The total number of ball pens in the heap = 144.
The number of defective pens = 20
The number of fine ball pens
= 144 – 20 = 124
(i) Let the probability of buying a pen be (A).
∵ We would like to buy a good pen
∴ The number of favourable outcomes of buying a ball pen = 124
Total number of all possible outcomes = 144
∴ P(A) = \(\frac { 124 }{ 144 }\) = \(\frac { 31 }{ 36 }\)
(ii) Let the event of not buying a ball pen be (\(\overline { A }\))
Then P(\(\overline { A }\)) = 1 – P(A)
= 1 – \(\frac { 31 }{ 36 }\) = \(\frac { 5 }{ 36 }\)