Amanufacturing company regularly conducts quality control checks at specified periods on the products it manufactures. historically, the failure rate for led light bulbs that the company manufactures is 5%. suppose a random sample of 10 led light bulbs is selected. what is the probability that a) none of the led light bulbs are defective? b) exactly one of the led light bulbs is defective? how would the company use this informationc) two or fewer of the led light bulbs are defective? d) three or more of the led light bulbs are not defective?

0.5987,0.3151,0.9885,0.0115

Step-by-step explanation:

Given that a  manufacturing company regularly conducts quality control checks at specified periods on the products it manufactures. Historically, the failure rate for LED light bulbs that the company manufactures is 5%

Let X be the no of defectives in the sample of 10

X is binomial since each bulb is independent of the other and there are only two outcomes

P(X=x) =

Using the above we calculate

a) P(X=0) =

b) P(X=1)  

c)

d)

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)