# Connected Component Problem - Probability

Components $1$ and $2$ are connected in parallel, so that subsystem works iff either $1$ or $2$ works;since $3$ and $4$ are connected in series, that subsystem works iff both 3 and 4 work. If components work independently of of one another and $P(\text{component works})=0.9$, calculate $P(\text{system works})$.

My try :

Call $1,2$ subsystem $A$, and $3,4$ subsystem $B$ \begin{align*} P[\text{system fails}] &= P[A\text{ fails}] + P[B \text{ fails}] - P[\text{both fail}] \\ &= (0.1)^2 + (1 - (0.9)^2) - (0.1)^2 (1-(0.9)^2) \\ &= 0.1981 \end{align*} Thus $$P[\text{system works}] = 1 - 0.1981 = 0.8019$$

• My try : call 1,2 subsystemA, and 3,4 subsystem B P[system fails] = P[A fails] + P[B fails] - P[both fail] = 0.1^2 + (1 - 0.9^2) - 0.1^2 *(1-0.9^2) = 0.1981 P[system works] = 1 - 0.1981 = 0.8019
– max
Feb 4 '16 at 22:42
• Formatting tips here.
– Em.
Feb 4 '16 at 22:51
• Your reasoning and result are correct. Feb 4 '16 at 23:10

I am wondering why this doesn't work.

Let $A$, $B$ be the events that these subsystems work, $A_i,B_i$ for $i=1,2$ be the event that component $i$ works, and I will use a bar to show that something doesn't work. Then, assuming the system will work if at least one subsystem works, \begin{align*} P(\text{System works})&=P(A\cup B)\\ &=P(A)+P(B)-P(A)P(B)\\ &=P(A_1\cup A_2)+P(B)-P(A)P(B)\\ &=P(A_1)+P(A_2)-P(A_1)P(A_2)+P(B_1)P(B_2)-P(A)P(B)\\ &=.9+.9-(.9)^2+(.9)^2-P(A)P(B)\\ &=0.99+ 0.81-(.99)(0.81)\\ &=0.9981 \end{align*}

• How is P(A1) = 0.99? It should be 0.81 right?
– max
Feb 4 '16 at 23:46
• @max [ Hehe That's my name :) ] Oooooh, thanks, I misread the probability. I will fix it.
– Em.
Feb 4 '16 at 23:51
• @max And no, they tell us the the probability that a component works is $.9$. $A_1$ is a component, thus $P(A_1) = .9$.
– Em.
Feb 4 '16 at 23:52
• @max I think you mean $P(B) = 0.81$. If so, then yes. I made an error.
– Em.
Feb 4 '16 at 23:57

You could technically go about this directly, since there are three cases to consider: the case where 1 works and 2 does not, the case where 1 does not work and 2 does, and the case where both work. 3 and 4 must always work, so we account for this probability in all cases. The first case is $(.9)(1-.9)(.9)^2$, the second case is the same and the third case is $(.9)(.9)(.9)^2.$ Hence, the answer should be $$2(.9)(1-.9)(.9)^2 + (.9)^4.$$

• What about when 1 and 2 doesn't work and 3 and 4 works? The system will still work right?
– max
Feb 4 '16 at 23:19