A Box Of 16 Components Contains 4 Defective Components. If 3 Components Are Drown From The Box What Is The Probability That They Are All Good. (I)If There Is Replacement (Ii)If There Is No Replacement

[godhardonsomu]

A box of 16 components contains 4 defective components. If 3 components are drown
from the box what is the probability that they are all good.
(i)If there is replacement
(ii)If there is no replacement

[woofgang]

do your own homework why don’t you?

[Buenchico]

The probability of the first component drawn being good is 12/16

If the good component is then replaced in the box (and that continues to happen) the probability of the 'good' outcome occurring three times in a row is therefore 12/16 x 12/16 x 12/16
[I'll leave you to do the cancelling and the multiplying!]

However if, after drawing a good component initially, there is no replacement then the probability of the second one drawn also being good will be 11/15 (because there are only 15 components left in the box, of which 11 are good)

Similarly, the probability of the last one being good will be 10/14,

So the probability of all three in a row being good is 12/16 x 11/15 x 10/14

[fiction-factory]

I agree with Buenchico's answer. The answer under the replacement option is easier to work out if you simplify 12/16 to 3/4- the solution is 3³/4³= 27/64

[godhardonsomu]

Thanks Buenchico

[fiction-factory]

Under the replacement option you'd need to shake the box (without damaging any)
i) works out as 0.421 (42.1%) to 3 significant figures
(ii) works out at 0.393 (39.3%) to 3 significant figures

I'm not immediately sure though why you'd want to use the replacement method in a case like this.

[fiction-factory]

^ okay, 0.422 for (i)- should have used a calculator