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trvptrav
03.03.2020 •
Mathematics
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?
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Ответ:
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!
Ответ:
1/2 of 60 is 30
so you expect an even number to be rolled 30 times