Light Bulb

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sandipgumtya: Master | Next Rank: 500 Posts; Posts: 126; Joined: Sat Jun 07, 2014 5:26 am; Thanked: 3 times

Light Bulb

by sandipgumtya » Mon Sep 07, 2015 10:41 pm

For each 6-month period during a light bulb's life span, the odds of it not burning out from over-use are half what they were in the previous 6-month period. If the odds of a light bulb burning out during the first 6-month period following its purchase are , what are the odds of it burning out during the period from 6 months to 1 year following its purchase?

A-5/27
B-2/9
C-1/3
D-4/9
E-2/3

Pl help how to solve this.

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Source: Beat The GMAT — Problem Solving | Mon Sep 07, 2015 10:41 pm

MartyMurray: Legendary Member; Posts: 2135; Joined: Mon Feb 03, 2014 9:26 am; Location: https://martymurraycoaching.com/; Thanked: 955 times; Followed by:140 members; GMAT Score:800

by MartyMurray » Tue Sep 08, 2015 12:09 am

sandipgumtya wrote:For each 6-month period during a light bulb's life span, the odds of it not burning out from over-use are half what they were in the previous 6-month period. If the odds of a light bulb burning out during the first 6-month period following its purchase are 1/3, what are the odds of it burning out during the period from 6 months to 1 year following its purchase?

A-5/27
B-2/9
C-1/3
D-4/9
E-2/3

The wording of this one is a little funny. I believe what the question is asking is really the following.

What is the probability that a particular bulb will burn out sometime during the second six months of use?

So we have the probability of it burning out in the first six months, 1/3.

To get through the first six months it has to not burn out, and so we need the probability of it doing that, which is 2/3.

The probability of it's not burning out in the second six months is half of what it was in the first six months. 1/2 * 2/3 = 1/3

So the probability of it's burning out in the second six months is 2/3.

So we are working with 2/3 probability that it will stick around for the first six months and then 2/3 probability that it will then burn out during the second six months.

2/3 * 2/3 = 4/9

So the correct answer is D.

Last edited by MartyMurray on Tue Sep 08, 2015 12:37 am, edited 1 time in total.

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Matt@VeritasPrep: GMAT Instructor; Posts: 2630; Joined: Wed Sep 12, 2012 3:32 pm; Location: East Bay all the way; Thanked: 625 times; Followed by:119 members; GMAT Score:780

by Matt@VeritasPrep » Tue Sep 08, 2015 12:36 am

Let's pretend that we turn on the lightbulb on January 1st.

We know that the probability that the bulb doesn't burn out before July 1st = 2/3. We'll call this Probability A.

We know that the probability that the bulb doesn't burn out between July 1st and December 31st (inclusive) = (1/2)*(2/3), or (1/3). We'll call this Probability B.

So the probability that the bulb burns out between July 1st and December 31st requires TWO things both to be true:

1:: No burnout before July 1st
2:: Burnout between July 1st and December 31st

In other words, Probability A * (1 - Probability B), or 2/3 * 2/3.

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Matt@VeritasPrep: GMAT Instructor; Posts: 2630; Joined: Wed Sep 12, 2012 3:32 pm; Location: East Bay all the way; Thanked: 625 times; Followed by:119 members; GMAT Score:780

by Matt@VeritasPrep » Tue Sep 08, 2015 1:49 am

Have to also say that I was impressed that this was a well-formulated problem. I thought it must have an error, since the probability of burnout in the second 6 was greater than the probability of burnout in the first 6. But it actually works! Here's a neat way of verifying that - but if you aren't a GMAT instructor or a math buff, you can skip the following.

The probability that the bulb eventually burns out must equal 1, so we should have the following equation:

P(Burnout first 6) + P(Burnout second 6) + P(Burnout third 6) + ... = 1

The probability of not burning out during any given six months, without factoring in the probability that the bulb has survived up to that point, is easy to find: 2/3 for the first six months, 1/3 for the second six, 1/6 for the third six, 1/12 for the fourth six, etc. Similarly, the probabilities of burning out, again not factoring in the dependencies, are 1/3, 2/3, 5/6, 11/12, etc.

