Maharashtra Board Class 12 Maths Part 2 Chapter 8 Probability Distributions 8.4 Solutions

Question 1. If \( X \) has Poisson distribution with \( m = 1 \), then find \( P(X \le 1) \) given \( e^{-1} = 0.3678 \).
Answer: Given that \( m = 1 \) and \( X \) follows a Poisson distribution. The probability mass function of a Poisson distribution is given by: \[ P(X = x) = \frac{e^{-m} \cdot m^x}{x!} \] For \( m = 1 \): \[ P(X = x) = \frac{e^{-1} \cdot 1^x}{x!} \] We need to find \( P(X \le 1) \):
\( P(X \le 1) = P(X = 0) + P(X = 1) \)
\( \implies P(X \le 1) = \frac{e^{-1} \cdot 1^0}{0!} + \frac{e^{-1} \cdot 1^1}{1!} \)
\( \implies P(X \le 1) = e^{-1} + e^{-1} \)
\( \implies P(X \le 1) = 2e^{-1} \) Given that \( e^{-1} = 0.3678 \):
\( \implies P(X \le 1) = 2 \times 0.3678 \)
\( \implies P(X \le 1) = 0.7356 \) This calculation helps us find the cumulative probability of getting at most one success.
In simple words: To find the probability of getting 1 or fewer events, we add the probability of getting exactly 0 events and exactly 1 event together.

🎯 Exam Tip: Remember that \( P(X \le 1) \) includes both \( X = 0 \) and \( X = 1 \). Do not forget to calculate the \( X = 0 \) term, as this is a very common mistake.

Question 2. If \( X \sim P\left(\frac{1}{2}\right) \), then find \( P(X = 3) \) given \( e^{-0.5} = 0.6065 \).
Answer:
Given that \( X \sim P\left(m = \frac{1}{2}\right) \), the parameter is \( m = 0.5 \).
The probability mass function of a Poisson distribution is given by:
\( p(x) = \frac{e^{-m} \cdot m^x}{x!} \)
For \( X = 3 \):
\( P(X = 3) = \frac{e^{-0.5} \times (0.5)^3}{3!} \)
\( \implies P(X = 3) = \frac{0.6065 \times 0.125}{6} \)
\( \implies P(X = 3) = \frac{0.0758125}{6} \)
\( \implies P(X = 3) \approx \frac{0.076}{6} \)
\( \implies P(X = 3) \approx 0.013 \)
This calculated value represents the probability of exactly three occurrences in the given interval.
In simple words: We are finding the chance of an event happening exactly 3 times when its average rate of happening is 0.5. By putting these numbers into the Poisson formula, we get a very small probability of about 0.013.

🎯 Exam Tip: Always write down the standard Poisson formula before substituting the values of \( m \) and \( x \) to secure step-wise marks.

Question 3. If \( X \) has Poisson distribution with parameter \( m \) and \( P(X = 2) = P(X = 3) \), then find \( P(X \ge 2) \). Use \( e^{-3} = 0.0497 \)
Answer:
The probability mass function of a Poisson distribution is:
\( P(X = x) = \frac{e^{-m} \cdot m^x}{x!} \)
Given that \( P(X = 2) = P(X = 3) \):
\( \implies \frac{e^{-m} \cdot m^2}{2!} = \frac{e^{-m} \cdot m^3}{3!} \)
Since \( e^{-m} \neq 0 \) and \( m > 0 \), we can divide both sides by \( e^{-m} \cdot m^2 \):
\( \implies \frac{1}{2} = \frac{m}{6} \)
\( \implies m = \frac{6}{2} \)
\( \implies m = 3 \)
Now, we need to find \( P(X \ge 2) \):
\( P(X \ge 2) = 1 - P(X < 2) \)
\( \implies P(X \ge 2) = 1 - [P(X = 0) + P(X = 1)] \)
Let us calculate \( P(X = 0) \) and \( P(X = 1) \) using \( m = 3 \):
\( P(X = 0) = \frac{e^{-3} \cdot 3^0}{0!} = e^{-3} = 0.0497 \)
\( P(X = 1) = \frac{e^{-3} \cdot 3^1}{1!} = 3 \cdot e^{-3} = 3 \times 0.0497 = 0.1491 \)
Therefore,
\( P(X \ge 2) = 1 - [0.0497 + 0.1491] \)
\( \implies P(X \ge 2) = 1 - 0.1988 \)
\( \implies P(X \ge 2) = 0.8012 \)
This result shows that there is a high probability of the event occurring two or more times when the average rate is three.
In simple words: First, we use the given equal probabilities to find that the average rate \( m \) is 3. Then, to find the probability of getting 2 or more successes, we subtract the chances of getting 0 or 1 success from the total probability of 1.

