Answer: [tex]\dfrac{63}{253}[/tex]
Step-by-step explanation:
Given : Number of good laptops =20
Number of defective ones=3
Total laptops =23
If 2 laptops were selected, then the probability that none of them was defective would be = [tex]\dfrac{^{20}C_2}{^{23}C_2}[/tex]
[tex]=\dfrac{\dfrac{20!}{2!18!}}{\dfrac{23!}{2!21!}}=\dfrac{19\times20}{23\times22}\\\\=\dfrac{190}{253}[/tex]
The probability that at least one will be defective= 1- P(none of them is defective)
[tex]=1-\dfrac{190}{253}=\dfrac{63}{253}[/tex]
Hence, the probability that at least one will be defective. [tex]=\dfrac{63}{253}[/tex]
