Let a random variable X follow Poisson distribution such that Prob (X = 1) = Prob (X= 2). The value of Prob (X = 3) is __________ (round off to 2 decimal places).

The correct answer is 0.18.

Step 1: Understand the Poisson Distribution Formula
For a random variable X following a Poisson distribution with parameter λ (where λ > 0 represents the average rate or mean number of occurrences), the probability mass function is given by:
$$P(X=k)=\frac{e^{-\lambda }{\lambda }^k}{k!}$$ where k is a non-negative integer (0, 1, 2, ...), and e is Euler's constant (approximately 2.718).

Step 2: Find the parameter λ using the given condition
We are given that the probability of X = 1 is equal to the probability of X = 2:
$$P(X=1)=P(X=2)$$ Substituting k = 1 and k = 2 into the Poisson probability formula:
$$\frac{e^{-\lambda }{\lambda }^1}{1!}=\frac{e^{-\lambda }{\lambda }^2}{2!}$$ Since 1! = 1 and 2! = 2, we can simplify this equation. Dividing both sides by the non-zero term e^-λ yields:
$$\lambda =\frac{{\lambda }^2}{2}$$ Rearranging the equation to solve for λ:
$${\lambda }^2-2\lambda =0$$ Factoring out λ:
$$\lambda (\lambda -2)=0$$ This gives two possible solutions: λ = 0 or λ = 2. Since the parameter λ in a Poisson distribution must be strictly greater than zero, we discard λ = 0. Therefore, we have:
λ = 2

Step 3: Calculate the probability P(X = 3)
Using our parameter value λ = 2, we find P(X = 3) by substituting k = 3 into the formula:
$$P(X=3)=\frac{e^{-2}{2}^3}{3!}$$ Since 2³ = 8 and 3! = 3 × 2 × 1 = 6:
$$P(X=3)=\frac{8\cdot e^{-2}}{6}=\frac{8}{6e^2}$$ Using the approximation e ≈ 2.718, we compute e² ≈ 7.389:
$$P(X=3)\approx \frac{8}{6\cdot 7.389}\approx \frac{8}{44.334}\approx 0.1804$$ Rounding off to two decimal places, we get 0.18.