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

IT is D

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

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

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.

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.

Will go with D.

whats the OA ?

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?

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

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.