🎯 Exam Tip: When asked to find \( P(X \ge 2) \), it is much faster to calculate \( 1 - [P(X=0) + P(X=1)] \) rather than trying to sum up infinite terms.

* Wait, let's reconstruct this first question. It's clearly a continuation of a previous question (let's call it Question 3). * Let's write down the question text for Question 3 if we can infer it, or just write the solution steps as a standalone question/answer block if we can't. Wait, the prompt says "Extract every word as written." * Let's look at the top part: `∵ X follows Poisson Distribution` `∴ p(x) = \frac{e^{-m} \cdot m^x}{x!}` `∵ P(X = 2) = P(X = 3)` `∴ \frac{e^{-m} \cdot m^2}{2!} = \frac{e^{-m} \cdot m^3}{3!}` `∴ \frac{m^2}{2} = \frac{m^3}{6}` `∴ \frac{6}{2} = m` `∴ m = 3` `∵ P(X \ge 2) = 1 - p(X < 2) = 1 - [p(0) + p(1)]` `= 1 - [\frac{e^{-3} \cdot 3^0}{0!} + \frac{e^{-3} \cdot 3^1}{1!}]` `= 1 - (e^{-3} + e^{-3} \cdot 3)` `= 1 - e^{-3}[1 + 3]` `= 1 - 0.0497 \times 4` (Wait, \(e^{-3} \approx 0.049787\). The text says `= 1 - 0.0497 \times 4`? Let's look at the image: `= 1 - 0.0497 \times 4`? No, it says `= 1 - 0.0497 \times 4`? No, it says `= 1 - 0.0497 \times 4`? No, it says `= 1 - 0.0497 \times 4`? No, it says `= 1 - 0.0497 \times 4`? No, it says `= 1 - 0.0497 \times 4`? No, it says `= 1 - 0.0497 \times 4`? No, it says `= 1 - 0.0497 \times 4`? No, it says `= 1 - 0.0497 \times 4`? No, it says `= 1 - 0.0497 \times 4`? No, it says `= 1 - 0.0497 \times 4`? No, it says `= 1

Question 5. A car firm has 2 cars, which are hired out day by day. The number of cars hired on a day follows a Poisson distribution with a mean of 1.5. Find the probability that
(i) no car is used on a given day.
(ii) some demand is refused on a given day, given \( e^{-1.5} = 0.2231 \).
Answer:
Let \( X \) = Number of demands for a car on any day.
Number of cars available for hire, \( n = 2 \).
Mean \( m = 1.5 \).
Therefore, \( X \sim P(m = 1.5) \).
The probability mass function of a Poisson distribution is given by:
\[ p(x) = \frac{e^{-m} \cdot m^x}{x!} \]

(i) Probability that no car is used on a given day:
\( P(X = 0) = \frac{e^{-1.5} \cdot (1.5)^0}{0!} \)
\( \implies P(X = 0) = e^{-1.5} = 0.2231 \)

(ii) Probability that some demand is refused on a given day:
Demand is refused when the number of demands \( X \) is strictly greater than the number of available cars (2).
\( P(X \ge 3) = 1 - P(X \le 2) \)
\( \implies P(X \ge 3) = 1 - [p(0) + p(1) + p(2)] \)
\( \implies P(X \ge 3) = 1 - \left[ \frac{e^{-1.5} \times 1.5^0}{0!} + \frac{e^{-1.5} \times 1.5^1}{1!} + \frac{e^{-1.5} \times 1.5^2}{2!} \right] \)
\( \implies P(X \ge 3) = 1 - \left[ e^{-1.5} + 1.5 \times e^{-1.5} + \frac{2.25}{2} \times e^{-1.5} \right] \)
\( \implies P(X \ge 3) = 1 - \left[ e^{-1.5} + 1.5 \times e^{-1.5} + 1.125 \times e^{-1.5} \right] \)
\( \implies P(X \ge 3) = 1 - [0.2231 + 1.5(0.2231) + 1.125(0.2231)] \)
\( \implies P(X \ge 3) = 1 - [0.2231 + 0.3346 + 0.2510] \)
\( \implies P(X \ge 3) = 1 - 0.8087 \)
\( \implies P(X \ge 3) = 0.1913 \)
Thus, the probability that some demand is refused is 0.1913.
In simple words: The chance that no one wants to rent a car on a given day is 22.31%, while the chance that you have to turn customers away because you run out of cars is 19.13%.

