A shipment of 500 bulbs contains 10% defective bulbs. If three bulbs a

Question

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A shipment of 500 bulbs contains 10% defective bulbs. If three bulbs are chosen randomly, what is the probability that exactly 2 of them are defective?

A.  \frac{1}{500}

B.  \frac{3}{500}

C.  \frac{3}{1000}

D.  \frac{9}{1000}

E.  \frac{27}{1000}

Abhisharma1808 · Answer

no. of cases = 1/10^2*1/9^1
3c2 is the situation=3*9/10^3
27/1000

E is the answer

andrenoble · Answer

I also arrived at E, but with a different apporoach:

(450C1*50C2) / 500C3 , which essentialy calculates number of cases of taking 1 good lamp of 450 and 2 defective lamps from 50 and dividing by the total number of cases.

I think that real probability will be a little higher than 27/1000, since the second defective lamp will not have the probability of 1/10 to be taken from the box, it will be, e.g. 49/499 or 49/498, for which the method with combinations accounts.

Bunuel , could you please comment on this?

Regor60 · Answer

The indicated answer works only if chosen means selected but not removed and available for the next draw

Posted from my mobile device

AmbAgg · Answer

The number of ways in selecting 2 defective and 1 non defective bulbs are  (\frac{1}{10})^2 * (\frac{9}{10})^1 
The number of ways to arrange 3 items with 2 similar items is given by  \frac{3!}{2!}  . (2 similar items are 2 defective bulbs)
Therefore   P(d) = \frac{3!}{2!} * (\frac{1}{10})^2 = 3 * \frac{9}{1000} = \frac{27}{1000}

In this we have assumed bulbs are being replaced. Even if bulbs are not being replaced, answer will be very close are we are selecting very less number of bulbs from a very large number of bulbs.
To confirm this,
  P(d) = \frac{3!}{2!} * \frac{(50*49*450)}{(500*499*498)} \approx{}\frac{25.641}{1000 } 
Hence in this case also we can see that closest answer is 27/1000 only

Jeemate · Answer

The question doesn't seem that hard, instead it confuses us because the word approximate isn't mentioned.
There are 50 defected and 450 non-defected pieces.
For the first pickup being non-defective: 450/500
For the second pickup being defective: 50/499 (one was picked up earlier)
For the third pickup being defective: 49/498 (2 were already picked up)
Now the order can be rearranged too so the number of ways this sequence can be rearranged: 3x2x1/2x1=3
(450x50x49/500x499x498)*3= 27/1000 approx. E

summerbummer · Answer

Bunuel  used to following approach. Let me know if it's correct?

(3 x 50C1 x 50C1 x 450C1) / (500C1 x 500C1 x 500C1) = 27 / 1000

50C1 because there are 50 defective bulbs and 450C1 because there are 450 working bulbs.

Maria240895chile · Answer

from 500 we have 0,1 defective =50 and 0,9 non-defective=450

we need  \frac{restriction}{all-options}

\frac{D*D*ND}{500*499*498}

\frac{50*49*450}{500*499*498}

simplify getting  \frac{49*45}{499*498} 
 \frac{49}{499}  is approximately 10
 \frac{45}{498}  is approximately 11

so we get  \frac{1}{10*11}=\frac{1}{110}

now, remember that we have 3 different arrangements for exactly 2 D
D;D;ND
D;ND;D
ND;D;D

so multiply  \frac{1}{110}*3  
 \frac{3}{110} is 0,027... which is the same as  \frac{27}{1000}

hope it helps

adewale223 · Answer

Probability of exactly 2 defective out of 3

50/500 x 49/499 x 450/498 x 3C2

1/10 x 1/10 x 9/10 x 3

Approximately 27/1000
GMAT quant instructor Lagos Nigeria

Posted from my mobile device

sujoykrdatta · Answer

Number of defectives = 10% of 500 = 50; non-defectives = 450
Number of ways of choosing any 3 bulbs = 500C3 = 500*499*498/3! = 500*499*83
Number of ways of choosing 2 defective and one non-defective = 50C2 * 450C1 = 50*49/2! * 450 = 50*49*225

Probability = (50*49*225)/(500*499*83) = 49*225/(10*499*83) = 225/(10*10*83) ~ 3/100 
Closest option is E

Answer E

8Harshitsharma · Answer

Poorly worded problem - doesn't state that replacement is allowed. Only upon assuming replacement is allowed will we obtain choice E 27/1000

SharukhB · Answer

A simpler way would be by doing the following

Probability of defective bulb = 0.1
Probability of normal bulb = 0.9

Therefore probability of Exactly 2 defective would be = 0.1 X 0.1 X 0.9 X 3 (because 3 cases B1 & B2 defective, B2 & B3 defective and finally B1 &B3 are defective)

Which give 0.027 = 27/1000 Hence E

WillHunting · Answer

I seriously challenge the marked answer E. I think (and I know I am right) that the right answer is D.

Lets see. 50 bulbs are defective but we are choosing randomly so we are not looking at the bulbs while choosing but randomly choosing 3 out of 500, in no particular order.

For our problem, we want 2 to be defective and one not defective.
Two (defective) bulbs chosen --> one at 50/500 probability and another at 49/499 probability.
One (non-defective) bulb chosen --> 450/498 probability

So, total probability is (50x49x450) / (500x499x498)
Now, lets make solving easy since we are looking for approximation, so, (1x1x9) / (10x10x10) = 9/1000. So, the answer is D.

***Just want to emphasis one thing since it's repeatedly been mentioned here incorrectly. THE ARRANGEMENT DOES NOT MATTER IN THIS CASE. You just pick 3 bulbs out of 500, it really does not matter if the first two are defected or last two. So, do not multiply by 3.

A shipment of 500 bulbs contains 10% defective bulbs. If three bulbs a

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