Please help to find the HCF and LCM of this question
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
