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
A man and a woman agree to meet at a certain location about 12:30 pm. If the man arrives at a time uniformly distributed between 12:15 pm and 12:45 pm, and if the woman independently arrives at a time uniformly distributed between 12:00 pm and 1 pm, find the probability that the first to arrive waits no longer than 5 minutes. What is the probability that the man arrives first?
Answer:
1/2
Step-by-step explanation:
What is the probability that the man arrives first?
Define random variables X And Y that marks the time at which the man and women arrives at a certain place.
Representing their precise times of arrival are 12: X and 12: Y.
It is given that X ~ Unif(15, 45) and Y Unif(0, 00) and that these variables are independent.
We are required to find the probability that the distance between these two arrivals will be longer than 5 minutes, ie.
Check the figure below
The probability that the man arrives first is P(X <Y).
But, observe the variables X and Y are symmetric around the same point 12: 30.
Hence we can say that none of them is at an advantage to arrive first.
Thus P(X <Y) = 1/2
