Don't be fooled on Christmas!

Probability Level 3

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. Then, the probability that only two tests are needed is:

\dfrac{1}{4}

\dfrac{1}{3}

\dfrac{1}{2}

\dfrac{1}{6}

\dfrac{3}{2}

None of these

1 solution

Akhil Bansal
Dec 25, 2015

P(E) = The probability that only two tests are needed = Probability that the first machine tests is faulty $\times$ probability that second tested machine is faulty.
$P(E) = \dfrac{2}{4} \times \dfrac{1}{3} = \boxed{\dfrac{1}{6}}$

Moderator note:

Not quite. As pointed out, you're missing the case where 2 good machines are found.

But when 1st n 2nd both are not faulted machines then also we can idenify the faulty ones as they will be the onesvwhich are left

raghav bagri - 5 years, 5 months ago

Thanks. I have updated the answer to $\frac{1}{3}$ .

In future, if you think that there is an error with the problem, you can report it and the creator / staff will look into it further.

Calvin Lin Staff - 5 years, 5 months ago

I agree with Raghav. If the first two machines tested are not faulty we could then identify the remaining two machines as the ones that are faulty. Thus the desired probability is $2 * \dfrac{1}{6} = \boxed{\dfrac{1}{3}}.$

Brian Charlesworth - 5 years, 5 months ago

I also followed the same logic and got wrong. I think u r right sir.

Shyambhu Mukherjee - 5 years, 5 months ago

Yes. It should be 1/3

gopal narayanan - 5 years, 5 months ago

Exactly same,+1!

Rohit Udaiwal - 5 years, 5 months ago

Hello Rohit, you are studying in which class?

Akhil Bansal - 5 years, 5 months ago

Hey! $9^{th}$ .And you?

Rohit Udaiwal - 5 years, 5 months ago

@Rohit Udaiwal – I am in $12^{th}$ ..
I am glad to see that you know so much at this age.. you are brilliant. With this pace,you can crack any competitive exam.. keep it up!

Akhil Bansal - 5 years, 5 months ago

@Akhil Bansal – Thanks!But I"m not that good!!There are many more concepts to explore(especially physics)!

Rohit Udaiwal - 5 years, 5 months ago

@Akhil Bansal – Sir,why don"t you join us on slack?

Rohit Udaiwal - 5 years, 5 months ago

@Rohit Udaiwal – I am left with only 2-3 months for jee exam and board and I am very weak in chemistry and also have to revise class 11th. So, I think chatting on slack in a waste of time. That's why....

Akhil Bansal - 5 years, 5 months ago

I too agree with Raghav.

We can also see it this way.

There are 24 ways for testing the machines out of which 8 are those in which either the faulty machines comes at first two places or last two places. Hence, the answer should be $\frac{1}{3}$ .

Pulkit Gupta - 5 years, 5 months ago

Don't be fooled on Christmas!

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