TIFR CSE 2012 | Part A | Question: 19

Question

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?

• A. $\left(\dfrac{6}{25}\right)$
• B. $\left(\dfrac{379}{400}\right)$
• C. $\left(\dfrac{1}{1200}\right)$
• D. $\left(\dfrac{1199}{1200}\right)$
• E. $\left(\dfrac{59}{60}\right)$

Answer 1

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.

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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} $,

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

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

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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: $B$

Comments

Nice answer !

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Another approach

I have labelled the links as 1,2,3.

It is clear that A and B will be disconnected only when all three of the links are disconnected.

So, it is easier to calculate the probability that A and B will be disconnected and then by taking the complement of this result we will get our answer.

Link 1 will be disconnected if either first point or second point or both of them are disconnected.

So, probability that link1 will be disconnected=$\frac{1}{5}+\frac{1}{5}-\frac{1}{25}(when\,both\,fail\,together)$ (by law of inclusion-exclusion) and this comes as $\frac{9}{25}$

Link 2 will be disconnected is straight forward with probability $\frac{1}{3}$

Link 3 will be disconnected when either first,second or both points are disconnected

Hence, probabiility of link3 being disconnected=$\frac{1}{4}+\frac{1}{4}-\frac{1}{16}=\frac{7}{16}$

So, probability that A and B are disconnected=P(Link1 failure)*P(Link2 failure)*P(Link3 Failure)

which is $\frac{9}{25}*\frac{1}{3}*\frac{7}{16}=\frac{21}{400}$

Hence, the probability that A and B are connected=$1-\frac{21}{400}=\frac{379}{400}$

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awesome anurag sir

Answer 2

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

Comments

P(A&B connected) = P(atleast 1 link functioning) = 1- P(No link functioning) = 1-[P(Ic)*P(IIc)*P(IIIc)]

P(Ic) = P(Either of two in failure) = 1-P(Both functioning) = 1-(4/5)2  = 1-16/25 = 9/25

P(IIc) = 1/3

P(IIIc) = P(Either of two in failure) = 1-P(Both functioning) = 1-(3/4)2 = 1-9/16 = 7/16

P(A&B connected) = 1-[P(Ic)*P(IIc)*P(IIIc)] = 1-[9/25*⅓*7/16] = 1-21/400 = 379/400

Answer 3

We need to find the probability that $A$ and $B$ are connected, which is denoted as $P(A \text{ and } B \text{ connected})$.

Let:

• Top Wire be $W_1$

• Middle Wire be $W_2$

• Bottom Wire be $W_3$

The terminals $A$ and $B$ will be connected if at least one of the wires is working.

$$P(A \text{ and } B \text{ connected}) = P(W_1 \text{ working} \cup W_2 \text{ working} \cup W_3 \text{ working})$$

Using the complement rule, this can be written as:

$$P(A \text{ and } B \text{ connected}) = 1 - P(\text{none of the wires are working})$$

$$P(A \text{ and } B \text{ connected}) = 1 - P(\overline{W_1} \cap \overline{W_2} \cap \overline{W_3})$$

Since the failure probabilities of the links are independent:

$$P(\overline{W_1} \cap \overline{W_2} \cap \overline{W_3}) = P(\overline{W_1}) \cdot P(\overline{W_2}) \cdot P(\overline{W_3})$$

For Wire 1 ($W_1$):

$$P(\overline{W_1}) = P(\text{point 1 not working} \cup \text{point 2 not working})$$

$$P(\overline{W_1}) = 1 - P(\text{point 1 working} \cap \text{point 2 working})$$

$$P(\overline{W_1}) = 1 - \left(P(\text{point 1 working}) \cdot P(\text{point 2 working})\right)$$

$$P(\overline{W_1}) = 1 - \left[1 - \frac{1}{5}\right] \cdot \left[1 - \frac{1}{5}\right]$$

$$P(\overline{W_1}) = 1 - \frac{4}{5} \cdot \frac{4}{5}$$

$$P(\overline{W_1}) = 1 - \frac{16}{25}$$

$$\therefore P(\overline{W_1}) = \frac{9}{25}$$

For Wire 2 ($W_2$):

$$P(\overline{W_2}) = \frac{1}{3}$$

For Wire 3 ($W_3$):

$$P(\overline{W_3}) = P(\text{point 1 not working} \cup \text{point 2 not working})$$

$$P(\overline{W_3}) = 1 - \left(P(\text{point 1 working}) \cdot P(\text{point 2 working})\right)$$

$$P(\overline{W_3}) = 1 - \left[1 - \frac{1}{4}\right] \cdot \left[1 - \frac{1}{4}\right]$$

$$P(\overline{W_3}) = 1 - \frac{3}{4} \cdot \frac{3}{4}$$

$$P(\overline{W_3}) = 1 - \frac{9}{16}$$

$$\therefore P(\overline{W_3}) = \frac{7}{16}$$

Calculating the final probability:

$$\therefore P(\text{none of the wires working}) = \frac{9}{25} \cdot \frac{1}{3} \cdot \frac{7}{16}$$

$$P(\text{none of the wires working}) = \frac{3}{25} \cdot \frac{7}{16} = \frac{21}{400}$$

Now, substitute this back into our original equation:

$$\therefore P(A \text{ and } B \text{ connected}) = 1 - \frac{21}{400}$$

$$P(A \text{ and } B \text{ connected}) = \frac{379}{400}$$

Therefore, the correct option is B.