1.

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))=0.1P(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 nnB).(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 calculateP(barB) directly from the elementarty outcomes of B.

Answer»

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Solution :(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


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