The probability mass function of a Poisson variate $X \sim \text{Poisson}(\lambda)$ is:

$$P(X=x) = \frac{e^{-\lambda} \lambda^x}{x!}, \quad x = 0, 1, 2, \dots$$

Given $P(X=1) = P(X=2)$:

$$\frac{e^{-\lambda} \lambda^1}{1!} = \frac{e^{-\lambda} \lambda^2}{2!}$$ $$\lambda e^{-\lambda} = \frac{\lambda^2 e^{-\lambda}}{2}$$

Dividing both sides by $\lambda e^{-\lambda}$ (since $\lambda \gt 0$):

$$1 = \frac{\lambda}{2} \implies \lambda = 2$$

Now, calculate $P(X=4)$:

$$P(X=4) = \frac{e^{-2} (2)^4}{4!} = \frac{e^{-2} \times 16}{24} = \frac{2e^{-2}}{3}$$

Karl Pearson's coefficient of correlation is defined as:

$$r = \frac{\text{Cov}(X, Y)}{\sigma_X \sigma_Y} = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sqrt{\sum (x_i - \bar{x})^2 \sum (y_i - \bar{y})^2}}$$

The numerator has units of $(\text{unit of } X) \times (\text{unit of } Y)$, and the denominator also has units of $(\text{unit of } X) \times (\text{unit of } Y)$. The units cancel out completely, making $r$ a pure, dimensionless number with 'No unit'. It is invariant to changes of origin and scale.

Calculation:
Total weight of all 6 bags = $6 \times 119 = 714\text{ kg}$.
Sum of the weights of the 5 known bags:

$$115 + 109 + 129 + 117 + 114 = 584\text{ kg}$$

Weight of the 6th bag:

$$\text{Weight} = 714 - 584 = 130\text{ kg}$$