SOLUTION: Given that p(A)= 1/4 p(B)= 1/5 p(C)= 1/2 Compute the following 1) p(A∪B) 2) p(A∩B) 3) p(A∪B∪C) IF ALL EVENTS ARE I) IND

Algebra ->  Probability-and-statistics -> SOLUTION: Given that p(A)= 1/4 p(B)= 1/5 p(C)= 1/2 Compute the following 1) p(A∪B) 2) p(A∩B) 3) p(A∪B∪C) IF ALL EVENTS ARE I) IND      Log On


   

Question 1131783: Given that
p(A)= 1/4
p(B)= 1/5
p(C)= 1/2

Compute the following
1) p(A∪B)
2) p(A∩B)
3) p(A∪B∪C)
IF ALL EVENTS ARE
I) INDEPENDENT
II) MUTUALLY EXCLUSIVE EVENTS
Note that:
Where p(A), p(B) and p(C) means probability of events A, B and C respectively.
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
Given that
p%28A%29=+1%2F4+
p%28B%29=+1%2F5+
p%28C%29=+1%2F2+
Compute the following
1)+p(A+∪ B)
2) p(A ∩ B)
3)+p (A ∪B ∪C )
IF ALL EVENTS ARE
I) INDEPENDENT
1)
+p(A+∪ B)=P(A)+P(B)-P(A ∩B)
+p(A+∪ B)=1%2F4%2B1%2F5-P(A &#p8745; B)
+p(A+∪ B)=5%2F20%2B4%2F20-1%2F20
+p(A+∪ B)=8%2F20
+p(A+∪ B)=2%2F5
+p(A+∪ B)=0.4
2)
p(A ∩ B) =P%28A%29*P%28B%29
p(A ∩ B)=%281%2F4%29%281%2F5%29=1%2F20
p(A ∩ B)=0.05
3)
p+(A∪B∪C) = p%28A%29+%2Bp%28B%29+%2Bp%28C%29+-p%28AB%29+-p%28AC%29-p%28BC%29+%2Bp%28ABC%29
p (A∪B∪C) =
p (A∪B∪C) =7%2F10
p (A∪B∪C) =0.7

II) MUTUALLY EXCLUSIVE EVENTS
the probability of A or B is the sum of the individual probabilities:
P(A+∪+B) = P%28A%29+%2B+P%28B%29
1)
p(A∪B) =1%2F4%2B1%2F5=5%2F20%2B4%2F20=9%2F20
p(A∪B) =0.45
When two events (call them "A" and "B") are Mutually Exclusive it is impossible for them to happen together: p(A ∩+B) = 0 as they have no element in common.

2) p(A ∩ B)=0
3) +p (A ∪B ∪C ) ->events are Mutually Exclusive
When two events are mutually exclusive, the probability of their union can be calculated with the addition rule.
The above discussion for two sets still holds. We can add together the probabilities of the individual sets A, B, and C, but in doing this we have double counted some elements.
The elements in the intersection of A and B have been double counted as before, but now there are other elements that have potentially been counted twice. The elements in the intersection of A and C and in the intersection of B and C have now also been counted twice. So the probabilities of these intersections must also be subtracted.
But have we subtracted too much? There is something new to consider that we did not have to be concerned about when there were only two sets. Just as any two sets can have an intersection, all three sets can also have an intersection. In trying to make sure that we did not double count anything, we have not counted at all those elements that show up in all three sets. So the probability of the intersection of all three sets must be added back in.
Here is the formula that is derived from the above discussion:
+p (A ∪B ∪C )= P%28A%29+%2B+P%28B%29+%2B+P%28C%29 - P(A ∩ B) - P(A ∩ C) - P(B ∩+C) + P(A ∩ B ∩ C)
+p (A ∪B ∪C ) =1%2F4%2B1%2F5%2B1%2F2-1%2F20-%281%2F4%29%281%2F2%29%2B%281%2F4%29%281%2F5%29%281%2F2%29
+p (A ∪B ∪C ) =4%2F5
+p (A ∪B ∪C ) =0.8