To find P(|X−Y|≤1), we need to consider the areas in the XY-plane where |X−Y|≤1.
Let's break it down into two cases:
1) When X ≤ Y:
In this case, we have 2 scenarios:
a) X ≤ Y ≤ X+1
b) X ≤ X-Y ≤ 1
2) When Y ≤ X:
In this case, we have 2 scenarios:
a) Y ≤ X ≤ Y+1
b) 0 ≤ X-Y ≤ 1
First, let's find the probability for case 1a:
P(X ≤ Y ≤ X+1) = ∫(from 3 to 8) ∫(from x to x+1)dydx
= ∫(from 3 to 8) (x+1-x) dx
= ∫(from 3 to 8) 1 dx
= (8-3)
= 5
Next, let's find the probability for case 1b:
P(X ≤ X-Y ≤ 1) = ∫(from 3 to 8) ∫(from 0 to x-y)dydx
= ∫(from 3 to 8) (x-y-0) dydx
= ∫(from 3 to 8) (x) dydx
= ∫(from 3 to 8) (x) dx
= [(x^2)/2] (from 3 to 8)
= [(8^2)/2] - [(3^2)/2]
= 32/2 - 9/2
= 16 - 4.5
= 11.5
Total probability for case 1: P(X ≤ Y ≤ X+1 or X ≤ X-Y ≤ 1) = 5 + 11.5 = 16.5
Now, let's find the probability for case 2a:
P(Y ≤ X ≤ Y+1) = ∫(from 3 to 8) ∫(from y to y+1)dxdy
= ∫(from 3 to 8) (y+1-y)dy
= ∫(from 3 to 8) (1)dy
= (8-3)
= 5
Lastly, let's find the probability for case 2b:
P(0 ≤ X-Y ≤ 1) = ∫(from 3 to 8) ∫(from y-1 to y) dxdy
= ∫(from 3 to 8) (y-y+1)dy
= ∫(from 3 to 8) (1)dy
= (8-3)
= 5
Total probability for case 2: P(Y ≤ X ≤ Y+1 or 0 ≤ X-Y ≤ 1) = 5 + 5 = 10
Therefore, the overall probability P(|X−Y|≤1) = P(X ≤ Y ≤ X+1 or X ≤ X-Y ≤ 1 or Y ≤ X ≤ Y+1 or 0 ≤ X-Y ≤ 1) = 16.5 + 10 = 26.5
Moving on to finding P(min(X,Y)≤6), we need to consider the areas in the XY-plane where the minimum of X and Y is less than or equal to 6.
Let's break it down into two cases:
1) When min(X, Y) = X:
In this case, we have 2 scenarios:
a) X ≤ Y ≤ 6
b) 6 ≤ Y ≤ X
2) When min(X, Y) = Y:
In this case, we have 2 scenarios:
a) Y ≤ X ≤ 6
b) 6 ≤ X ≤ Y
For case 1a, P(X ≤ Y ≤ 6) = 5 (same as above)
For case 1b, P(6 ≤ Y ≤ X) = 0 (since X is uniformly distributed on [3,8], it cannot be less than 6)
For case 2a, P(Y ≤ X ≤ 6) = 5 (same as above)
For case 2b, P(6 ≤ X ≤ Y) = 0 (since X is uniformly distributed on [3,8], it cannot be less than 6)
Therefore, the overall probability P(min(X,Y)≤6) = P(X ≤ Y ≤ 6 or Y ≤ X ≤ 6) = 5 + 5 = 10
Suppose that X is uniformly distributed on the interval [3,8], and that given X=x, Y is uniformly distributed on the interval [0,x].
P(|X−Y|≤1) =
P(min(X,Y)≤6) =
1 answer