simple but tricky

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tishan: Junior | Next Rank: 30 Posts; Posts: 14; Joined: Tue Aug 26, 2008 9:15 pm; Location: INDIA

by tishan » Mon Oct 27, 2008 8:13 am

A certain portfolio consisted of 5 stocks, priced at $20, $35, $40, $45 and $70, respectively. On a given day, the price of one stock increased by 15%, while the price of another decreased by 35% and the prices of the remaining three remained constant. If the average price of a stock in the portfolio rose by approximately 2%, which of the following could be the prices of the shares that remained constant?

A) 20, 35, 70

B) 20, 45, 70

C) 20, 35, 40

D) 35, 40, 70

E) 35, 40, 45

please help.....unable to solve
2)If the probability of rain on any given day in city x is 50% what is the probability it with rain on exactly 3 days in a five day period?

3)A set of data consists of the following 5 numbers: 0,2,4,6, and 8. Which two numbers, if added to create a set of 7 numbers, will result in a new standard deviation that is close to the standard deviation for the original 5 numbers?

A). -1 and 9
B). 4 and 4
C). 3 and 5
D). 2 and 6
E). 0 and 8
ans D

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Source: Beat The GMAT — Problem Solving | Mon Oct 27, 2008 8:13 am

smohsin: Senior | Next Rank: 100 Posts; Posts: 30; Joined: Wed Jun 11, 2008 7:07 pm; Location: TO

by smohsin » Thu Nov 13, 2008 3:42 pm

Please post the answers....OA?

going big!

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Ossa: GMAT Instructor; Posts: 30; Joined: Sun Oct 26, 2008 5:42 pm; Thanked: 5 times

Re: simple but tricky

by Ossa » Fri Nov 14, 2008 7:49 pm

Hi,

Since one stock rose by 15% and another decreased by 35% and the average of the stocks rose, this means that the difference between the two stocks must be big and that is why a small increase in one stock mitigated the 35% decrease in the other. Without calculating, the two most likely stocks then are 70, and 20. If you chose the last option and caclulate the increase it would be 1.7%, so this is the right answer.

Good luck,

tishan wrote:A certain portfolio consisted of 5 stocks, priced at $20, $35, $40, $45 and $70, respectively. On a given day, the price of one stock increased by 15%, while the price of another decreased by 35% and the prices of the remaining three remained constant. If the average price of a stock in the portfolio rose by approximately 2%, which of the following could be the prices of the shares that remained constant?

A) 20, 35, 70

B) 20, 45, 70

C) 20, 35, 40

D) 35, 40, 70

E) 35, 40, 45

please help.....unable to solve
2)If the probability of rain on any given day in city x is 50% what is the probability it with rain on exactly 3 days in a five day period?

3)A set of data consists of the following 5 numbers: 0,2,4,6, and 8. Which two numbers, if added to create a set of 7 numbers, will result in a new standard deviation that is close to the standard deviation for the original 5 numbers?

A). -1 and 9
B). 4 and 4
C). 3 and 5
D). 2 and 6
E). 0 and 8
ans D

Ossa Elhadary, PhD, CISA, PMP
Math Specialist
Test Prep New York
maximize your score, minimize your stress

www.testprepny.com
[email protected]

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EricLien9122: Master | Next Rank: 500 Posts; Posts: 106; Joined: Thu May 22, 2008 10:50 am; Thanked: 3 times

by EricLien9122 » Mon Nov 17, 2008 9:12 pm

2. Rain,Rain,Rain, N.Rain,N.Rain

probability of exactly 3 days of rain= (# of ways with 3 days of rain)/(total number of possibility of a raining day in 5 days)

(5!)/(2!3!)/(5!)=1/(2*2*3)=1/12

I think this is the right answer, please correct me if i made a mistake.

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Rashmi1804: Master | Next Rank: 500 Posts; Posts: 139; Joined: Sun Nov 09, 2008 12:04 am; Thanked: 3 times; GMAT Score:620

by Rashmi1804 » Wed Mar 04, 2009 6:49 am

there is also probability associated with each day of rain.

so the formula is ncr.(p)^r.(1-p)^(n-r)

so it should be [5!/2!3! (1/2)^3(1/2)^2] = 5/16

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keeyu2: Junior | Next Rank: 30 Posts; Posts: 16; Joined: Mon Mar 09, 2009 10:56 am

by keeyu2 » Tue Mar 10, 2009 6:56 am

Rashmi1804 wrote:there is also probability associated with each day of rain.

so the formula is ncr.(p)^r.(1-p)^(n-r)

so it should be [5!/2!3! (1/2)^3(1/2)^2] = 5/16

5C3 is for the number of days it will rain = 10
The total number of outcome for a binomial is 2^5 =32

10/32 = 5/16

Try this problem that it will rain 3 days in a row giving the same information as your question.

