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fewer than half are defective

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fewer than half are defective

by sanju09 » Sat Feb 21, 2009 4:42 am
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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by sureshbala » Sat Feb 21, 2009 6:22 am
Given that n<5.

Statement I: It is given that nC2/10C2 = 1/15
i.e nC2 = 3. Hence n = 3

Statement II: It is given that n(10-n)/10C2 = 7/15
i.e n(10-n) = 21. Hence n =3.

So D must be the answer
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by rahul.s » Mon Jan 11, 2010 9:23 am
sureshbala wrote:Given that n<5.

Statement I: It is given that nC2/10C2 = 1/15
i.e nC2 = 3. Hence n = 3

Statement II: It is given that n(10-n)/10C2 = 7/15
i.e n(10-n) = 21. Hence n =3.

So D must be the answer
Hi Suresh,

Could you please explain the second statement a little more in detail? Why is it n(10-n)/10C2? I didn't quite understand.
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by sanju09 » Tue Jan 12, 2010 12:12 am
rahul.s wrote:
sureshbala wrote:Given that n<5.

Statement I: It is given that nC2/10C2 = 1/15
i.e nC2 = 3. Hence n = 3

Statement II: It is given that n(10-n)/10C2 = 7/15
i.e n(10-n) = 21. Hence n =3.

So D must be the answer
Hi Suresh,

Could you please explain the second statement a little more in detail? Why is it n(10-n)/10C2? I didn't quite understand.



If n bulbs are defective then (10 - n) bulbs are fit; and if two bulbs were drawn, total ways are 10C2. For the required probability as in st (2), there are n ways of drawing a defective bulb and (10 - n) ways of drawing a fit bulb, and so the probability = n (10 - n)/10C2. It's alright, rahul!!
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by rahul.s » Tue Jan 12, 2010 2:28 am
I got it. Thank you Sanju :)
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by Stuart@KaplanGMAT » Tue Jan 12, 2010 1:15 pm
sanju09 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.
We can also use the probability formula to solve.

Probability = # of desired outcomes / total # of possibilities

For multiple events, we MULTIPLY the individual probabilities.

(1) We start with n defectives and 10 total bulbs, so the probability that the first is defective is n/10.

We now are left with (n-1) defective bulbs and 9 total bulbs, so the probability that the second is defective is (n-1)/9.

So:

n/10 * (n-1)/9 = 1/15. Since we know that n is non-negative, we can solve for n: sufficient.

(2) Two cases satisfy what we need: 1 defective, 2 normal; 1 normal, 2 defective.

When we have alternative probabilities, we ADD them together.

So:

n/10 * (10-n)/9 + (10-n)/10 * n/9 = 7/15. Since we know that n is non-negative, we can solve for n: sufficient.

Now, if we're truly data sufficiency Grandmasters, we don't need to set up these equations; we recognize that we can set up solvable equations for both statements and therefore each is sufficient alone.
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by sanju09 » Wed Jan 13, 2010 2:40 am
Sure Stuart! Thanks
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by GMATMadeEasy » Mon Jan 18, 2010 5:34 am
Stuart Kovinsky wrote:
sanju09 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.
We can also use the probability formula to solve.

Probability = # of desired outcomes / total # of possibilities

For multiple events, we MULTIPLY the individual probabilities.

(1) We start with n defectives and 10 total bulbs, so the probability that the first is defective is n/10.

We now are left with (n-1) defective bulbs and 9 total bulbs, so the probability that the second is defective is (n-1)/9.

So:

n/10 * (n-1)/9 = 1/15. Since we know that n is non-negative, we can solve for n: sufficient.

(2)
Two cases satisfy what we need: 1 defective, 2 normal; 1 normal, 2 defective.
When we have alternative probabilities, we ADD them together.

So:

n/10 * (10-n)/9 + (10-n)/10 * n/9 = 7/15. Since we know that n is non-negative, we can solve for n: sufficient.

Now, if we're truly data sufficiency Grandmasters, we don't need to set up these equations; we recognize that we can set up solvable equations for both statements and therefore each is sufficient alone.
Isn't one case is enough i.e. One defective and second non defective.. as order wont matter in this case ?
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by Stuart@KaplanGMAT » Mon Jan 18, 2010 10:31 am
GMATMadeEasy wrote: Isn't one case is enough i.e. One defective and second non defective.. as order wont matter in this case ?
For statement (2), order does matter.

