Question 938149
{{{3/120=1/40}}} of the bulbs are defective,
so the probability of Ben's first pick being defective is {{{1/40}}},
and the probability of Ben's first pick being good is {{{1-1/40=39/40}}} .
 
a) If Ben's first pick is defective,
he will have 119 light bulbs for his second choice,
and 2 of those will be defective.
So, the probability of picking a second defective bulb after picking the first defective bulb is
{{{2/119}}} .
All in all, the probability that Ben will pick 2 defective bulbs is
{{{(1/40)*(2/119)=1/(20*119)=1/2380}}} .
That is about {{{0.00042="0.042%"}}} .
 
On the other hand, if Ben's first pick is a good bulb,
he will have 119 bulbs to choose from, 3 of them defective, along with {{{119-3=116}}} good bulbs.
So, the probability of Ben picking a second good bulb after picking the first good bulb is {{{116/119}}} .
All in all, the probability that Ben will pick 2 good bulbs is
{{{(39/40)*(116/119)=39*116/(40*119)=39*58/(20*119)=2262/2380=1131/1190}}} .
 
b) If Ben's chosen bulbs are neither both defective, nor both good bulbs, Ben will have picked exactly 1 defective bulb.
The probability of that is {{{1-1/2380-2262/2380=(2380-1-22622)/2380=117/2380}}} .
That is about {{{0.049="4.9%"}}} .
 
Another way to get at the part b) answer is adding up
the probability of first bulb defective, second bulb good, plus
the probability of first bulb good, second bulb defective.
That would be
{{{(1/40)(117/119)+(39/40)(3/119)=117/(40*119)+39*3/(40*119)=117/(40*119)+117/(40*119)=117/(20*119)=117/2380}}}
 
When two ways to solve the problem yield the same result, I am happy.