Question

[ Solved by @reemii and @thomas5267 ] A cute probability problem

Answer 1

am not good at probability...aawwnn:"(

Answer 2

i'd say 2/3

Answer 3

Probability is always between 0 and 1.
2/3 is less than 1, so I'd say it isn't a bad guess ;)

Answer 4

yes ser

Answer 5

Do you get a numerical answer?

Answer 6

answer is a function of p and n

Answer 7

Is it p(n-1)?

Answer 8

Nope, but it seems you're on right track.. I'll post the options, one sec

Answer 9

Not sure haven't checked, is it: \(n^2p^n\)?

Answer 10

here are the options

1)\(1-[np^{n-1}(1-p)]^{n} \)

2) \((1-p^n)^{n^2} \)

3) \(1-(1-p^n)^{n^2} \)

4) \(1-n^2p^n\)

Answer 11

option number 2, my final answer.

Answer 12

mine looks pretty similar to 4 so that'll be my guess buuuut I won't really be satisfied until I figure it out haha so don't spoil it. I'll think about it mroe.

Answer 13

Okay, I won't say anything :)

Answer 14

are you sure (3) has this \(n^2\) exponent?

Answer 15

Nice question btw

Answer 16

Yes, options 2 and 3 are compliments of each other

Answer 17

I only find \((1-(1-p^n)^n)^n\) . so i'lll wait for the answer.. maybe this is equivalent to one of your options but I don't see it..

Answer 18

Could you please explain how you arrived at

Answer 19

your answer seems very close to the options\[(a^b)^b = a^{b^2}\]

Answer 20

Yep me too. \(\left[1-\left(1-p^n\right)^n\right]^n\).

Answer 21

Okay, lamp glows = every sheet works : \(P(lamps\: glows) = P(all\: sheets\: work) = P(sheet\:1\:works)^n\). ( by independence)

Sheet 1:
\(P(sheet\: 1\:works) =\) P(at least one line is ok) = 1 - P(no line is ok) = 1 - \(P(line\: 1\: is\: not\: ok)^n\)

and line 1:
P(line 1 not ok) = 1 - P(line 1 ok) = 1 - \(p^n\).

Answer 22

I fold things up and I arrive at \((1 - (1 - p^n)^n)^n\).

Answer 23

Put it this way, the probability of 1 line working is \(p^n\). The probability of n lines failing is \(\left(1-p^n\right)^n\), so the probability of at least one line working is \(1-\left(1-p^n\right)^n\). Since all n sheet must work, the probability of the circuit working is \(\left[1-\left(1-p^n\right)^n\right]^n\).

Answer 24

Awesome! I completely agree that is the correct answer, textbook options are wrong.

Answer 25

Is there some sort of massive cancellation?

Answer 26

I guess what you have is the best simplified form possible

Answer 27

It seems like the textbook is indeed wrong. Numerical verification shows that none of the answer equals mine and reemii's.

Answer 28

It turns out the chances of the circuit working is really small. Even with p=0.5 and n=3 the probability is only 0.0359625.

Answer 29

I guess the author confused himself. 3 is the closest since if the probability of the circuit working is \(1-\left(1-p^n\right)^{n^2}\) then the probability of the circuit not working would be \(\left(1-p^n\right)^{n^2}\). This can be interpreted as all lines failing. Clearly if all lines fail then the circuit won't work but there exists cases such that despite not all lines fail the circuit still fails. I have no idea what 1 meant and why 4 is here. Suppose the switches all worked perfectly then for n>1 the probability as calculated by 4 is larger than 1?????

Answer 30

I mean the probability is smaller than 0 for option 4 if n>1 and p=1.

Answer 31

Yeah absolutely! we would get #3 option if the sheets were in parallel, instead of series

Answer 32

Yep if the sheets were in parallel #3 is the correct one.

Answer 33

You might want to add an warning to not look past the questionable options you gave.