ISI2017-MMA-21

Question

There are four machines and it is known that exactly two of them are faulty. They are tested one by one in a random order till both the faulty machines are identified. The probability that only two tests are required is

• A. $\left(\dfrac{1}{2}\right)$
• B. $\left(\dfrac{1}{3}\right)$
• C. $\left(\dfrac{1}{4}\right)$
• D. $\left(\dfrac{1}{6}\right)$

Comments

@sukanya :given answer is wrong.

correct answer will be $\frac{1}{3}$

---

$\frac{2C1*1C1}{4C2} = \frac{2}{6} = \frac{1}{3}$

---

Out of 4 , it's given that exactly two are faulty therefore two are non faulty
now, ways to permute them will be 4!/(2!*2!) = 6

so total sample space contain 6 items now probab. that in two test we detect the faulty machines = (probab. of detecting two faulty m/c in 1st two tests) OR (probab. of detecting two unfaulty m/c in 1st two tests)
                            
                      = (1/6) + (1/6)
                      = (2/6)
                      =(1/3)

Answer 1

There will be two ways -

1. When the first test & the second test both identified the faulty machines

OR

2. When the first test & the second test both identified the non-faulty machines (because then we know that the remaining two machines are faulty)
    

1. Probability of the first machine tested is faulty = $\dfrac{\text{Number of favourable outcome}}{\text{Total number of outcomes}} = \dfrac{2}{4}$ $\big[\text{∵ as there are 2 faulty machines out of 4 machines}\big]$

Now, there are $3$ machines left as $1$ faulty machine has already been identified.

Probability (second machine tested is faulty in the second test ) = $\dfrac{\text{Number of favourable outcome}}{\text{Total number of outcomes}} = \dfrac{1}{3}$

$\color{green}{\text{Probability that only two test is needed}} =$ $\color{blue}{\text{Probability of the first machine tested is faulty}} \times$ $\color{blue}{\text{ Probability of the second machine tested is faulty in the second test}}$

$\qquad = \dfrac{2}{4}\times \dfrac{1}{3}$

$\qquad = \dfrac{1}{2}\times \dfrac{1}{3}$

$\qquad = \color{Gold}{\dfrac{1}{6}}$

2.  Probability of the first machine tested is Not-faulty = $\dfrac{\text{Number of favourable outcome}}{\text{Total number of outcomes}} = \dfrac{2}{4}$ $\big[\text{∵ as there are 2 Non-faulty machines out of 4 machines}\big]$

Now, there are $3$ machines left as $1$ Non-faulty machine has already been identified.

Probability (second machine tested is Not-faulty in the second test ) = $\dfrac{\text{Number of favourable outcome}}{\text{Total number of outcomes}} = \dfrac{1}{3}$

$\color{green}{\text{Probability that only two test is needed}} =$ $\color{blue}{\text{Probability of the first machine tested is Not-faulty}} \times$ $\color{blue}{\text{ Probability of the second machine tested is Not-faulty in the second test}}$

$\qquad = \dfrac{2}{4}\times \dfrac{1}{3}$

$\qquad = \dfrac{1}{2}\times \dfrac{1}{3}$

$\qquad = \color{Gold}{\dfrac{1}{6}}$

Now, $\color{Orange}{\text{Required Probability}}$ = $\color{purple}{\text{Case 1) OR Case 2)}}$

$\qquad \qquad = \dfrac{1}{6}+\dfrac{1}{6}$

$\qquad \qquad = \dfrac{2}{6}$

$\qquad \qquad = \color{purple}{\dfrac{1}{3}}$

Correct Answer: $B$

Comments

@sukanya :given answer is wrong.

correct answer will be $\frac{1}{3}$

---

Only two tests are required so they can be like this=

P(getting first machine  tested is faulty)*P(getting second machine tested is faulty) OR

   P(getting first machine tested is non faulty)*P( getting second machine tested is non faulty)

=2/4*1/3 + 2/4 * 1/3

=1 /3

---

Answer should be $\frac{1}{3}$ only.
Note that it's given in the question -
 "They are tested one by one in a random order ${\color{Magenta} {till}}$ both the faulty machines are identified."

So 2 cases are there- FF and NN.
similar question

Answer 2

possible cases

F= faulty N= Non faulty

FF,FN,NF,NN

if we get FF or NN we can say we found 2 Faulty machine

so (F,F) = $\frac{2}{4}*\frac{1}{3}$

and (N,N) =  $\frac{2}{4}*\frac{1}{3}$

Total probability = $\frac{1}{3}$

Answer 3

$\frac{2}{4} * \frac{1}{3} = \frac{1}{6}$

coorect me if i'm wrong

Answer 4

There are 4 machines $M_{1},M_{2},M_{3},M_{4}$

Here say $M_{3},M_{4}$ are faulty

So, we can select it either by $M_{3},M_{4}$ or  $M_{4},M_{3}$ =$2$ ways

Now among 4 machines we can select 2 in $\binom{4}{1}\times \binom{3}{1}$ ways=$12$ ways

So, total probability that only $2$ test cases required to get both mchines are faulty is $\frac{2}{12}=\frac{1}{6}$

Comments

what is P(faulty *faulty) means?

It means item is faulty and test is also faulty.

But we need item is faulty and test must be non faulty

right?

---

probability of first faulty machine  is given 2/4

probability of now second faulty machine can be be given as 1/3

as two events are independent  given as 2/4*1/3*2 as two faulty machines can be tested in 2 ways here

total probability as 1/3  

am i right

---

@srestha I think this approach is not correct.

We have considered ordering of machines M1,M2,M3 and M4. So even in denominator we will have to consider the orderings.

No. of ways of arranging M1,M2,M3,M4 = 4! = 24

No. of ways in which the faulty machines(M3 and M4) appear as the first 2 machines

or non-faulty machines(M1 and M2) appear as the first 2 machines

= 2 * {2!} + 2 * {2!} .........(cases: M3,M4,_,_  M4,M3,_,_  or M1,M2,_,_  M2,M1,_,_)

= 4 + 4 = 8

Hence the required probability = 8/24 = 1/3

Let me know if anything is wrong here. :)

Answer 5

Without loss of generality. There are $C(4,2)$ = $6$ ways to identify two of the machines.  Only two tests are required. Thus,  $p$= $2/6$=$1/3$

Answer 6

They are tested one by one in a random order till both the faulty machines are identified. The probability that only two tests are required that's mean the first to machine which was selected is faulty then total way to chose 2 machine from 4 machine is = 4 Choose 2 =6

fav case than choose the faulty machine 2C1 ways.

then prob=2/6=1/3