Answer:
The required probability = 0.9345
Step-by-step explanation:
Given that:
p = 0.17
n = 6
The required probability that two or less light bulbs are defective = (P<2)
(P<2) = P(X=0) +P(X=1) +P(X=2)
[tex](P<2) =\bigg [ ({^6C_0}\times 0.17^0 \times (1-0.17)^{6-0})+ ({^6C_1}\times 0.17^1 \times (1-0.17)^{6-1}) +( {^6C_2}\times 0.17^2 \times (1-0.17)^{6-2}) \bigg ][/tex]
[tex](P<2) =\bigg [ \dfrac{6!}{0!(6-0)!}\times 0.17^0 \times (1-0.17)^{6})+ (\dfrac{6!}{1!(6-1)!} \times 0.17^1 \times (1-0.17)^{5}) +( \dfrac{6!}{2!(6-2)! }\times 0.17^2 \times (1-0.17)^{4}) \bigg ][/tex]
[tex]\mathbf{(P<2) = 0.9345}[/tex]
