## Solution :

Required Probability = P(X = 0) + P(X = 1)

= \(e^{-5}\over 0!\).\(5^0\) + \(e^{-5}\over 1!\).\(5^1\)

= \(e^{-5}\) + 5\(e^{-5}\) = \(6\over {e^5}\)

Required Probability = P(X = 0) + P(X = 1)

= \(e^{-5}\over 0!\).\(5^0\) + \(e^{-5}\over 1!\).\(5^1\)

= \(e^{-5}\) + 5\(e^{-5}\) = \(6\over {e^5}\)