SOLUTION: If P(A)=0.5, P(B)=0.5 and P(A \textrm{ and } B)= 0.05, find the following probabilities:
a) P(A \textrm{ or } B) =
b) P( \textrm{not }A ) =
c) P( \textrm{not }B ) =
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-> SOLUTION: If P(A)=0.5, P(B)=0.5 and P(A \textrm{ and } B)= 0.05, find the following probabilities:
a) P(A \textrm{ or } B) =
b) P( \textrm{not }A ) =
c) P( \textrm{not }B ) =
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Question 1171689: If P(A)=0.5, P(B)=0.5 and P(A \textrm{ and } B)= 0.05, find the following probabilities:
a) P(A \textrm{ or } B) =
b) P( \textrm{not }A ) =
c) P( \textrm{not }B ) =
d) P( A \textrm{ and } (\textrm{not }B) ) =
e) P( \textrm{not }(A \textrm{ and } B)) = Answer by math_tutor2020(3816) (Show Source):
You can put this solution on YOUR website!
Draw out a Venn diagram with circles A and B overlapping. Both of these circles are inside a rectangle representing the universal set (U).
Let's say there are 100 people. Place 50 of them in circle A and 50 in circle B. This produces P(A) = 50/100 = 0.5 and P(B) = 0.5 for the same reasons.
Place 5 in the overlapping region to indicate P(A and B) = 5/100 = 0.05
Note how 50-5 = 45 people are in circle A, but outside circle B. A similar situation happens with B as well.
There are 45+5+45 = 95 people in either circle, or both. This leaves 100-95 = 5 outside both circles, but inside the universal set.
We have this Venn diagram
Use that diagram to answer the questions below
P(A or B) = (45+5+45)/100 = 95/100 = 0.95
We add up all the values in the circles, then divide by 100
P(not A) = (45+5)/100 = 50/100 = 0.50
We're adding everything that isn't in circle A
P(not B) = (45+5)/100 = 50/100 = 0.50
Same idea as P(not A), but now add up everything that isn't in circle B.
P(A and not B) = 45/100 = 0.45
Only the value 45 is in circle A but outside circle B
P(not(A and B)) = (45+45+5)/100 = 95/100 = 0.95
We add up the 45's outside the shared overlapped region, plus that 5 outside both circles.