Now we bring in the history of the bulb. In order for it to burn out in the second six months, it must have survived the first six. So we have (Survive 1st)*(Not Survive 2nd), or the first number on the red list times the second number on the blue list. Likewise, for it not to burn out in the third six months, we need (Survive 1st) * (Survive 2nd) * (Not Survive 3rd), which is the product of the first TWO numbers on the red list and the third number on the green list.

This means our equation is

(Not Survive 1st) + (Survive 1st)(Not Survive 2nd) + (Survive 1st)(Survive 2nd)(Not Survive 3rd) + (Survive 1st)(Survive 2nd)(Survive 3rd)(Not Survive 4th) + ..., or

(1/3) + (2/3)(2/3) + (2/3)(1/3)(5/6) + (2/3)(1/3)(1/6)(11/12) + ..., or

(1/3) + (2/3)Â¹(1/2)â�°*(1 - (1/3)(1/2)â�°) + (2/3)Â²(1/2)Â¹(1 - (1/3)(1/2)Â¹) + (2/3)Â³(1/2)Â³(1 - (1/3)(1/2)Â²) + ..., or

Î£ from k=0 to k=âˆž of (2/3)áµ� * (1/2)áµ�â�½áµ�â�»Â¹â�¾â�½Â½â�¾ * (1 - (1/3)(1/2)áµ�â�»Â¹)

To my absolute astonishment, this sum converges to 1.

I've been teaching the GMAT for almost six years and answering student questions on random problems for nearly that entire time, and I don't think I've ever been this surprised. This problem looked so hinky that I would've bet the farm that it didn't work, but it does. (Good news for my farm!)

Whoever wrote this question, I owe you an apology and a cheap but tasty craft beer of your choice.

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MartyMurray: Legendary Member; Posts: 2135; Joined: Mon Feb 03, 2014 9:26 am; Location: https://martymurraycoaching.com/; Thanked: 955 times; Followed by:140 members; GMAT Score:800

by MartyMurray » Tue Sep 08, 2015 5:22 am

Matt@VeritasPrep wrote:I've been teaching the GMAT for almost six years and answering student questions on random problems for nearly that entire time, and I don't think I've ever been this surprised. This problem looked so hinky that I would've bet the farm that it didn't work, but it does. (Good news for my farm!)

That is some way cool analysis Matt. A similar question about the probabilities crossed my mind so fast that I barely even noticed it, and I am glad to see the answer.

On the other hand, I believe that given the direction of your bet, the farm would be gone, which would not be good news for your farm, unless what you are saying is that it would be better off without you.

Marty Murray
Perfect Scoring Tutor With Over a Decade of Experience
MartyMurrayCoaching.com
Contact me at [email protected] for a free consultation.

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pm; Location: Vancouver, BC; Thanked: 5254 times; Followed by:1268 members; GMAT Score:770

by Brent@GMATPrepNow » Tue Sep 08, 2015 5:27 am

For each 6-month period during a light bulb's life span, the odds of it not burning out from over-use are half what they were in the previous 6-month period. If the odds of a light bulb burning out during the first 6-month period following its purchase are 1/3, what are the odds of it burning out during the period from 6 months to 1 year following its purchase?

5/27
2/9
1/3
4/9
2/3

I'm not too crazy about the wording of this question, since there's a big difference between ODDS and PROBABILITY

Probability = (favorable outcomes)/ (total outcomes)
Odds = the ratio of favorable outcomes to unfavorable outcomes

So, for example, if we randomly select a number from {3, 5, 7, 9, 11}, then . . .
- The PROBABILITY of selecting a prime number = 4/5
- The ODDS in favor of selecting a prime number = 4:1

I don't think I've ever seen the GMAT use the term "odds" (other than in the context of even and odd integers).

Having said all of that, if we take the question and replace "odds" with "probability" then the above solutions are great.

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com