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In a box, 20 bulbs are kept and some of them are defective. 2 bulbs

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In a box, 20 bulbs are kept and some of them are defective. 2 bulbs  [#permalink]

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New post 05 Mar 2019, 21:43
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Difficulty:

  65% (hard)

Question Stats:

56% (02:23) correct 44% (02:09) wrong based on 93 sessions

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In a box, 20 bulbs are kept and some of them are defective. 2 bulbs are selected randomly. Is the probability that both selected bulbs are defective less than 0.4?

    Statement 1. More than 50% of the bulbs are defective.
    Statement 2. Less than 60% of the bulbs are defective.


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Re: In a box, 20 bulbs are kept and some of them are defective. 2 bulbs  [#permalink]

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New post 06 Mar 2019, 07:33
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EgmatQuantExpert wrote:
In a box, 20 bulbs are kept and some of them are defective. 2 bulbs are selected randomly. Is the probability that both selected bulbs are defective less than 0.4?

    Statement 1. More than 50% of the bulbs are defective.
    Statement 2. Less than 60% of the bulbs are defective.


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IF YOU FIND MY SOLUTION TO BE HELPFUL, PLEASE GIVE ME KUDOS

(1) More than 50% of bulbs are defective. This means that more than 10 of the 20 bulbs are defective. Then the probability that two bulbs in a row will be defective is >10/20*9/20 = > 90/400 which is a little less than .25. So we can say the probability two bulbs are defective is > about .2 Of course there are several values >.2 that are less than and greater than .4 NS

(2) Less than 60% of bulbs are defective. This means that < 12/20 are defective. Then the probability both bulbs are defective is <12/20 * 11/20 < 132/400, which is a little more than .25, lets say .3. Therefore the probability that both bulbs are defective < .3, which of course <.4 Sufficient

The answer is B
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Re: In a box, 20 bulbs are kept and some of them are defective. 2 bulbs  [#permalink]

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New post 07 Mar 2019, 04:06
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Hello,

Greetings for the day!

This is a question of medium difficulty level. The statements have been framed in such a way that if you do not examine them carefully, you will end up falling for the trap answer which is Option C.

Let us now look at how to solve this question:
Of the 20 bulbs, let ‘x’ bulbs be defective and ‘y’ be non-defective.
If 2 bulbs are selected at random, the total possible outcomes will be 20C2.
Now, 20C2 = (20 x 19)/2 = 190.
Number of ways to select two bulbs which are defective = xC2, which also represents the favourable outcomes for the event defined in the question.

Therefore, Probability (Both bulbs are defective) = xC2 /190 The question asks us if this probability is less than 0.4.
Now, 0.4 = 2/5.

Hence, we are trying to ascertain if (xC2 / 190) < (2/5) which simplifies to whether
xC2 < 76.

xC2 can be less than 76 only if x is less than 13. Now, this is the data that we have to look for in the statements.

Statement I says that more than 50% of the bulbs were defective. This means that x can be 11 or higher. This does not conclusively tell us if x is less than 13 or not. Hence, statement I alone is insufficient.

Statement II says that less than 60% of the bulbs are defective. This means that x is 12 or lesser. If x is 12 or lesser, it is definitely lesser than 13, which is what we were trying to ascertain. Hence, statement II alone is sufficient.

Hence, the correct answer option is B.

In such DS questions on probability, it is important to analyse the question as much as possible and create a mathematical situation wherein you can use the data given in the statements with ease, as we did in this question. If you do not utilize the question data to your advantage, you will end up resorting to plugging in values which is not the ideal method for such questions.

Hope this helps!
Cheers.
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Re: In a box, 20 bulbs are kept and some of them are defective. 2 bulbs  [#permalink]

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New post 07 Mar 2019, 22:31

Solution


Given:
In this question, we are given
    • In a box, 20 bulbs are kept and some of them are defective.
    • 2 bulbs are randomly selected from the box.

To find:
We need to determine
    • If the probability that both selected bulbs are defective is less than 0.4 or not.

Approach and Working:
Let us assume that among 20 bulbs, N number of bulbs are defective.
    • Therefore, the probability that 2 selected balls will be defective = \(^NC_2/^{20}C_2 = \frac{N(N – 1)}{2} * \frac{2}{19 * 20} = \frac{N(N – 1)}{19 * 20}\)

Now, if the probability is to be less than 0.4, then we can say
    • \(\frac{N(N – 1)}{19 * 20} < 0.4\)
    Or, N(N – 1) < 152

As N is an integer, N must be less than 13.
Hence, we need to know whether N < 13 or not.

Analysing Statement 1
As per the information given in statement 1, more than 50% of the bulbs are defective.
    • Therefore, we can say N > 10
    • However, we cannot conclude whether N < 13 or not.

Hence, statement 1 is not sufficient to answer the question.

Analysing Statement 2
As per the information given in statement 2, less than 60% of the bulbs are defective.
    • Therefore, we can say that N < 12
    • As N < 12, we can definitely say N < 13.

Hence, statement 2 is sufficient to answer the question.

Hence, the correct answer is option B.

Answer: B
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Re: In a box, 20 bulbs are kept and some of them are defective. 2 bulbs  [#permalink]

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New post 11 May 2019, 11:41
[/color]Always approach such type qns from option point of view.
QN says whether p(xc2)<0.4?
option-1 says (>50% are defective),implying either 11c2/20c2,which is <0.3,however on the contrary if we take 19c2/20c2,which is more than >0.4(so insuficient to answer)
option-2 says <60% defective,implying <12 are defective.now take probability,which is lets take max value less than 12 as 11,11c2/20c2,which is less than <0.4)so sufficient.

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Re: In a box, 20 bulbs are kept and some of them are defective. 2 bulbs   [#permalink] 11 May 2019, 11:41
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