223,106 views
15 votes
15 votes
A changes

16. By accident, 6 burned out bulbs have been mixed in with 16 good ones, Ken is replacing old bulbs in his house. If he selects two bulbs at random from the box of 22, what is the
probability they both work?

A changes 16. By accident, 6 burned out bulbs have been mixed in with 16 good ones-example-1
User Mle
by
3.0k points

2 Answers

19 votes
19 votes

Answer: Choice C) 40/77

=================================================

Step-by-step explanation:

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.

------------

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.
User Spawn
by
2.9k points
5 votes
5 votes

Answer: 8/11

Step-by-step explanation:

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.

User Lars Lau Raket
by
2.7k points