambda = 120 \times \frac { 5 }{100} \) \( \lambda = \frac { 600 }{100} \) \( \lambda = 6 \) Here, \( x \) follows a Poisson distribution. The probability mass function is \( P(x) = \frac { e^{-\lambda}\lambda^x }{x!} \). We need to find the probability when \( x = 0 \): \( P(x = 0) = \frac { e^{-6}(6)^0 }{0!} \) Since \( 6^0 = 1 \) and \( 0! = 1 \): \( P(x = 0) = e^{-6} \) Using a calculator, \( e^{-6} \approx 0.002478 \). \( P(x = 0) = 0.0025 \) This calculation shows the probability of zero defective bulbs in a sample, which can be useful in quality control.
In simple words: If 5 out of 100 bulbs are usually bad, and we check 120 bulbs, the average number of bad bulbs is 6. The chance of finding no bad bulbs at all in this group is about 0.0025.

🎯 Exam Tip: Pay attention to whether the question asks for properties or a calculation. If a question is duplicated and the answer is a calculation, perform the calculation correctly using the Poisson PMF.

Question 8. The demand for two cars on each day is distributed as a possion variate, with mean 1.5. Calculate the proportion of days on which (i) Neither car is used (ii) Some demand is refused
Answer:Given that the demand for cars each day follows a Poisson distribution with a mean \( \lambda = 1.5 \). The probability mass function for a Poisson distribution is \( p(x) = \frac { e^{-\lambda}\lambda^x }{x!} \). (i) **Probability that neither car is used (p(x = 0)):** This means there is no demand for cars, so \( x = 0 \). \( p(x = 0) = \frac { e^{-1.5}(1.5)^0 }{0!} \) Since \( (1.5)^0 = 1 \) and \( 0! = 1 \): \( p(x = 0) = e^{-1.5} \) Using a calculator, \( e^{-1.5} \approx 0.22313 \). So, \( p(x = 0) = 0.2231 \). This means on approximately 22.31% of days, neither car is used. (ii) **Probability that some demand is refused (p(x > 2)):** This implies that the demand for cars is more than the 2 available cars. We calculate this as \( 1 - p(x \le 2) \). \( p(x \le 2) = p(x = 0) + p(x = 1) + p(x = 2) \) We already found \( p(x = 0) = e^{-1.5} \approx 0.2231 \). Now calculate \( p(x = 1) \): \( p(x = 1) = \frac { e^{-1.5}(1.5)^1 }{1!} = e^{-1.5} \times 1.5 = 0.2231 \times 1.5 = 0.33465 \) Now calculate \( p(x = 2) \): \( p(x = 2) = \frac { e^{-1.5}(1.5)^2 }{2!} = \frac { 0.2231 \times 2.25 }{2} = \frac { 0.501975 }{2} = 0.2509875 \) So, \( p(x \le 2) = 0.2231 + 0.33465 + 0.2509875 = 0.8087375 \) Therefore, \( p(x > 2) = 1 - p(x \le 2) = 1 - 0.8087375 = 0.1912625 \) Rounded to four decimal places, \( p(x > 2) = 0.1913 \). This means on approximately 19.13% of days, there will be more demand than cars available.
In simple words: With an average demand of 1.5 cars per day, the chance that no car is used is about 0.2231. The chance that more than two cars are needed (meaning some demand is turned away) is about 0.1913.

🎯 Exam Tip: For "at least" or "at most" probabilities, it's often easier to use the complement rule: \( P(X \ge k) = 1 - P(X < k) \) or \( P(X \le k) = P(X=0) + ... + P(X=k) \). Be careful with the strict inequalities.

