A shipment of 150 television sets contains 3 defective units. Find the probability of the vending company receiving (a) no defective units (b) all defective units and (c) at least one good unit?

How many TV's is the vending company getting? Is that given as well?

For instance, it might read, "A shipment of 150 television sets contains 3 defective units. If the vending company buys 10, find the probability of them receiving (a) no defective units (b) all defective units and (c) at least one good unit?".

No it just says

A shipment of 150 TV's contains 3 defective units. Find the probability of the vending company receiving (a) no defective units (b) all defective units and (c) at least one good unit

so they want three answers

Not much to it then. Wouldn't the prob. Of all defects be 3/150?

How many are not defects? Then, the prob. Of no defects would be \frac{\text{number not defective}}{150}

For 'at least one' defect, find the prob. Of no defects and subtract from 1.

That's what I had tried... or at least I think so..

The answers they give me are

a) 0.941 b)0.00000181 c)0.999998

a. No defects: \frac{147}{150}

b. all defects: \frac{3}{150}

I still think there is a piece of info missing from this problem as I mentioned earlier.

For 'at least one' good one, find the prob. Of no good ones and subtract from 1. But that is part a.

There is definitely something missing. I can tell from the answers you were given.

Thank youuu tons of help!

Your problem should be:

"A shipment of 150 television sets contains 3 defective units. If the vending company buys 3, find the probability of the vending company receiving (a) no defective units (b) all defective units and (c) at least one good unit?

all defects: \frac{C(3,3)C(147,0)}{C(150,3)}=.00000181

no defects: \frac{C(147,3)C(3,0)}{C(150,3)}=.9408