probability Q.

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godemol: Junior | Next Rank: 30 Posts; Posts: 18; Joined: Sat Jun 20, 2009 6:06 pm; Thanked: 1 times; GMAT Score:720

by godemol » Tue Aug 11, 2009 5:35 pm

A box contains 10 light bulbs, fewer than half of which are defective. Two bulbs are to be drawn simultaneously from the box. If n of the bulbs in box are defective, what is the value of n?

(1) The probability that the two bulbs to be drawn will be defective is 1/15.
(2) The probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15.

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Source: Beat The GMAT — Data Sufficiency | Tue Aug 11, 2009 5:35 pm

prindaroy: Master | Next Rank: 500 Posts; Posts: 113; Joined: Thu Jul 16, 2009 11:23 am; Thanked: 15 times; GMAT Score:730

by prindaroy » Tue Aug 11, 2009 9:19 pm

are you sure it's simultaneously?

hmm...I would go with A, because equation 1 is solvable when you set up;

(x/10)(x/10) = 1/15, but you get x as sqrt(6.66667) and i'm pretty sure you can't pick part of a bulb

statement 2 or equation 2;

you end up with two really weird numbers since it's

(x/10)(10-x/10) = 7/15

so i would say statement 2 is not possible at all

It's either A or E

but i'm inclined to go with A

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real2008: Legendary Member; Posts: 527; Joined: Thu May 01, 2008 12:06 am; Thanked: 7 times

by real2008 » Wed Aug 12, 2009 1:11 am

IT is D

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tohellandback: Legendary Member; Posts: 752; Joined: Sun May 17, 2009 11:04 pm; Location: Tokyo; Thanked: 81 times; GMAT Score:680

by tohellandback » Wed Aug 12, 2009 1:26 am

IMO D:

1)the number of defective bulbs can be 2,3,4

total number of ways to select 2 bulbs=10C2=45
2C2=1
3C2=3
4C2=6
only 3/45=1/15
so 3 bulbs are efective.
SUFF

2)
again 10C2=45
defective bulbs can be 1,2,3,4
if 1,
prob= 1*9/45=1/5
if 2,
2C1* 8C1/45= 16/45

if 3,
3C1*7C1/45=7/15
if4,

4C1*6C1/45=8/15

SUFF

The powers of two are bloody impolite!!

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pralhadesh: Newbie | Next Rank: 10 Posts; Posts: 5; Joined: Tue Aug 11, 2009 3:35 am

by pralhadesh » Wed Aug 12, 2009 1:31 am

HI,

It is given that fewer than half the bulbls are defective.

from (1)

Number of ways in which 10 bulbs can be drawn is 10C2 = 45

The probaility of drawing tow bulbs is therefore = 3

and 3 = number oy ways in which three bulbs can be drawn at random or 3C2, hence sufficient

from (2), one bulb is defetive and one is not

the combinations available are

Defecetive Non defencetive ways

4 6 24
3 7 21
2 8 16
1 9 9

from the above acombinations and data in statement 2,

the chances that one bul is defective and one is not is 7 / 15 or 21 / 45

hence suffficient

Therfore the answer is D

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herjari: Newbie | Next Rank: 10 Posts; Posts: 5; Joined: Sat Feb 23, 2008 1:36 pm; GMAT Score:650

by herjari » Wed Aug 12, 2009 6:57 am

prindaroy wrote:are you sure it's simultaneously?

(x/10)(x/10) = 1/15, but you get x as sqrt(6.66667) and i'm pretty sure you can't pick part of a bulb

You are making a mistake. The equation has to be:

(x/10)[(x-1)/9] = 1/15

Then, just do your algebraic magic and X = 3 deffective bulbs.

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vikram_k51: Master | Next Rank: 500 Posts; Posts: 208; Joined: Sat Jan 31, 2009 11:32 am; Location: Mumbai; Thanked: 2 times

by vikram_k51 » Wed Aug 12, 2009 12:21 pm

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lav: Master | Next Rank: 500 Posts; Posts: 116; Joined: Wed Feb 18, 2009 4:06 am; Thanked: 6 times

by lav » Thu Aug 13, 2009 2:44 am

whats the OA ?

Kid in Verbal

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ket: Senior | Next Rank: 100 Posts; Posts: 93; Joined: Sun Mar 15, 2009 9:07 am; Thanked: 7 times

by ket » Thu Aug 13, 2009 4:46 am

I would go like prindaroy - with slight difference

1. probability of simultaneously drawing 2 defective would be

n/10 x (n-1)/9 = 1/15

After one solves quadratic equation the only positive result for n is 3 - therefore SUFFICIENT
2. probability of drawing one defective and anothe non-defective

n/10 X (10-n)/9 = 7/15 the resulting quadratic equation can not be solved .. therefore INSUFFICIENT

So answer is IMO A

What I don't get why ppl chose to use Combination Formulas... Isn't it more long way to solve? What's the OA?

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tohellandback: Legendary Member; Posts: 752; Joined: Sun May 17, 2009 11:04 pm; Location: Tokyo; Thanked: 81 times; GMAT Score:680

by tohellandback » Thu Aug 13, 2009 6:55 am

ket wrote:I would go like prindaroy - with slight difference

1. probability of simultaneously drawing 2 defective would be

n/10 x (n-1)/9 = 1/15

After one solves quadratic equation the only positive result for n is 3 - therefore SUFFICIENT
2. probability of drawing one defective and anothe non-defective

n/10 X (10-n)/9 = 7/15 the resulting quadratic equation can not be solved .. therefore INSUFFICIENT

So answer is IMO A

What I don't get why ppl chose to use Combination Formulas... Isn't it more long way to solve? What's the OA?

n/10 X (10-n)/9 = 7/15
this is wrong.

It should be
n/10 * (10-n)/9 + (10-n)/10 * n/9= 7/15

now solve and get n

The powers of two are bloody impolite!!

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maihuna: Legendary Member; Posts: 1578; Joined: Sun Dec 28, 2008 1:49 amFollowed by: 9 members; Thanked: 82 times; GMAT Score:720

Re: probability Q.

by maihuna » Thu Aug 13, 2009 7:28 am

godemol wrote:A box contains 10 light bulbs, fewer than half of which are defective. Two bulbs are to be drawn simultaneously from the box. If n of the bulbs in box are defective, what is the value of n?

(1) The probability that the two bulbs to be drawn will be defective is 1/15.
(2) The probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15.

1. Let n = no of defective bulbs
so nC2/10C2 = 1/15 or n(n-1)*!2*!8/!10*2 = 1/15
or n(n-1)*15 = 10*9
or n(n-1) = 2*3 => n =3

2. Again let n are defective so 10-n are non defective.

so nC1*(10-n)C1/10C2 = 7/15
or n*(10-n)*!2/!10*!8 = 7/15
or n(10-n)*2/10*9 = 7/15
or n(10-n)*15 = 7*9*5
or n(10-n) = 7*3

or n = 3

so D.

Charged up again to beat the beast

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