Total light bulbs = 24
Defective bulbs = 2
Non defective bulbs = 22
The probability of selecting 10 bulbs with two defective bulbs
We will use combination formula which states:
[tex]^nC_r = \frac{n!}{r!(n-r)!}[/tex]
= [tex]\frac{^2C_2 \times ^{22}C_8}{^{24}C_{10}}[/tex]
= [tex]\frac{10 \times 9}{24 \times 23}[/tex]
= [tex]\frac{90}{552}[/tex]
= 0.16
Therefore, the probability that both defective bulbs will be selected is 0.16