We want exactly 1 to be defective, and there are two ways that can happen:

1) 1st defective, 2nd just fine; and
2) 1st just fine, 2nd defective.

Each of these cases is a separate event and must be factored in to our calculations.
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by GMATMadeEasy » Mon Jan 18, 2010 4:03 pm
Stuart Kovinsky wrote:
GMATMadeEasy wrote: Isn't one case is enough i.e. One defective and second non defective.. as order wont matter in this case ?
For statement (2), order does matter.

We want exactly 1 to be defective, and there are two ways that can happen:
1) 1st defective, 2nd just fine; and
2) 1st just fine, 2nd defective.
Each of these cases is a separate event and must be factored in to our calculations.
Isn't it same as drawing two cards out of which one is ace and another is jack.

4/52* 4/51

Here we dont multiply this by two or order does not matter.
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by faysal204 » Wed Jun 30, 2010 3:29 pm
Stuart Kovinsky wrote:
GMATMadeEasy wrote: Isn't one case is enough i.e. One defective and second non defective.. as order wont matter in this case ?
For statement (2), order does matter.

We want exactly 1 to be defective, and there are two ways that can happen:

1) 1st defective, 2nd just fine; and
2) 1st just fine, 2nd defective.

Each of these cases is a separate event and must be factored in to our calculations.
Here,the question asks that the two bulbs be drawn simultaneously(bulbs are not drawn one after another).So how
do we calculate this type of probability?
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by Stuart@KaplanGMAT » Thu Jul 01, 2010 12:31 am
faysal204 wrote:
Here,the question asks that the two bulbs be drawn simultaneously(bulbs are not drawn one after another).So how
do we calculate this type of probability?
As long as you're making selections without replacement, you can treat simultaneous draws exactly as you'd treat sequential draws.

For example:
There are 5 brown socks and 5 blue socks in a drawer. If Bob randomly draws a sock then randomly draws a second sock without replacing the first, what's the probability that both are brown?
and
There are 5 brown socks and 5 blue socks in a drawer. If Bob reaches into the drawer with both hands and randomly grabs two socks, what's the probability that both are brown?
are the exact same question.
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by Stuart@KaplanGMAT » Thu Jul 01, 2010 12:34 am
GMATMadeEasy wrote:
Stuart Kovinsky wrote:
GMATMadeEasy wrote: Isn't one case is enough i.e. One defective and second non defective.. as order wont matter in this case ?
For statement (2), order does matter.

We want exactly 1 to be defective, and there are two ways that can happen:
1) 1st defective, 2nd just fine; and
2) 1st just fine, 2nd defective.
Each of these cases is a separate event and must be factored in to our calculations.
Isn't it same as drawing two cards out of which one is ace and another is jack.

4/52* 4/51

Here we dont multiply this by two or order does not matter.
Actually, your calculation is wrong. It should be:

8/52 * 4/51,

since on our first pick, we're happy with either an ace or a jack. Once we've made that first selection, there remain 4 desired outcomes for the second pick.

Of course, you'll note that:

8/52 * 4/51 = 4/52 * 4/51 * 2,

so we are in fact multiplying by 2 in your example as well.
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by dream700 » Thu Jul 01, 2010 5:15 am
@ Stuart Kovinsky

The qn does state that Two bulbs are to be drawn simultaneously...

hence, I guess the order their defect will not matter...


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by gmat1011 » Thu Jul 01, 2010 9:34 am
Stuart -

Is there any trick to get some idea as to whether a quadratic equation will get you + numbers or non square root numbers as an answer...

In the bulbs question, if we were to set up the quad equations for 1 and 2 and solve and finally arrive at 3 for each, it took me a good 3.5 minutes!

The thing is sometimes after solving you get two negative numbers or something and you are forced to conclude that the stem is insufficient.... so I am not always 100 pc sure as to whether I can merely rely on the ability to set up quad equations as an indicator for sufficiency

Since you mentioned: data sufficiency Grandmasters don't need to set up equations; just want to know what grandmasters look for to recognize that solvable equations for both statements can be set up... Any practical pointers wrt quad equations vs. ability to recognize sufficiency would be greatly appreciated!!

Many thanks....
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