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

Could not understand this. Can you please elaborate a little.

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

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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

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

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.

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

Bunuel

may you explain to me why are we arranging bulbs and taking 3 cases? I have seen questions similar to this for example: A bag has 5 red balls and 7 white balls. Three balls are drawn randomly from the bag (without replacement). What is the probability that two of the balls drawn are white and one is red? Here, we are not arranging balls just choosing 3 balls. How this question is different from the ball question coz there are 2 types of bulbs defective and non-defective similar to two types of balls, red and white? Thank you in advance for your help

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


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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