Can someone please help me :(
1 answer:
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Answer:
Step-by-step explanation:
Probability distribution of this type is a vicariate distribution because it specifies two random variable X and Y. We represent the probability X takes x and Y takes y by
f(x,y) = P((X=x,Y=y) ,
and given that the random variables are independent the joint pmf isbgiven by:
f(x,y) = fX(x) .fY(y) and this gives the table (see attachment)
(b) the required probability is given by considering X=0,1 and Y =0,1
f(x,y) = sum{(P(X ≤ 1, Y ≤ 1) }
= [0.1 +0.2 ] × [0.1 + 0.3]
= 0.3 × 0.4
= 0.12
Now we find
fX(x) = sum{[f(x.y)]} for y =0, 1 and similarly for fY(y)= sum{[f(x,y)]} for x=0, 1
fX(x) = sum{[f(x,y)]}= 0.1+0.3
= 0.4
fY(y) = sum{[f(x.y)]} = 0.1 + 0.2
= 0.3
Since fY(y) × fX(x) = 0.4 × 0.3
= 0.12
Hence f(x,y) = fX(x) .fY(y), the events are independent
(c) the required event is give by: [P(X<=1, PY<=1)]
= P(X<=1) . P(Y<=1)
= [sum{[f(x.y]} over Y=0,1] × [sum{[f(x.y)]}, over X= 0, 1
= 0.4 × 0.3
= 0.12
(d) the required event is given by the idea that: either The South has no bed and the North has or the the North has no bed and the South has. Let this event be A
A =sum[(P(X = 1<= x<= 4, Y =0)] + sum[P(X =0 Y= 1 <= x <= 3)]
A = [0.02 + 0.03 + 0.02 + 0.02] + [0.03 + 0.04 + 0.02]
A = [0.09] + [ 0.09]
A = 0.18
Pls note the sum of the vertical column X=2 is 0.3
<u>Annotation</u>
General formula for distance-time-velocity relationship is as following
d = v × t
The velocity of the first car will be v₁, the time is 2 hours, the distance will be d₁.
The velocity of the second car will be v₂, the time is 2 hours, the distance will be d₂.
One of them traveling 5 miles per hour faster than the others. That means the velocity of the first car is 5 miles per hour more than the velocity of the second car.
v₁ = v₂ + 5 (first equation)
The distance of the two cars after two hours will be 262 miles apart. Because they go to opposite direction, we could write it as below.
d₁ + d₂ = 262 (second equation)
Plug the d-v-t relationship to the second equation
d₁ + d₂ = 262
v₁ × t + v₂ × t = 262
v₁ × 2 + v₂ × 2 = 262
2v₁ + 2v₂ = 262
Plug the v₁ as (v₂+5) from the first equation
2v₁ + 2v₂ = 262
2(v₂ + 5) + 2v₂ = 262
2v₂ + 10 + 2v₂ = 262
4v₂ + 10 = 262
4v₂ = 252
v₂ = 252/4
v₂ = 63
The second car is 63 mph fast.
Find the velocity of the first car, use the first equation
v₁ = v₂ + 5
v₁ = 63 + 5
v₁ = 68
The first car is 68 mph fast.
Answer
General formula for distance-time-velocity relationship is as following
d = v × t
The velocity of the first car will be v₁, the time is 2 hours, the distance will be d₁.
The velocity of the second car will be v₂, the time is 2 hours, the distance will be d₂.
One of them traveling 5 miles per hour faster than the others. That means the velocity of the first car is 5 miles per hour more than the velocity of the second car.
v₁ = v₂ + 5 (first equation)
The distance of the two cars after two hours will be 262 miles apart. Because they go to opposite direction, we could write it as below.
d₁ + d₂ = 262 (second equation)
Plug the d-v-t relationship to the second equation
d₁ + d₂ = 262
v₁ × t + v₂ × t = 262
v₁ × 2 + v₂ × 2 = 262
2v₁ + 2v₂ = 262
Plug the v₁ as (v₂+5) from the first equation
2v₁ + 2v₂ = 262
2(v₂ + 5) + 2v₂ = 262
2v₂ + 10 + 2v₂ = 262
4v₂ + 10 = 262
4v₂ = 252
v₂ = 252/4
v₂ = 63
The second car is 63 mph fast.
Find the velocity of the first car, use the first equation
v₁ = v₂ + 5
v₁ = 63 + 5
v₁ = 68
The first car is 68 mph fast.
Answer