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

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

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New post 16 Mar 2018, 08:51
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A
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  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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A shipment of 500 bulbs contains 10% defective bulbs. If three  [#permalink]

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New post 16 Mar 2018, 11:15
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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
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Re: A shipment of 500 bulbs contains 10% defective bulbs. If three  [#permalink]

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New post 18 Mar 2018, 06:45
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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
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Re: A shipment of 500 bulbs contains 10% defective bulbs. If three  [#permalink]

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New post 29 Mar 2018, 06:25
1
ANSWER:E
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
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A shipment of 500 bulbs contains 10% defective bulbs. If three  [#permalink]

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New post 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
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A shipment of 500 bulbs contains 10% defective bulbs. If three   [#permalink] 04 Oct 2019, 00:49

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

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