A sample space consists of 9 elementary events *E*_{1}, *E*_{2}, *E*_{3}, ..., *E*_{9} whose probabilities are

P(*E*_{1}) = P(*E*_{2}) = 0.08, 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}) = 0.07

Suppose *A* = {*E*_{1}, *E*_{5}, *E*_{8}}, *B* = {*E*_{2}, *E*_{5}, *E*_{8},* E*_{9}}

Calculate \[P\left( \bar{ B} \right)\] from P(*B*), also calculate \[P\left( \bar{ B } \right)\] directly from the elementary events of \[\bar{ B } \] .

#### Solution

Let *S* be the sample space of the elementary events.*S* = {*E*_{1}, *E*_{2}, *E*_{3}, ..., *E*_{9}}

Given:*A* = {*E*_{1}, *E*_{5}, *E*_{8}}*B* = {*E*_{2}, *E*_{5}, *E*_{8},* E*_{9}}

P(*E*_{1}) = P(*E*_{2}) = 0.08, 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}) = 0.07

\[P\left( B \right) = 1 - P\left( B \right) = 1 - 0 . 32 = 0 . 68\] [From (i)]

Also, we know that \[\bar{ B } \]= *S *− *B* = {*E*_{1}, *E*_{3}, *E*_{4}, *E*_{6},* E*_{7}}

∴ \[P\left( \bar{B} \right)\] = P(*E*_{1}) + P(*E*_{3}) + P(*E*_{4}) + P(*E*_{6}) + P(*E*_{7})

= 0.08 + 0.1 + 0.1 + 0.2 + 0.2

= 0.68

#### Notes

The solution of the problem is provided by taking P(*E*_{5}) = 0.1. This information is missing in the question as given in the book.