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

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Senior PS Moderator
Joined: 26 Feb 2016
Posts: 3253
Location: India
GPA: 3.12
A shipment of 500 bulbs contains 10% defective bulbs. If three  [#permalink]

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16 Mar 2018, 08:51
1
6
00:00

Difficulty:

85% (hard)

Question Stats:

50% (02:42) correct 50% (02:58) wrong based on 70 sessions

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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. $$\frac{27}{1000}$$

Source: Experts Global

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Manager
Joined: 28 Jan 2018
Posts: 50
Location: Netherlands
Concentration: Finance
GMAT 1: 710 Q50 V36
GPA: 3
A shipment of 500 bulbs contains 10% defective bulbs. If three  [#permalink]

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16 Mar 2018, 11:15
1
1
Since the number of bulb is too big, i have to use approximation to arrive to the conclusion

Probability that exactly 2 bulb will be broken: 50/500 * 49/499 * 448/498 = 0.1 * 0.1 * 0.9 = 0.009

Since 2 broken light bulb can be arrange in 3!/2! way, the final answer will be: 0.09 * 3 = 0.027 => Answer E
Senior Manager
Joined: 31 Jul 2017
Posts: 499
Location: Malaysia
GMAT 1: 700 Q50 V33
GPA: 3.95
WE: Consulting (Energy and Utilities)
Re: A shipment of 500 bulbs contains 10% defective bulbs. If three  [#permalink]

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18 Mar 2018, 06:45
1
pushpitkc wrote:
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. $$\frac{27}{1000}$$

Source: Experts Global

Number of Defective Bulbs = 50
Number of Non-Defective Bulbs = 450
Ways of Selecting Exactly 2 Ddefective Bulbs & 1 non-Defective Bulb
= $$50C2 * 450C1$$
Total Number of Ways = $$500C3.$$

Probability = $$50C2 * 450C1 / 500C3.$$ = 0.027
Intern
Joined: 27 May 2015
Posts: 6
Schools: ISB '18
Re: A shipment of 500 bulbs contains 10% defective bulbs. If three  [#permalink]

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29 Mar 2018, 06:25
1
Probability of bulb being defective: 1/10
Probability of bulb being non- defective: 1-1/10=9/10
No. of ways of choosing 2 Defective and one Non Defective Bulb: 3!/2!=3
Probability required: 3*(1/10)*(1/10)*(9/10)=27/1000
Senior Manager
Joined: 15 Feb 2018
Posts: 454
A shipment of 500 bulbs contains 10% defective bulbs. If three  [#permalink]

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04 Oct 2019, 00:49
There are 50 defective bulbs. There are 450 non-defective bulbs. We want to choose 2 defective bulbs and 1 non-defective bulb. We use combinatorics

nCr=n!/(n-r)!r!

$$\frac{(50C2·450C1)}{500C3}$$

$$=\frac{\frac{50!}{48!·2!}+\frac{450!}{449!·1!}}{\frac{500!}{497!·3!}}$$

$$=\frac{\frac{50·49·48!}{48!·2!}+\frac{450·449!}{449!·1!}}{\frac{500·499·498·497!}{497!·3!}}$$

The numerators and denominators cancel out

$$=\frac{\frac{50·49}{2!}+\frac{450}{1!}}{\frac{500·499·498}{3!}}$$

$$=\frac{\frac{50·49}{2·1}+\frac{450}{1}}{\frac{500·499·498}{3·2·1}}$$

$$=\frac{\frac{50·49·450}{2}}{\frac{500·499·498}{3·2}}$$

$$=\frac{50·49·450·3·2}{500·499·498·2}$$

$$=\frac{50}{500} · \frac{49}{499} · \frac{450}{498} · \frac{3}{1}$$

The first two fractions are approximately 1/10 and the third fraction is approximately 9/10

$$=\frac{1}{10} · \frac{1}{10} · \frac{9}{10} · \frac{3}{1}$$

$$= \frac{1·1·9·3}{10^3}$$

$$=\frac{27}{1000}$$

hoang
It should be 450/498 as none of the non-defective bulbs have been chosen yet
A shipment of 500 bulbs contains 10% defective bulbs. If three   [#permalink] 04 Oct 2019, 00:49