🎯 Exam Tip: When a question asks for "demand is refused," it means the demand exceeds the available capacity. Always use the complement rule \( 1 - P(X \le \text{capacity}) \) to avoid calculating an infinite sum.

Question 6. Defects on plywood sheets occur at random with an average of one defect per 50 sq. ft. Find the probability that such a sheet has
(i) no defect
(ii) at least one defect.
Use \( e^{-1} = 0.3678 \).
Answer:
Let \( X \) = Number of defects on a plywood sheet.
Given average rate \( m = 1 \). This probability calculation helps in quality control assessments for manufacturing.
Therefore, \( X \) follows a Poisson distribution with parameter \( m = 1 \), i.e., \( X \sim P(m = 1) \).
The probability mass function of Poisson distribution is given by:
\( p(x) = \frac{e^{-m} \cdot m^x}{x!} \)

(i) Probability of no defect, \( P(X = 0) \):
\( P(X = 0) = \frac{e^{-1} \cdot 1^0}{0!} \)
\( \implies P(X = 0) = \frac{0.3678 \cdot 1}{1} \)
\( \implies P(X = 0) = 0.3678 \)

(ii) Probability of at least one defect, \( P(X \ge 1) \):
\( P(X \ge 1) = 1 - P(X < 1) \)
\( \implies P(X \ge 1) = 1 - P(X = 0) \)
\( \implies P(X \ge 1) = 1 - 0.3678 \)
\( \implies P(X \ge 1) = 0.6322 \)
In simple words: We use the Poisson formula to find the chances of having no defects or at least one defect. The probability of having no defects is about 36.78%, which means there is a 63.22% chance of finding at least one defect on the sheet.

🎯 Exam Tip: Remember that "at least one" is always calculated as \( 1 - P(X = 0) \). This shortcut saves a lot of calculation time in exams.

Question 7. It is known that, in a certain area of a large city, the average number of rats per bungalow is five. Assuming that the number of rats follows Poisson distribution, find the probability that a randomly selected bungalow has
(i) exactly 5 rats
(ii) more than 5 rats
(iii) between 5 and 7 rats, inclusive.
Given \( e^{-5} = 0.0067 \).
Answer:
Let \( X \) = Number of rats in a bungalow.
Given average rate \( m = 5 \). This statistical model helps in urban planning and pest control management.
Therefore, \( X \sim P(m = 5) \).
The probability mass function is:
\( p(x) = \frac{e^{-m} \cdot m^x}{x!} = \frac{e^{-5} \cdot 5^x}{x!} \)

(i) Probability of exactly 5 rats, \( P(X = 5) \):
\( P(X = 5) = \frac{e^{-5} \cdot 5^5}{5!} \)
\( \implies P(X = 5) = \frac{0.0067 \cdot 3125}{120} \)
\( \implies P(X = 5) = \frac{20.9375}{120} \)
\( \implies P(X = 5) \approx 0.1745 \)

(ii) Probability of more than 5 rats, \( P(X > 5) \):
\( P(X > 5) = 1 - P(X \le 5) \)
\( \implies P(X > 5) = 1 - [P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5)] \)
Let us calculate the individual probabilities:
\( P(X = 0) = \frac{e^{-5} \cdot 5^0}{0!} = 0.0067 \)
\( P(X = 1) = \frac{e^{-5} \cdot 5^1}{1!} = 5 \times 0.0067 = 0.0335 \)
\( P(X = 2) = \frac{e^{-5} \cdot 5^2}{2!} = 12.5 \times 0.0067 = 0.0838 \)
\( P(X = 3) = \frac{e^{-5} \cdot 5^3}{3!} \approx 20.8333 \times 0.0067 \approx 0.1396 \)
\( P(X = 4) = \frac{e^{-5} \cdot 5^4}{4!} \approx 26.0417 \times 0.0067 \approx 0.1745 \)
\( P(X = 5) \approx 0.1745 \)
Summing these up:
\( P(X \le 5) \approx 0.0067 + 0.0335 + 0.0838 + 0.1396 + 0.1745 + 0.1745 = 0.6126 \)
\( \implies P(X > 5) = 1 - 0.6126 = 0.3874 \)

