Explanation.
The question is incomplete. However, the complete part of the question simply states that: What is the probability that the man arrives first?
Answer:
Therefore, we can say that the probability that the man will arrive first is half that is 1/2 while the probability that no one will wait more than five times is 1/6
Step-by-step explanation:
We assume the man and the woman arrive at 12:X and 12:Y respectively (1:00 is 12:60). Also, the space of X and Y are [15, 45] and [0, 60] respectively, and fₓ (x) ≡ 1 / (45 − 15) = 1/30, fy (y) ≡ 1 / (60 − 0) = 1 / 60.
So the joint pdf of X and Y is
f(x, y) = 1 / 30 x 1 / 60
= 1 / 1800
where ( x , y ) ∈ [15, 45] × [0, 60].
Consequentially,
P(| X − Y | ≤ 5) = ∫ Iim (45)(15) ∫ lim (x+5)(x-5) f (x,y)dydx
= 1 / 1800 ∫ Iim (45)(15) y | ₓ₋₅ˣ⁺⁵ dx
= 30 x 10 / 1800
300 / 1800
= 1 / 6
Hence,
P( X < Y ) = ∫ Iim (45)(15) ∫ lim (60)(x) f (x,y) dydx
= 1 / 1800 ∫ Iim (45)(15) y | ⁶⁰ₓ dx
= 1 / 1800 ∫ Iim (45)(15) ( 60 - x ) dx
= 60 x − x ²/2 ÷ 1800 ║⁴⁵₁₅
= 1 / 2
