Answer:
the required probability is 0.0357
Explanation:
Given the data in the question;
X-Exp( λ = 40/60 = 0.6667
Pdf: f( x ) = λe^(-λx), 0 < x
so
Cdf: P( X < x ) = 1 - e^(-λe)
( X ≤ x ) = 1 - e^(-λe), for x > 0 ⇒ p( X > x ) = e^(-λe)
now to wait at least 5 minutes before seen another
p(X > 5 ) = e^(-λe)
we substitute
p(X > 5 ) = e^(-0.6667 × 5)
p(X > 5 ) = e^(-3.3335)
p(X > 5 ) = 0.035668 ≈ 0.0357
Therefore, the required probability is 0.0357