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2,552 views
23 votes
23 votes

An electric circuit between two terminals $A$ and $B$ is shown in the figure below, where the numbers indicate the probabilities of failure for the various links, which are all independent.

What is the probability that $A$ and $B$ are connected?

  1. $\left(\dfrac{6}{25}\right)$
  2. $\left(\dfrac{379}{400}\right)$
  3. $\left(\dfrac{1}{1200}\right)$
  4. $\left(\dfrac{1199}{1200}\right)$
  5. $\left(\dfrac{59}{60}\right)$
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2 Answers

Best answer
27 votes
27 votes

Lets define the following events:

$P$ : Uppermost link is working.

$Q$: Middle link is working.

$R$: Lowermost link is working.

$W$ : Terminals $A$ & $B$ are connected.


From the given information, we can calculate the probabilities of events $P$, $Q$ and $R$ as follow:

(as I already consumed letter (capital) $P$ denoting an event so here by I will use letter (small) $p$ to denote probabilities,

e.g. $p\left ( X \right )$ denotes probability of happening of event $X$).

$\qquad p\left ( P \right ) = \dfrac{4}{5} \times \dfrac{4}{5} = \dfrac{16}{25}$,

$\Rightarrow\; p\left ( \widehat{P} \right ) = 1 - \dfrac{16}{25} = \dfrac{9}{25}$,

$\qquad p\left ( Q \right ) = \dfrac{2}{3} $,

$\Rightarrow\; p\left ( \widehat{Q} \right ) = \dfrac{1}{3} $,

$\qquad p\left ( R \right ) = \dfrac{3}{4} \times \dfrac{3}{4} = \dfrac{9}{16}$,

$\Rightarrow\; p\left ( \widehat{R} \right ) = 1 - \dfrac{9}{16} = \dfrac{7}{16}$.


The situation here can be represented using the following Venn Diagram:

Here Red region denotes the event $\widehat{W}$ in which $A \& B$ are disconnected.

Different gradients of Green colour represent the $W$ in which terminals $A \& B$ are connected.

We have to find $p\left ( W \right )$

Events $W$ and $\widehat{W}$ are mutually exclusive & totally exhaustive,

since either terminals $A$ and $B$ will be connected or they will be disconnected.

So we can write $p\left ( W \right )$ in terms of $p\left ( \widehat{W} \right )$ as follow:

$ p\left ( W \right )= 1 - p\left ( \widehat{W} \right )$.

Also it can be seen that 

$ p\left ( \widehat{W} \right )= p\left ( \widehat{P} \right ) \cap \left ( \widehat{Q} \right ) \cap \left ( \widehat{R} \right )$ where $\widehat{P}$, $\widehat{Q}$ and $\widehat{R}$ are independent events.

That is terminals $A$ & $B$ will be disconnected only when all of the links will fail simultaneously.

Using independence we can write,

$p\left ( \widehat{W} \right ) = p\left ( W \right )= p\left ( \widehat{P} \right ) \times \left ( \widehat{Q} \right ) \times \left ( \widehat{R} \right )$,

$\Rightarrow p\left ( \widehat{W} \right )= \dfrac{9}{25} \times \dfrac{1}{3} \times \dfrac{7}{16} = \dfrac{21}{400}$

$\Rightarrow p\left ( W \right )= 1 -  \dfrac{21}{400} = \dfrac{379}{400}$.

Correct Answer: $B$

19 votes
19 votes

If component are connected in Series(Both working) then we can say Probability of A connected to B
P(A ∩ B) = P(A)P(B)
If component are connected in Parallel(any one among two working) then we can say Probability of A connected to B
P(A U B) = P(A)+P(B) - P(A∩B)

Probability of top wire success P(A ∩ B) = P(A)P(B) = $\frac{16}{25}$
Probability of middle wire success P(A)  = $\frac{2}{3}$
Probability of lowest wire success P(A ∩ B) = $\frac{9}{16}$

Probability of circuit success = P(A U B U C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(A ∩ C) - P(B ∩ C) + P(A ∩ B ∩ C)
                                                                   =$\frac{16}{25} + \frac{2}{3} + \frac{9}{16} - \frac{16}{25}* \frac{2}{3} - \frac{16}{25}*\frac{9}{16} - \frac{2}{3}* \frac{9}{16} + \frac{16}{25}*\frac{2}{3}* \frac{9}{16}$
 
                                                                    = $\frac{379}{400}$ OPTION B

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