Probability of a system

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cooltafel

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The assignment description:
Consider a system consisting of 6 components: C1, C2, C3, C4, C5 and C6. The system works
if at least 3 components work and at least 1 of the components C1 and C2 works. Each
component works, independently of all the other components, with probability 1/2. Let A
be the event that at least 3 components work and let B be the event that at least 1 of the
components C1 and C2 works.

a) Determine P(A), P(B), P(A or B) and P(A and B).
b) ...
c) ...

Where I got stuck:
I don't fully understand the description, is it that event A consists of C3, C4, C5 and C6 and B consists of C1 and C2?
Or is it that event A is C1, C2, C3, C4, C5 and C6 and event B is C1 and C2?

I have tried to draw it, but it didn't give a more clear view.
Sketch:
----C3----C4----C5----C6----C1----
--------------------------------C2----
But then you I don't count event A?

Could someone help with the drawing of the system?
I can't calculate the P(A) or P(B) without the drawing.

Thanks in advance for your effort and time!
 
No, A is the event that at least 3 components work. So A {c1,c2c3, c1c2c4, c1c2c5, c1c2c6, c1c3c4, c1c3c5, c1c3c6, c1c4c5, c1c4c6, c1c5c6, c2c3c4,....}
There are 6C3 elements in A!!
 
Thank you for your reply Jomo!
I see it now.
That would make then 6^3= 216 possibilities for event A.

Is it then correct that the probability of A is:
P (A) = (C and C and C) = (1-1/2)*1/2 + (1-1/2)*1/2 + (1-1/2)*1/2 - 1/8 = 5/8
And that
P (B) = 1/2 + 1/2 + 1/4 = 3/4
?
 
What does at least three mean to you in this problem?

P(c1, c2 and c3 works) = P(c1 works) AND P(c2 works) AND P(c3 works) = P(c1 works)*P(c2 works)* P(c3 works)

p(B) = P( just c1 works OR just c2 works OR c1 and c2 works) = .......

AND means to multiply. OR means to add !!!!!!!


Please try again!
 
Again thank you for your reply!

I tried it again:
P (A) = 1/2×1/2×1/2 = 1/8
P (B) = 1/2 + 1/2 - 1/4 = 3/4

For P (B) I used the following formula: P (C1 OR C2 OR (C1 AND C2)) = P(C1) + P(C2) - P (C1 AND C2)

Since if I would add the values for C1, C2 and (C1 AND C2) the result was will be greater than 1.

I hope it is correct now.
Thank you in advance for replying!
 
OR does not mean to subtract.
Think under what conditions will just c1 work!
 
cooltafel said:
The assignment description:
Consider a system consisting of 6 components: C1, C2, C3, C4, C5 and C6. The system works
if at least 3 components work and at least 1 of the components C1 and C2 works. Each
component works, independently of all the other components, with probability 1/2. Let A
be the event that at least 3 components work and let B be the event that at least 1 of the
components C1 and C2 works.
a) Determine P(A), P(B), P(A or B) and P(A and B).
Click to expand...
If you flip a coin six times how many of the \(2^6=64\) outcomes contain at least three heads? That is the model for your event \(A\).
How many of the total outcomes have a head in either the first or second position? That is the model for your event \(B\).
Can you discuss the probability \(\mathcal{P}(A\cap B)~?\)
 
p(B) = P( just c1 works OR just c2 works OR c1 and c2 works) = P( just c1 works) + P( just c2 works) + P( c1 and c2 works)

=P( just c1 works and c2 does not) + P( just c2 works and c2 does not) + P( c1 and c2 works)

=P( just c1 works)*P( c2 does not) + P( just c2 works)*P(c2 does not) + P( c1works)P(c2 works)

= (1/2)*(1/2) + (1/2)*(1/2) + (1/2)*(1/2) = 3(1/4) = 3/4
 
Thank you for your replies!

@Jomo Thank you for your last reply. I see it now, my formula was wrong. I did not take into account whether the other one isn't working.

@pka Thank you too for your reply!
P(A∩B) = P(A) * P(B) = 1/8 * 3/4 = 3/32
P(A∪B) = P(A) + P(B) =1/8 + 3/4 = 7/8
 
cooltafel said:
@pka Thank you too for your reply!
P(A∩B) = P(A) * P(B) = 1/8 * 3/4 = 3/32
P(A∪B) = P(A) + P(B) =1/8 + 3/4 = 7/8
Click to expand...
I urge you to consider the coin model. There are three cases
Component one works, two does not, and at least two of the remaining four work.
Component one does not works, two does work, and at least two of the remaining four work.
Component one works, two also does , and at least one of the remaining four works.
All the cases are pairwise disjoint because they differ in the first two components.
As I urge we can think in terms of a coin. We have six components each of which are working or not.
That is binary giving \(2^6=64\) possible states. We want three of the components working, one of which must \(C_1\text{ or }C_2\).
Each of case 1 &2 has eleven outcomes, while case 3 has fifteen outcomes. The total is thirty-seven.
The system has the probability is \(\dfrac{37}{64}\) of working.
 
cooltafel said:
Thank you for your replies!

@Jomo Thank you for your last reply. I see it now, my formula was wrong. I did not take into account whether the other one isn't working.
Click to expand...
The hardest part of computing p(E) is to understand what event E really means. You need to be careful and precise. At some point it will become natural for you. For now think a lot about whether or not you understand E perfectly.
 
pka said:
I urge you to consider the coin model. There are three cases
Component one works, two does not, and at least two of the remaining four work.
Component one does not works, two does work, and at least two of the remaining four work.
Component one works, two also does , and at least one of the remaining four works.
All the cases are pairwise disjoint because they differ in the first two components.
As I urge we can think in terms of a coin. We have six components each of which are working or not.
That is binary giving \(2^6=64\) possible states. We want three of the components working, one of which must \(C_1\text{ or }C_2\).
Each of case 1 &2 has eleven outcomes, while case 3 has fifteen outcomes. The total is thirty-seven.
The system has the probability is \(\dfrac{37}{64}\) of working.
Click to expand...

Thank you for this elaborate explanation!
Unfortunately, the probability theory part is just a small part of the whole course, so we are rushing through it.
I am trying to understand what you did, but it's a bit hard.
How can I represent it in formulas? Perhaps I see it then better.
 
cooltafel said:
Unfortunately, the probability theory part is just a small part of the whole course, so we are rushing through it. I am trying to understand what you did, but it's a bit hard. How can I represent it in formulas? Perhaps I see it then better.
Click to expand...
Let's suppose that we play a coin toss. A coin is tossed six times and the results are recorded.
Because the outcome is heads or tails ( binary) there would be \(2^6=64\) possible results.
You win if your six tosses include at least three heads among which the first or second must be a head. (here or can be both).
Can you see that this setup is exactly the same as the component system in your post?
My reply #10 explains why there are \(37\) of the \(64\) ways to win or that many ways for the system components to function.
 
Although I stayed away from using pea's suggestion (not sure why) he is correct. His thinking is the best way to understand the problems.

Many times when you have a probability problem you think about it in an entirely different way that the author posed the problem but this equivalent to your way of thinking about it.
 
Thank you for your replies, I worked it out.
It's more clear to me now.

Thanks again!
 
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