Total number of pens `=144`
Number of defective pens, `=20`
So, number of good pens, ` = 144-20 = 124`
Now, Nuri will buy a pen if it is good.
(i)So, probability Nuri will buy a pen = `124/144 = 31/36`

(ii) Probability of Nuri not buying a pen `= 20/144 = 5/36`

No. of good pens = 144 – 20 = 24

No. of detective pens = 20

Total no. of possible outcomes = 144 {total no pens}

(i) E ⟶ event of buying pen which is good.

No. of favourable outcomes = 124 {124 good pens}

P(E) = (No.of favorable outcomes)/(Total no.of possible outcomes)

P(E) = 124/144 = 31/36

(ii) Bar  E⟶ event of not buying a pen which is bad P(E) + P(Bar E) = 1

P(E)+P(Bar E)=1

P(Bar E) = 1 – P(E)

=1−31/36 = 5/36

Given : 20 out of 144 are defective i.e., no. of defective ball pens = 20 

no. of good ball pens = 144 – 20 = 124 

∴ Total outcomes in drawing a ball pen at random = 144. 

i) Sudha buys it if it is not defective / a good one. No. of outcomes favourable to a good pen = 124. 

∴ Probability of buying it

= \(\frac{No.\,of\,favourable\,outcomes}{Total\,no.\,of\,outcomes}\)

\(=\frac{124}{144}=\frac{31}{36}\)

ii) Sudha will not buy it-if it is a defective pen 

No. of outcomes favourable to a defective pen = 20 

∴ Probability of not buying it

= \(\frac{No.\,of\,favourable\,outcomes}{Total\,no.\,of\,outcomes}\)

\(=\frac{20}{144}=\frac{5}{36}\)

!! (not buying) = 1 – P (buying) 

= 1 – \(\frac{31}{36}=\frac{5}{36}\)