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Reader: Junior | Next Rank: 30 Posts; Posts: 20; Joined: Sat Mar 07, 2009 4:15 pm; Thanked: 2 times

by Reader » Thu Mar 12, 2009 12:52 pm

keeyu2 wrote:
Rashmi1804 wrote:there is also probability associated with each day of rain.

so the formula is ncr.(p)^r.(1-p)^(n-r)

so it should be [5!/2!3! (1/2)^3(1/2)^2] = 5/16
5C3 is for the number of days it will rain = 10
The total number of outcome for a binomial is 2^5 =32

10/32 = 5/16

Try this problem that it will rain 3 days in a row giving the same information as your question.

I still don't get it, could someone explan more in details? Or point me to a link?

Thanks a lot.

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keeyu2: Junior | Next Rank: 30 Posts; Posts: 16; Joined: Mon Mar 09, 2009 10:56 am

by keeyu2 » Tue Mar 17, 2009 10:55 am

Reader wrote:
keeyu2 wrote:
Rashmi1804 wrote:there is also probability associated with each day of rain.

so the formula is ncr.(p)^r.(1-p)^(n-r)

so it should be [5!/2!3! (1/2)^3(1/2)^2] = 5/16
5C3 is for the number of days it will rain = 10
The total number of outcome for a binomial is 2^5 =32

10/32 = 5/16

Try this problem that it will rain 3 days in a row giving the same information as your question.
I still don't get it, could someone explan more in details? Or point me to a link?

Thanks a lot.

5 Days you can have 2 outcomes per day. Rain (1/2 is the proability) or No Rain (1/2 is the proability). So the total outcome is 1/2^5 or 1/2*1/2*1/2*1/2*1/2 = 1/32 = it will rain or not rain for one of the five days.

You can have 10 combinations of 3 days rain and 2 days no rain for the five days. ( 5C3 ). = 10

RRRNN, NNRRR, NRRRN, NRNRR, NRRNR, RNNRR, RNRRN, RRNNR, RNRNR,
RRNRN. = 10 events

Proability is
The Number Of Ways Event Can Occur (10) / The Total Number Of Possible Outcomes(32)

-------------------------------------------------------------------------------------
For it to rain 3 days in a row.

Probability You can have 5 raining days in a row which is 1/32 (RRRRR)
Probability You can have 4 raining days in a row which is 4/32 (RRRRN, NRRRR,RNRRR,RRRNR)
Probability You can have 3 raining days in a row which is 3/32 (NRRRN,RRRNN,NNRRR)
1+4+3 = 8 Ways it can occur / 32 total outcomes.

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Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

by Stuart@KaplanGMAT » Tue Mar 17, 2009 4:14 pm

Reader wrote:I still don't get it, could someone explan more in details? Or point me to a link?

Thanks a lot.

You can try this thread: https://www.beatthegmat.com/coin-flip-qu ... 17911.html. The thread discusses both coin flip and pseudo-coin flip questions. The question in this thread falls into the latter category.

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course

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gmat_2010: Senior | Next Rank: 100 Posts; Posts: 50; Joined: Thu Apr 09, 2009 10:33 am

by gmat_2010 » Sun Aug 02, 2009 2:17 pm

This is how this Bernoulli's Trial came:

You need exactly k nos. of success in n trials.

Hence, you need k success nos. and n - k nos. of failures, all should occur together.

Probability of success in a single trial is p. Hence probability of failure is 1-p

Hence, probability is

(p^k)*((1-p)^(n-k))

But, we have to select when these k successes occur in the sequence of trials. The no. of ways to do so is nCk = (n!/k!(n-k)!)

Hence, final answer:

(n!/(k!(n-k)!))*(p^k)*((1-p)^(n-k))

p = 1/2 here, n = 5, k = 3

Key in the values:

(5!/3!2!)*((1/2)^5)
=10*(1/32)
=5/16

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swathi8388: Junior | Next Rank: 30 Posts; Posts: 18; Joined: Tue Nov 29, 2011 1:17 am; Thanked: 1 times

by swathi8388 » Tue Jul 02, 2013 3:33 am

Sadly i never really understood the concpet behind binomial distribution ,just memorized the formula for questions which have 1/2 probability of either outcomes's occurance.

so here is how i solve these gmat questions :

ways to select 3 options out of 5 = 5C3
total probability of either occuring = 1/2 * 1/2 * 1/2 * 1/2 * 1/2 = (1/2)^5

so the probability of 3 days of rain in 5 = 5C3 * (1/2)^5 = 5/16

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