This is because there are a total of 22 bulbs. 16 of those bulbs work, giving us the fraction: 16/22. If you simplify 16/22 by dividing the numerator and denominator by 2, you get 8/11.
There are 16 working bulbs out of 6+16 = 22 bulbs total.
The probability of randomly selecting a working bulb is 16/22
After that first bulb is selected and not put back, the probability of randomly selecting another working bulb is 15/21. Take note that I subtracted 1 from each part of the original fraction.
So we get the answer of
(16/22)*(15/21) = 240/462 = 40/77 which is choice C.
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Extra info:
Choice A is only true if Ken puts the first selection back. You would compute (16/22)*(16/22) = 64/121. However, it sounds like he's not doing replacement. So whatever is selected is not put back. This is why I ruled out choice A.
Choice B is ruled out as well because 16/22 = 8/11 refers to the probability of one working bulb (instead of 2 in a row)
It's not clear how the fraction of choice D is formed, but we can rule it out because choice C is the answer.
This is distributive property and to do distributive property, you distribute what is outside of the parenthesis to the numbers inside the parenthesis.
