In an experiment, A, B, C, andD are events with probabilitiesP[A UB] = 5/8,P[A] =3/8,

P[C ∩D] = 1/3, andP[C] =1/2. Furthermore, Aand B are disjoint, whileC and D areindependent.

(a) Find P[A∩ B],P[B],P[A ∩Bc], andP[A UBc].
(b) Are A andB independent?
(c) FindP[D],P[C ∩Dc],P[Cc∩ Dc],andP[C|D].
(d) Find P[CU D] andP[C UDc].
(e) Are C andDcindependent?

Step-by-step explanation:

Hello!

You have 4 events A, B, C and D

With probabilities:

P(A∪B)= 5/8

P(A)= 3/8

P(C∩D)= 1/3

P(C)= 1/2

A and B are disjoint events, this means that there are no shared elements between then and their intersection is void, symbolically A∩B= ∅, in consequence, these events are mutually exclusive.

C and D are independent events, this means that the occurrence of one of them does not affect the probability of occurrence of the other one in two consecutive repetitions.

a.

i. P(A∩B)= 0

⇒ Since A and B are disjoint events, the probability of their intersection is zero.

ii. A and B are mutually exclusive events, this means that P(A∪B)= P(A)+P(B)

⇒ From this expression, you can clear the probability of b as P(B)= P(A∪B)-P(A)= 5/8-3/8= 1/4

iii. If Bc is the complementary event of B, its probability would be P(Bc)= 1 - P(B)= 1 - 1/4= 3/4. If the events A and B are mutually exclusive and disjoint, it is logical to believe that so will be the events A and Bc, so their intersection will also be void:

P(A∩Bc)= 0

vi.P(A∪Bc)= P(A) + P(Bc)= 3/8+3/4= 9/8

b.

If A and B are independent then the probability of A is equal to the probability of A given B, symbolically:

P(A)= P(A/B)

P(A)= 3/8

P(A) ≠ P(A/B) ⇒ A and B are not independent.

c.

i. P(D) ⇒ Considering C and D are two independent events, then we know that P(C∩D)= P(C)*P(D)

Then you can clear the probability of D as:

P(D)= P(C∩D)/P(C)= (1/3)/(1/2)= 2/3

ii. If Dc is the complementary event of D, then its probability is P(Dc)= 1 - P(D) = 1 - 2/3= 1/3

P(C∩Dc)= P(C)*P(Dc)= (1/2)*(1/3)= 1/6

iii. Now Cc is the complementary event of C, its probability is P(Cc)= 1 - P(C)= 1 - 1/2= 1/2

P(Cc∩Dc)= P(Cc)*P(Dc)= (1/2)*(1/3)= 1/6

vi. and e.

P(C)=1/2

As you can see the P(C)=P(C/D) ⇒ This fact proves that the events C and D are independent.

d.

i. P(C∪D)= P(C) + P(D) - P(C∩D)= 1/2 + 2/3 - 1/3= 5/6

ii. P(C∪Dc)= P(C) + P(Dc) - P(C∩Dc)= 1/2 + 1/3 - 1/6= 2/3

I hope it helps!

the answer is b

step-by-step explanation:

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3