Question

John is putting together a toy chest for his daughter. The toy chest was shipped in a box. The box contains 12 parts, of which 3 parts are defective. Two parts are selected at random without replacement.

a) Find the probability that both parts are defective.

b) Find the probability that both parts are not defective.

c) Find the probability that at least one part is defective.

Please show how you arrive at your answers.

Answer 1

a)

There are 3 parts are defective in 12 total parts in box, this means remaining 9 parts are not defective.

Both defective parts are chosen from 3 defective parts from the box by ³C₂ ways.

Total number of ways to select 2 parts from 12 available parts by ¹²C_{2 .}

P( 2 defective parts) =  ³C₂ / ¹²C_{2 .}

In general ⁿC_r = n! / [ ( n-r)! * r! ]

P( 2 defective parts) =  ³C₂ / ¹²C_{2 .}

= 3 / 66

= 0.0455

b)

Both non defective parts are chosen from 9 non-defective parts by ⁹C₂ ways.

Total number of ways to select 2 parts from 12 available parts by ¹²C₂

( 2 non defective parts) =  ⁹C₂ / ¹²C₂

= 36 / 66

= 0.5455

c)

P( At least one part is defective) = 1 - P( Both parts are non-defective)

= 1 - 0.5455

= 0.4545