A sample space consists of 9 elementary outcomes outcomes `E_(1), E_(2)`,…, `E_(9)` whose probabilities are:
P(E_(1))=P(E_(2)) = 0.09, P(E_(3))=P(E_(4))=P(E_(5))=0.1`
`P(E_(6)) = P(E_(7)) = 0.2, P(E_(8)) = P(E_(9)) = 0.06`
If `A = {E_(1), E_(5), E_(8)}, B= {E_(2), E_(5), E_(8), E_(9)}` then
(a) Calculate P(A), P(B), and P(A `nn` B).
(b) Using the addition law of probability, calculate `P(A uu B)`.
(c ) List the composition of the event `A uu B`, and calculate `P(A uu B)` by adding the probabilities of the elementary outcomes.
(d) Calculate `P(barB)` from P(B), also calculate `P(barB)` directly from the elementarty outcomes of B.

Explanation

(a) `P(A) = P(E_(1)) + P(E_(5)) + P(E_(8))`
`=0.09 + 0.1 + 0.06 = 0.25`
(b) `P(B) = P(E_(2)) + P(E_(5)) + P(E_(8)) = P(E_(9))`
`= 0.09 + 0.1 + 0.06 + 0.06 = 0.31`
`P(A uu B) = P(A) + P(B) - P(A nn B)`
Now, `A nn B = {E_(5), E_(8)}`
`therefore P(A nn B) = P(E_(5)) + P(E_(8)) = 0.1 + 0.06 = 0.16`
`therefore P(A uu B) = 0.25 + 0.31 - 0.16 = 0.40`
(c ) `A uu B = {E_(1), E_(2), E_(5), E_(8), E_(9)}`
`P(A uu B) = P(E_(1)) + P(E_(2)) + P(E_(5)) + P(E_(8)) + P(E_(9))`
`=0.09 + 0.09 + 0.1 +0.06 + 0.06 = 0.40`
(d) `because P(bar(B)) = 1- P(B) = 1 - 0.31 = 0.69`
and `bar(B) = {E_(1), E_(3), E_(4), E_(6), E_(7)}`
`therefore P(bar(B)) = P(E_(1)) + P(E_(3)) + P(E_(4)) + P(E_(6)) + P(E_(7))`
= 0.09 + 0.1 + 0.1 + 0.2 + 0.2 = 0.69