Answer:
The correct option is d.
Step-by-step explanation:
It is given that the box contains 20 light bulbs, of which 5 are defective.
The possible ways of selecting r items from total n.
[tex]^nC_r=\frac{n!}{r!(n-r)!}[/tex]
The total possible ways to select 4 light bulbs form 20 bulbs is
[tex]^{20}C_4=4845[/tex]
We have to find the probability that at most 2 of them are defective.
Total possibility of selecting at most 2 defective bulbs.
Total = 0 bulb is defective + 1 bulb is defective +2 bulb is defective
[tex]Possibility=^{15}C_4\times ^{5}C_0+^{15}C_3\times ^{5}C_1+^{15}C_2\times ^{5}C_2[/tex]
[tex]Possibility=1365\times 1+455\times 5+105\times 10=4690[/tex]
The probability of selecting at most 2 defective bulbs is
[tex]P=\frac{\text{Required possibility}}{\text{Total possibility}}[/tex]
[tex]P=\frac{4690}{4845}[/tex]
Divide the numerator and denominator by 5.
[tex]P=\frac{938}{969}[/tex]
The correct option is d.
Answer:
Step-by-step explanation:
938/969 is right just took test got a 5/5