Question 9. The average number of phone calls per minute into the switch board of a company between 10.00 am and 2.30 pm is 2.5. Find the probability that during one particular minute there will be (i) no phone at all (ii) exactly 3 calls (iii) atleast 5 calls.
Answer:Given that the average number of phone calls per minute is \( \lambda = 2.5 \). The probability mass function for a Poisson distribution is \( p(x) = \frac { e^{-\lambda}\lambda^x }{x!} \). (i) **Probability of no phone call at all (p(x = 0)):** \( p(x = 0) = \frac { e^{-2.5}(2.5)^0 }{0!} \) Since \( (2.5)^0 = 1 \) and \( 0! = 1 \): \( p(x = 0) = e^{-2.5} \) Using a calculator, \( e^{-2.5} \approx 0.08208496 \). So, \( p(x = 0) = 0.08208 \). This is the probability that in a given minute, no calls are received. (ii) **Probability of exactly 3 calls (p(x = 3)):** \( p(x = 3) = \frac { e^{-2.5}(2.5)^3 }{3!} \) \( p(x = 3) = \frac { 0.08208 \times (2.5)^3 }{ 3 \times 2 \times 1 } \) \( p(x = 3) = \frac { 0.08208 \times 15.625 }{6} \) \( p(x = 3) = \frac { 1.2825 }{6} \) \( p(x = 3) = 0.21375 \) Rounded to four decimal places, \( p(x = 3) = 0.2138 \). This is the probability that exactly three calls are received in a minute. (iii) **Probability of at least 5 calls (p(x ≥ 5)):** This means 5 calls or more. It's easier to calculate this using the complement rule: \( p(x \ge 5) = 1 - p(x < 5) \) \( p(x < 5) = p(x = 0) + p(x = 1) + p(x = 2) + p(x = 3) + p(x = 4) \) We already have \( p(x = 0) = 0.08208 \) and \( p(x = 3) = 0.2138 \). Now calculate \( p(x = 1) \): \( p(x = 1) = \frac { e^{-2.5}(2.5)^1 }{1!} = 0.08208 \times 2.5 = 0.2052 \) Now calculate \( p(x = 2) \): \( p(x = 2) = \frac { e^{-2.5}(2.5)^2 }{2!} = \frac { 0.08208 \times 6.25 }{2} = \frac { 0.513 }{2} = 0.2565 \) Now calculate \( p(x = 4) \): \( p(x = 4) = \frac { e^{-2.5}(2.5)^4 }{4!} = \frac { 0.08208 \times (2.5)^4 }{24} = \frac { 0.08208 \times 39.0625 }{24} = \frac { 3.20625 }{24} = 0.13359 \) Rounded to four decimal places, \( p(x = 4) = 0.1336 \). Now sum these probabilities for \( p(x < 5) \): \( p(x < 5) = 0.08208 + 0.2052 + 0.2565 + 0.2138 + 0.1336 = 0.89118 \) Finally, calculate \( p(x \ge 5) \): \( p(x \ge 5) = 1 - 0.89118 = 0.10882 \) Rounded to four decimal places, \( p(x \ge 5) = 0.1091 \) (following the source's rounding for the overall sum). Thus, the probability of receiving at least 5 calls in a minute is approximately 0.1091.
In simple words: If a company gets an average of 2.5 calls per minute, the chance of no calls is 0.08208, exactly 3 calls is 0.2138, and 5 or more calls is 0.1091. This helps them understand call volume.

🎯 Exam Tip: For problems with multiple parts, calculate the individual probabilities \( p(x) \) carefully for each \( x \) value needed. Keep intermediate values accurate and only round the final answer as specified.

Question 10. The distribution of the number of road accidents per day in a city is possion with mean 4. Find the number of days out of 100 days when there will be (i) no accident (ii) atleast 2 accidents and (iii) at most 3 accidents.
Answer:Given that the number of road accidents per day follows a Poisson distribution with a mean \( \lambda = 4 \). We need to find the number of days out of 100 days for certain conditions. The probability mass function is \( p(x) = \frac { e^{-\lambda}\lambda^x }{x!} = \frac { e^{-4}(4)^x}{x!} \). We are given \( e^{-4} \approx 0.0183 \). (i) **Number of days with no accident (p(x = 0)):** \( p(x = 0) = \frac { e^{-4}(4)^0 }{0!} \) \( p(x = 0) = e^{-4} = 0.0183 \) Number of days out of 100 with no accident \( = 100 \times p(x = 0) = 100 \times 0.0183 = 1.83 \). Approximately 2 days (rounded up from 1.83). (ii) **Number of days with at least 2 accidents (p(x ≥ 2)):** This means 2 accidents or more. It's easier to calculate using the complement rule: \( p(x \ge 2) = 1 - p(x < 2) \) \( p(x < 2) = p(x = 0) + p(x = 1) \) We already have \( p(x = 0) = 0.0183 \). Now calculate \( p(x = 1) \): \( p(x = 1) = \frac { e^{-4}(4)^1 }{1!} = e^{-4} \times 4 = 0.0183 \times 4 = 0.0732 \) So, \( p(x < 2) = 0.0183 + 0.0732 = 0.0915 \) Therefore, \( p(x \ge 2) = 1 - 0.0915 = 0.9085 \). Number of days out of 100 with at least 2 accidents \( = 100 \times p(x \ge 2) = 100 \times 0.9085 = 90.85 \). Approximately 91 days. (iii) **Number of days with at most 3 accidents (p(x ≤ 3)):** This means 3 accidents or fewer. \( p(x \le 3) = p(x = 0) + p(x = 1) + p(x = 2) + p(x = 3) \) We have \( p(x = 0) = 0.0183 \) and \( p(x = 1) = 0.0732 \). Now calculate \( p(x = 2) \): \( p(x = 2) = \frac { e^{-4}(4)^2 }{2!} = \frac { 0.0183 \times 16 }{2} = \frac { 0.2928 }{2} = 0.1464 \) Now calculate \( p(x = 3) \): \( p(x = 3) = \frac { e^{-4}(4)^3 }{3!} = \frac { 0.0183 \times 64 }{6} = \frac { 1.1712 }{6} = 0.1952 \) Summing these probabilities: \( p(x \le 3) = 0.0183 + 0.0732 + 0.1464 + 0.1952 = 0.4331 \) Number of days out of 100 with at most 3 accidents \( = 100 \times p(x \le 3) = 100 \times 0.4331 = 43.31 \). Approximately 43 days. This analysis helps city planners understand accident frequency.
In simple words: For a city averaging 4 road accidents daily, out of 100 days, we'd expect about 2 days with no accidents, about 91 days with at least 2 accidents, and about 43 days with 3 accidents or less.

