Thee number of persons joining a cinema ticket counter is a minute has Poisson distribution with parameter 6. Find the probability that
i) no one joins the queue in a particular minute
ii) two or more persons join the queue is a minute.

Answer

To solve the problem, we will use the Poisson distribution formula, which is given by:

\[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \]

where:
- \( P(X = k) \) is the probability of \( k \) events occurring in a fixed interval,
- \( \lambda \) is the average rate (mean) of occurrences,
- \( e \) is the base of the natural logarithm (approximately equal to 2.71828),
- \( k \) is the number of occurrences (which can be 0, 1, 2, ...).

In this case, the parameter \( \lambda = 6 \).

### Step 1: Calculate the probability that no one joins the queue in a particular minute (i.e., \( k = 0 \)).

Using the Poisson formula:

\[ P(X = 0) = \frac{e^{-6} \cdot 6^0}{0!} \]

Since \( 6^0 = 1 \) and \( 0! = 1 \):

\[ P(X = 0) = e^{-6} \cdot 1 = e^{-6} \]

### Step 2: Calculate the probability that two or more persons join the queue in a minute.

This can be calculated using the complement rule:

\[ P(X \geq 2) = 1 - P(X < 2) = 1 - (P(X = 0) + P(X = 1)) \]

#### Step 2.1: Calculate \( P(X = 1) \).

Using the Poisson formula again:

\[ P(X = 1) = \frac{e^{-6} \cdot 6^1}{1!} = e^{-6} \cdot 6 \]

#### Step 2.2: Combine the probabilities.

Now we can find \( P(X < 2) \):

\[ P(X < 2) = P(X = 0) + P(X = 1) = e^{-6} + e^{-6} \cdot 6 \]

Factoring out \( e^{-6} \):

\[ P(X < 2) = e^{-6}(1 + 6) = 7e^{-6} \]

#### Step 2.3: Final calculation for \( P(X \geq 2) \).

Now we can find \( P(X \geq 2) \):

\[ P(X \geq 2) = 1 - P(X < 2) = 1 - 7e^{-6} \]

### Final Answers:

1. The probability that no one joins the queue in a particular minute is \( P(X = 0) = e^{-6} \).
2. The probability that two or more persons join the queue in a minute is \( P(X \geq 2) = 1 - 7e^{-6} \).