For each of the last 100 days at a certain factory, Light A w

 

­Dude at least double-check once whether you have pasted the problem properly. This leads people to waste their precious time trying to solve an incorrectly copied problem.

The highlighted part should be: for exactly 40 of the last 100 days at least one of the lights was on,

Thanks

Well, my apologies if I offended you. It was just a feedback because 10 odd people had attempted the problem then and I felt the need to point this out because almost all of them had gotten it wrong but didn’t post anything to update.

I am still learning the know-hows of the forum, so I am unable to contri by posting questions. But I have been posting answers lately. So to each their own.

Thanks and my regards

Posted from my mobile device

­Got this in a practice test. Had skipped it then.

Jotting my approach here- 
If A was off, B was also off. (Trap trap- doesnt imply that if B was off, A was also off)
If B was on, A was also on. (Trap trap- doesnt imply that if A was on, B was also on)

Out of the last 100 days,
20 days- both were on.
Remaining- 20 days.
Cant be the case that B was on. Had it been so, A would also have been on and had that been the case, the number of days both were on would have been 20+20=40. So Only A was on during these 20 days.

Total days A was on: 40
Total days B was on: 20

Required probability: 
P(A)= 40/100
P(B)= 20/100

MartyMurray KarishmaB Can you please help me on this question ?

Posted from my mobile device

­For each of the last 100 days at a certain factory, Light A was either on or off for that day, and likewise for Light B. Each day that Light A was off, Light B was also off. Each day that Light B was on, Light A was also on. In addition, for exactly 40 of the last 100 days, at least one of the lights was on, and for exactly 20 of the last 100 days, both of the lights were on.

Assume one of the last 100 days is chosen at random. Select for P(A) the probability that on the chosen day Light A was on, and select for P(B) the probability that on the chosen day Light B was on. Make only two selections, one in each column.­

Information we are given:

100 days

Each day that Light A was off, Light B was also off.

Each day that Light B was on, Light A was also on.

We need to notice a few key things about the above two statements:

- They mean the same thing. After all, if B is off when A is off, then if B is on, A is on.

- They are about only when both lights are on or off.

- They indicate that, whenever B is on, both are on.

- They don't say that whenever A is on, B is on. In other words, given those statements, A can be on by itself.

for exactly 40 of the last 100 days, at least one of the lights was on

for exactly 20 of the last 100 days, both of the lights were on

This question may seem hard, but if we see some key things, answering it becomes relatively straightforward.

The keyword "exactly" in "exactly 20 days" tells us that, on 20 days and no more than 20 days, both were on.

Thus, since both are on when B is on, B could have been on only on the 20 days on which both were on. 

Then, the 20 days must overlap the 40 days since, when "both of the lights are on," it's also true that "at least one of the lights was on."

Thus, since B can't be on when A is not on but there's no information indicating that A can't be on when B is not on, A must have been on on the other 20 days when B was not on, in addition to being on on the 20 days when both were on.

So, A was on for 20 + 20 = 40/100 days, and B was on for 20/100 days, and this question turned out to be more of a logic question than a math question. 

0.08

0.20

0.30

0.40

0.52

Select 0.40 for P(A) and 0.20 for P(B).

Correct answer: 0.40, 0.20­

tough one! got tripped up by assuming that A on = B on; and A off=B off. Gotta be so careful!