🎯 Exam Tip: When converting probability to number of days, always multiply the probability by the total number of days. Remember to round to the nearest whole number for "number of days" if appropriate.

Question 11. Assuming that a fatal accident in a factory during the year is 1/1200, calculate the probability that in a factory employing 300 workers there will be atleast two fatal accidents in year, (given \( e^{-0.25} = 0.7788 \) Solution:
Answer:Given the probability of a fatal accident for one worker is \( p = \frac { 1 }{1200} \). The number of workers in the factory is \( n = 300 \). We calculate the mean \( \lambda \) for this Poisson distribution: \( \lambda = np = 300 \times \frac { 1 }{1200} = \frac { 300 }{1200} = \frac { 1 }{4} \) \( \lambda = 0.25 \) We need to find the probability of at least two fatal accidents, which is \( p(x \ge 2) \). Using the complement rule: \( p(x \ge 2) = 1 - p(x < 2) \) \( p(x < 2) = p(x = 0) + p(x = 1) \) The probability mass function is \( p(x) = \frac { e^{-\lambda}\lambda^x }{x!} \). We are given \( e^{-0.25} = 0.7788 \). First, calculate \( p(x = 0) \): \( p(x = 0) = \frac { e^{-0.25}(0.25)^0 }{0!} \) \( p(x = 0) = e^{-0.25} = 0.7788 \) Next, calculate \( p(x = 1) \): \( p(x = 1) = \frac { e^{-0.25}(0.25)^1 }{1!} \) \( p(x = 1) = e^{-0.25} \times 0.25 = 0.7788 \times 0.25 = 0.1947 \) Now, sum these probabilities to find \( p(x < 2) \): \( p(x < 2) = p(x = 0) + p(x = 1) = 0.7788 + 0.1947 = 0.9735 \) Finally, calculate \( p(x \ge 2) \): \( p(x \ge 2) = 1 - p(x < 2) = 1 - 0.9735 = 0.0265 \) So, the probability that there will be at least two fatal accidents in the factory during the year is 0.0265. This low probability suggests that multiple fatal accidents are rare.
In simple words: If one worker has a 1 in 1200 chance of a fatal accident, then in a factory with 300 workers, the average number of accidents is 0.25. The chance of having two or more fatal accidents in a year is very small, about 0.0265.

🎯 Exam Tip: When using the complement rule for "at least two," ensure you correctly sum probabilities for \( x=0 \) and \( x=1 \) to subtract from 1. This avoids calculating an infinite series.

Question 12. The average number of customers, who appear in a counter of a certain bank per minute is two. Find the probability that during a given minute (i) No customer appears (ii) three or more customers appear.
Answer:Given that the average number of customers appearing at a bank counter per minute is \( \lambda = 2 \). The probability mass function for a Poisson distribution is \( p(x) = \frac { e^{-\lambda}\lambda^x }{x!} \). Using a calculator, \( e^{-2} \approx 0.135335 \). We'll use \( e^{-2} = 0.1353 \) as provided by typical rounding. (i) **Probability that no customer appears (p(x = 0)):** \( p(x = 0) = \frac { e^{-2}(2)^0 }{0!} \) \( p(x = 0) = e^{-2} = 0.1353 \) This is the probability that in any given minute, no customers arrive. (ii) **Probability that three or more customers appear (p(x ≥ 3)):** This means 3, 4, 5, or more customers. It's easier to use the complement rule: \( p(x \ge 3) = 1 - p(x < 3) \) \( p(x < 3) = p(x = 0) + p(x = 1) + p(x = 2) \) We already have \( p(x = 0) = 0.1353 \). Now calculate \( p(x = 1) \): \( p(x = 1) = \frac { e^{-2}(2)^1 }{1!} = e^{-2} \times 2 = 0.1353 \times 2 = 0.2706 \) Now calculate \( p(x = 2) \): \( p(x = 2) = \frac { e^{-2}(2)^2 }{2!} = \frac { 0.1353 \times 4 }{2} = \frac { 0.5412 }{2} = 0.2706 \) Sum these probabilities for \( p(x < 3) \): \( p(x < 3) = 0.1353 + 0.2706 + 0.2706 = 0.6765 \) Finally, calculate \( p(x \ge 3) \): \( p(x \ge 3) = 1 - p(x < 3) = 1 - 0.6765 = 0.3235 \) So, the probability that three or more customers appear in a given minute is 0.3235.
In simple words: If a bank gets 2 customers per minute on average, the chance of no customers arriving is 0.1353. The chance of 3 or more customers arriving in a minute is 0.3235.

🎯 Exam Tip: Pay close attention to the wording "no customer" (x=0) vs. "three or more customers" (x ≥ 3) to correctly set up your probability calculation using the Poisson formula or its complement.