A lot consists of 144 ball pens of which 20 are defective and the othe

1.	A lot consists of 144 ball pens of which 20 are defective and the others are good. The shopkeeper draws one pen at random and gives it to Sudha. What is the probability that (i) She will buy it? (ii) She will not buy it?
Answer» 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 – \(\frac{31}{36}=\frac{5}{36}\)

A lot consists of 144 ball pens of which 20 are defective and the others are good. The shopkeeper draws one pen at random and gives it to Sudha. What is the probability that (i) She will buy it? (ii) She will not buy it?

Answer»

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 – \(\frac{31}{36}=\frac{5}{36}\)

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