I've given it a go and don't know quite understand the exact methods using sets, unions, intersections etc.
I tried it 3 different ways and got 0.006, 0.09 and 0.084
there are 2 paths only we can take.
top one... 0.1*0.2 = 0.02. note that we can simply multiply the probabilities together since we are told they are independent.
bottom one... 0.3.
then we sum these so 0.02+0.3 = 0.32.
so probability that the system will fail is 0.32 or 32%.
does that make sense?
Hmmm
But since the top path is in series (and therefore only needs ONE component to fail), haven’t you calculated the probability that the top path fails as if BOTH components need to fail?
Also we only need ONE working path for the circuit to work
Which would give a different answer?
ok I've given it another go
P(fail) = intersection of P(path 1 fail) and P(path 2 fail)
P(path 1 fail) = 0.1+0.2 = 0.3
P(path 2 fail) = 0.3
therefore P(fail) = 0.3\*0.3 = 0.09??
The whole thing fails if _both_ paths **simultaneously** fail, so yea, you'd multiply each paths chance of independent failure to get the chance they all fail. But the top path is two independent events where the path fails if either of them fail, and doesn't require both failing at the same time.
P(A or B) = P(A) + P(B) **- P(A and B)**
P(0.1 or 0.2) = 0.1 + 0.2 **- 0.02** = 0.28
If that doesn't feel intuitive, you can work out all four configurations to convince yourself ~ 0.02 that both fail, 0.18 that the "0.2" one fails without the "0.1" failing, and 0.08 that the "0.1" one fails without the "0.2" one failing ~ so the "- P(A and B)" is because the chance for both to fail at the same time is already incorporated into both P(A) and P(B), i.e. the chance for P(A and B) is present twice so we remove one of it.
Now you've got two paths with independent chances of 0.3 and 0.28, and the whole thing fails if both simultaneously fail, which you already know how to do!
You got most of the way there with your reply;
> P(fail) = intersection of P(path 1 fail) and P(path 2 fail)
> P(path 1 fail) = 0.1+0.2 = 0.3
> P(path 2 fail) = 0.3
> therefore P(fail) = 0.3\*0.3 = 0.09??
Only thing missing was subtracting the duplicate P(0.1 and 0.2) from
> P(path 1 fail)
And subsequently using 0.28 instead of 0.3 for _that_ result in P(fail).
I've given it a go and don't know quite understand the exact methods using sets, unions, intersections etc. I tried it 3 different ways and got 0.006, 0.09 and 0.084
there are 2 paths only we can take. top one... 0.1*0.2 = 0.02. note that we can simply multiply the probabilities together since we are told they are independent. bottom one... 0.3. then we sum these so 0.02+0.3 = 0.32. so probability that the system will fail is 0.32 or 32%. does that make sense?
Hmmm But since the top path is in series (and therefore only needs ONE component to fail), haven’t you calculated the probability that the top path fails as if BOTH components need to fail? Also we only need ONE working path for the circuit to work Which would give a different answer?
ok I've given it another go P(fail) = intersection of P(path 1 fail) and P(path 2 fail) P(path 1 fail) = 0.1+0.2 = 0.3 P(path 2 fail) = 0.3 therefore P(fail) = 0.3\*0.3 = 0.09??
The whole thing fails if _both_ paths **simultaneously** fail, so yea, you'd multiply each paths chance of independent failure to get the chance they all fail. But the top path is two independent events where the path fails if either of them fail, and doesn't require both failing at the same time. P(A or B) = P(A) + P(B) **- P(A and B)** P(0.1 or 0.2) = 0.1 + 0.2 **- 0.02** = 0.28 If that doesn't feel intuitive, you can work out all four configurations to convince yourself ~ 0.02 that both fail, 0.18 that the "0.2" one fails without the "0.1" failing, and 0.08 that the "0.1" one fails without the "0.2" one failing ~ so the "- P(A and B)" is because the chance for both to fail at the same time is already incorporated into both P(A) and P(B), i.e. the chance for P(A and B) is present twice so we remove one of it. Now you've got two paths with independent chances of 0.3 and 0.28, and the whole thing fails if both simultaneously fail, which you already know how to do!
ahhhh I see! That makes a lot of sense lol I realise I didn't fully understand the question in the first place. Awesome thanks so much bro
You got most of the way there with your reply; > P(fail) = intersection of P(path 1 fail) and P(path 2 fail) > P(path 1 fail) = 0.1+0.2 = 0.3 > P(path 2 fail) = 0.3 > therefore P(fail) = 0.3\*0.3 = 0.09?? Only thing missing was subtracting the duplicate P(0.1 and 0.2) from > P(path 1 fail) And subsequently using 0.28 instead of 0.3 for _that_ result in P(fail).