Question

a retailer of electronic equipment received 6 VCRs from the manufacturer. Three of the VCRs were damaged in the shipment. The retailer sold two VCRs to two customers.

A) what is the probability that both customers received damaged VCRs?

B) What is the probability that one of the two customers received a defective VCR?

Answer 1

A)

Number of ways to select 2 defective VCRs from 3 defective VCRs = ³C₂

Total number of ways to select 2 VCRs from 6 VCRs = ⁶C₂

Therefore,

P(Both customers received damaged VCRs) = ³C₂ / ⁶C₂

= 0.2

B)

Number of ways to select 1 defective VCR from 3 defective VCRs =  ³C₁

Number of ways to select remaing 1 VCRs from 3 non-defective VCRs =  ³C₁

Total number of ways to select 2 VCRs from 6 VCRs = ⁶C₂

P(One of the two customers received a defective VCR) = ³C₁ * ³C₁ / ⁶C₂

= 0.6