One of 5 pens is faulty. They are tested one at a time, what is the pr

Question

One of 5 pens is faulty. They are tested one at a time, what is the probability that the faulty pen is the 3rd one tested ?

A. 12/125
B. 1/5
C. 2/5
D. 3/5
E. 4/5

CrackverbalGMAT · Answer

P(First Good Pen) = 4/5

P(Second good pen) = 3/4

P(3rd pen is faulty) = 1/3

The required probability = 4/5 * 3/4 * 1/3 = 1/5

Option B

Arun Kumar

jrk23 · Answer

Can't we say that if third pen was defected than first 2 is also defective. thus, probability = 4/5 * 3/4 * 1/3 (I mean after checking 1st pen 4 left so probabilty of being defective after testing first is 3/4) and same with third.

In above case how question would be written... if p= 4/5 * 3/4 * 1/3

Thanks

CrackverbalGMAT · Answer

Hi. In that case, the probability of getting a faulty pen will always be 1/5, whether it comes 1st, 2nd 3rd 4th or 5th

chetan2u · Answer

The answer will be 1/5.
You have made an error in 4/5 * 4/5 *1/5. It should be 4/5 *3/4 *1/3=1/5
The probability will keep changing and will not remain constant. This would be the case if we were replacing the pen after testing.

chetan2u · Answer

Logically 
Each pen should have the same probability of being defective. Thus 1/5 is the probability

Solution
The probability of first pen being good =4/5
The probability of the 2nd pen being good =3/4, as there are only 4 pens left now. 
The probability of the third pen being defective =1/3, as there are only 3 pens left and one is defective 
P= \frac{4}{5}*\frac{3}{4}*\frac{1}{3}=\frac{1}{5}

B

CrackverbalGMAT · Answer

Hi chetan2u. Basically, irrespective when the faulty pen comes (whether 1st, 2nd, 3rd, 4th or 5th), the probability will always be 1/5, right?

jrk23 · Answer

Can't we say that if third pen was defected than first 2 is also defective. thus, probability = 4/5 * 3/4 * 1/3 (I mean after checking 1st pen 4 left so probabilty of being defective after testing first is 3/4) and same with third.

In above case how question would be written... if p= 4/5 * 3/4 * 1/3

Thanks

That mean my thought was right.

Answer is 1/5 not 16/25...... m i right?

Hi. In that case, the probability of getting a faulty pen will always be 1/5, whether it comes 1st, 2nd 3rd 4th or 5th

wishmasterdj · Answer

We go with the constraints in the question.

First pick - non defective (1 out 4 non defective) - Probability= 4/5
Second pick - again non defective (1 out of 3 non defective left, as one got picked in the first pick) - Probability = 3/4
Third pick - defective pick (1 out of 3 remaining, as the other 2 are free from defects) - Probability = 1/3

So for this sequential scenario, we just multiply the individual event probabilities 
4/5 * 3/4 * 1/3

One of 5 pens is faulty. They are tested one at a time, what is the pr

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One of 5 pens is faulty. They are tested one at a time, what is the pr

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