Atoy manufacturer inspects boxes of toys before shipment. each box contains 19 toys. the inspection procedure consists of randomly selecting three toys from the box. if one or more of the toys are defective, the box is not shipped. suppose that a given box has two defective toys. what is the probability that it will be shipped?

The probability that a box with two defective toys will be shipped is 0.29825

Step-by-step explanation:

The first toy picked has a 17 in 19 chance of not being defective

The second toy picked has a 16 in 18 chance of not being defective

The third toy picked has a 15 in 17 chance of not being defective

The probability that the toy is not defective, and will be shipped

P(toy not defective and shipped) = (17/19) *(16/18)*(15/17)

                                                       = (16*15)/(19*18)

                                                        = 240/342

                                                        = 0.70175

The question said, suppose that a given box has two defective toys, what is the probability that it will be shipped?

P( toy defective and not shipped) = 1 - P(toy not defective and shipped)

                                                        = 1-0.70175

                                                        =  0.29825

The probability that a box with two defective toys will be shipped is 0.29825

fa tha first one you gonna add y to both sides

it look like this

x-y+y=6+y

x=y+6

therefore your answer is

x =y+6

fa tha second one

you gonna flip tha equation

it gonna look like x-4y=y

then add 4 to both sides

it gonna look like tha

x-4 +4 =y +4

so therefore tha answer is x=y+4

step-by-step explanation: