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In a bulb factory there are different kinds of bulbs, what is the

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New post 08 Sep 2011, 17:13
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In a bulb factory there are different kinds of bulbs, what is the probability that a bulb chosen randomly is a halogen?

(1) There are three times as many halogens than any other bulb in the factory.
(2) The ratio between the halogen to all the other bulbs is 2 to 7.
[Reveal] Spoiler: OA

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New post 08 Sep 2011, 18:16
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b. probability for halogen bulb = 2/ (7+2) = 2/9. sufficient.

a there can be 2 bulbs such as halogen and mercury based.

H = 3 M = 1 probability = 3/4

there can be 3 bulbs such as Halogen,Mercury and Sulfur based.

H= 6 M = 1 S = 1. probability = 6/ 8.

not sufficient.

Thus B it is.
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New post 10 Sep 2011, 21:34
I am getting D as the answer.

There are three times halogen bulbs than any other bulbs in the factory means

if other bulbs total to a value of X then there are 3X halogen bulbs

total number of bulbs = 4x and probability = 3/4


in the above explanation 6/8 = 3/4

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Re: In a bulb factory there are different kinds of bulbs, what is the [#permalink]

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New post 28 Sep 2011, 02:43
manishgeorge wrote:
I am getting D as the answer.

There are three times halogen bulbs than any other bulbs in the factory means

if other bulbs total to a value of X then there are 3X halogen bulbs

total number of bulbs = 4x and probability = 3/4


in the above explanation 6/8 = 3/4

but we don't know how many different types of bulbs are there.

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Re: In a bulb factory there are different kinds of bulbs, what is the [#permalink]

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New post 01 Oct 2011, 01:40
mohan514 wrote:
10. In a bulb factory there are different kinds of bulbs, what is the
probability that a bulb chosen randomly is a halogen?
(1) There are three times as many halogens than any other bulb in the
factory.
(2) The ratio between the halogen to all the other bulbs is 2 to 7.


We want HB/TB = Probability of selecting a halogen bulb
where HB = Halogen Bulb
and TB = Total Bulbs

(1) HB = 3 x any other bulb,
which bulb?
No sufficient

(2) HB/TB -HB = 2/7
TB/HB - 1 = 7/2 => TB/HB = 9/2 => HB/TB = 2/9
Sufficient

Hence (B)
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Re: In a bulb factory there are different kinds of bulbs, what is the [#permalink]

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New post 05 Nov 2017, 07:19
This approach used for proving S1 insufficient is correct, but I guess incorrect numbers are used in the highlighted part. This would be the right approach

S1: There are three times as many halogens than any other bulb in the factory.

    Case 1: Assume there can be 2 type of bulbs - halogen and mercury based. Let H = 3 and M = 1. probability = 3/4

    Case 2: Assume there can be 3 type of bulbs - Halogen, Mercury, and Sulfur based. Let H= 3, M = 1, and S = 1. probability = 3/ 5.

Two cases with two different answer. Hence S1 is insufficient.

amit2k9 wrote:
b. probability for halogen bulb = 2/ (7+2) = 2/9. sufficient.

a there can be 2 bulbs such as halogen and mercury based.

H = 3 M = 1 probability = 3/4

there can be 3 bulbs such as Halogen,Mercury and Sulfur based.

H= 6 M = 1 S = 1. probability = 6/ 8.

not sufficient.

Thus B it is.

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Re: In a bulb factory there are different kinds of bulbs, what is the   [#permalink] 05 Nov 2017, 07:19
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