You start with 7 total and because you have already picked twice so you now have 5 left.
It wants the probability of the second defective light so that means of the two first picked one was good and one was defective.
Now you have 3 good lights and 2 defective lights.
the probability of picking a defective light would be 2/5 =0.4
Need some help please!
A box contains four good light bulbs and three defective ones. Bulbs are selected one at a time (without replacement). Find the probability that the second defective bulb is found on the third selection.
2 answers
let g be good bulb
let b be bad bulb
so your events are:
gbb ---> (4/7)(3/6)(2/5) = 4/35
bgb ---> (3/7)(4/6)(2/5) = 4/35
prob(as stated) = 4/35 + 4/35 = 8/35 or appr .229
let b be bad bulb
so your events are:
gbb ---> (4/7)(3/6)(2/5) = 4/35
bgb ---> (3/7)(4/6)(2/5) = 4/35
prob(as stated) = 4/35 + 4/35 = 8/35 or appr .229
1 answer