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simple but tricky

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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by smohsin » Thu Nov 13, 2008 3:42 pm
Please post the answers....OA?
going big!
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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
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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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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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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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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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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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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.
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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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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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