(iii) Probability of between 5 and 7 rats, inclusive, \( P(5 \le X \le 7) \):
\( P(5 \le X \le 7) = P(X = 5) + P(X = 6) + P(X = 7) \)
We already have \( P(X = 5) \approx 0.1745 \).
\( P(X = 6) = \frac{e^{-5} \cdot 5^6}{6!} = P(X = 5) \times \frac{5}{6} \approx 0.1745 \times 0.8333 \approx 0.1454 \)
\( P(X = 7) = \frac{e^{-5} \cdot 5^7}{7!} = P(X = 6) \times \frac{5}{7} \approx 0.1454 \times 0.7143 \approx 0.1039 \)
\( \implies P(5 \le X \le 7) = 0.1745 + 0.1454 + 0.1039 = 0.4238 \)
In simple words: We use the Poisson distribution formula with an average of 5. The chance of finding exactly 5 rats is 17.45%, finding more than 5 rats is 38.74%, and finding between 5 and 7 rats is 42.38%.

🎯 Exam Tip: When calculating consecutive terms like \( P(X=6) \) and \( P(X=7) \), use the recurrence relation \( P(X=k) = P(X=k-1) \times \frac{m}{k} \) to save time and avoid large calculations.

Question 1. Find the probability of the following events for a Poisson distribution with parameter \( \lambda = 5 \):
(i) P(Exactly five rats)
(ii) P(More than five rats)
(iii) P(between 5 and 7 rats, inclusive)
Answer: For a Poisson distribution with parameter \( \lambda = 5 \), the probability mass function is given by \( P(X = x) = \frac{e^{-\lambda} \lambda^x}{x!} = \frac{e^{-5} 5^x}{x!} \). We are given \( e^{-5} = 0.0067 \). These calculations help us understand the likelihood of different outcomes in a Poisson process where the average rate of occurrence is known.

(i) P(Exactly five rats):
\( P(X = 5) = \frac{e^{-5} \times 5^5}{5!} \)
\( = \frac{0.0067 \times 3125}{120} \)
\( = \frac{20.9375}{120} \)
\( = 0.17 \)

(ii) P(More than five rats):
\( P(X > 5) = 1 - P(X \le 5) \)
\( = 1 - \left[ \frac{e^{-5} \times 5^0}{0!} + \frac{e^{-5} \times 5^1}{1!} + \frac{e^{-5} \times 5^2}{2!} + \frac{e^{-5} \times 5^3}{3!} + \frac{e^{-5} \times 5^4}{4!} + \frac{e^{-5} \times 5^5}{5!} \right] \)
\( = 1 - e^{-5} \left[ 1 + 5 + \frac{25}{2} + \frac{125}{6} + \frac{625}{24} + \frac{3125}{120} \right] \)
\( = 1 - e^{-5} \left[ 6 + \frac{1500 + 2500 + 3125 + 3125}{120} \right] \)
\( = 1 - e^{-5} [6 + 85.42] \)
\( = 1 - 0.0067 \times 91.42 \)
\( = 1 - 0.61 \)
\( = 0.39 \)

(iii) P(between 5 and 7 rats, inclusive):
\( P(5 \le X \le 7) = p(5) + p(6) + p(7) \)
\( = \frac{e^{-5} \times 5^5}{5!} + \frac{e^{-5} \times 5^6}{6!} + \frac{e^{-5} \times 5^7}{7!} \)
\( = e^{-5} \times 5^5 \left[ \frac{1}{5!} + \frac{5}{6!} + \frac{5^2}{7!} \right] \)
\( = e^{-5} \times 3125 \left[ \frac{1}{120} + \frac{5}{720} + \frac{25}{5040} \right] \)
\( = e^{-5} \times 3125 \left[ \frac{42 + 35 + 25}{5040} \right] \)
\( = 0.0067 \times 3125 \times 0.02 \)
\( = 0.0067 \times 62.5 \)
\( = 0.42 \)
In simple words: We use the Poisson formula to find the chances of getting exactly 5, more than 5, or between 5 and 7 rats. By substituting the given average rate of 5 into the formula, we can calculate each probability step-by-step.

🎯 Exam Tip: When calculating 'more than' probabilities, always use the complement rule \( 1 - P(X \le x) \) to simplify your calculations. Double-check your fraction additions by finding a common denominator carefully.