Question

There are two defective pencils in a pack of dozen pencils. If 3 pencils are taken at random, find the probability that (i) at the most one pencil is defective, (ii) two pencils are defective.​

Answer 1

Answer:

(i) $\frac{21}{22}$

(ii) $\frac{1}{22}$

Explanation:

(ii) There are 2 pencils defective in a pack of 12. If you happened to choose the first one, there would be $\frac{2}{12}$ chance that you would choose the defective one. If you succeed in choosing it, there would now be 1 pencil defective in a pack of 11. And your chance of choosing a second defective pencil would be $\frac{1}{11}$, if you succeed again, your chance of choosing a non-defective pencil on your third try would be a sure event

In other words you would have a $\frac{1}{11}$ of $\frac{2}{12}$ chance of choosing both the defective ones, which is

$\frac{1}{11} \times \frac{2}{12} = \frac{1}{66}$

But this order isn't the only way to achieve the desired outcome, you can choose the defective ones in the first and second try, second and third try, or first and third try all with the same $\frac{1}{66}$ probability. So

Net probability = $\frac{1}{66}\times3=\frac{1}{22}$

(i) Probability of choosing at the most 1 defective pencil = 1 - probability of choosing two defective ones = $1 - \frac{1}{22} = \frac{21}{22}$

Hope my answer has helped. Take care :)

Answer 2

There are 2 pencils defective in a pack of 12. If you happened to choose the first one, there would be  chance that you would choose the defective one. If you succeed in choosing it, there would now be 1 pencil defective in a pack of 11. And your chance of choosing a second defective pencil would be , if you succeed again, your chance of choosing a non-defective pencil on your third try would be a sure event

In other words you would have a  of  chance of choosing both the defective ones, which is

But this order isn't the only way to achieve the desired outcome, you can choose the defective ones in the first and second try, second and third try, or first and third try all with the same  probability. So

Net probability =

(i) Probability of choosing at the most 1 defective pencil = 1 - probability of choosing two defective ones = 1 - $\frac{1}{22}$ = $\frac{21}{22}$.