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S96-15

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S96-15  [#permalink]

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New post 16 Sep 2014, 01:51
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  35% (medium)

Question Stats:

66% (01:33) correct 34% (01:46) wrong based on 50 sessions

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Every trading day, the price of CF Corp stock either goes up by $1 or goes down by $1 with equal likelihood. At the end of 5 trading days, what is the probability that the price of CF Corp stock is up by exactly $3 from its initial price?

A. \(\frac{1}{16}\)
B. \(\frac{1}{8}\)
C. \(\frac{5}{32}\)
D. \(\frac{9}{32}\)
E. \(\frac{3}{8}\)

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Re S96-15  [#permalink]

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New post 16 Sep 2014, 01:51
1
Official Solution:

Every trading day, the price of CF Corp stock either goes up by $1 or goes down by $1 with equal likelihood. At the end of 5 trading days, what is the probability that the price of CF Corp stock is up by exactly $3 from its initial price?

A. \(\frac{1}{16}\)
B. \(\frac{1}{8}\)
C. \(\frac{5}{32}\)
D. \(\frac{9}{32}\)
E. \(\frac{3}{8}\)


The daily change in CF Corp's stock price can be compared to a coin flip. Heads - the price goes up by $1. Tails - the price goes down by $1. Moreover, the coin is "fair": that is, each daily outcome is equally possible (meaning that the chance of heads is 50%, and the chance of tails is 50%). This also means that any particular sequence of flips is equally probable. As a result, our probability calculation is simplified. We just count the 5-day sequences that give us a $3 increase, then we compare that number to the total number of 5-day sequences.

To go up exactly $3, we need exactly 4 "up" days (heads) and 1 "down" day (tails). We don’t care about the order in which these days come. So we need to count the possible arrangements of 4 heads and 1 tails. We can simply list these out:

HHHHT

HHHTH

HHTHH

HTHHH

THHHH

The only question is what day of the week the "down" day falls on, so there are 5 possibilities for a $3 increase. Alternatively, we can use the combinations or anagrams method to calculate \(\frac{5!}{4!} = 5\).

Now, we need to count all the possible 5-day sequences of flips. Since each day can have 2 outcomes (H or T), we have \(2 \times 2 \times 2 \times 2 \times 2 = 32\) total possible outcomes over 5 days.

Finally, to compute the probability of a $3 increase over 5 days, we divide 5 successful outcomes by 32 total possibilities (of equal weight) to get \(\frac{5}{32}\).


Answer: C
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Collection of Questions:
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Re: S96-15  [#permalink]

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New post 20 Jul 2016, 04:20
1
Bunuel wrote:
Official Solution:

Every trading day, the price of CF Corp stock either goes up by $1 or goes down by $1 with equal likelihood. At the end of 5 trading days, what is the probability that the price of CF Corp stock is up by exactly $3 from its initial price?

A. \(\frac{1}{16}\)
B. \(\frac{1}{8}\)
C. \(\frac{5}{32}\)
D. \(\frac{9}{32}\)
E. \(\frac{3}{8}\)


The daily change in CF Corp's stock price can be compared to a coin flip. Heads - the price goes up by $1. Tails - the price goes down by $1. Moreover, the coin is "fair": that is, each daily outcome is equally possible (meaning that the chance of heads is 50%, and the chance of tails is 50%). This also means that any particular sequence of flips is equally probable. As a result, our probability calculation is simplified. We just count the 5-day sequences that give us a $3 increase, then we compare that number to the total number of 5-day sequences.

To go up exactly $3, we need exactly 4 "up" days (heads) and 1 "down" day (tails). We don’t care about the order in which these days come. So we need to count the possible arrangements of 4 heads and 1 tails. We can simply list these out:

HHHHT

HHHTH

HHTHH

HTHHH

THHHH

The only question is what day of the week the "down" day falls on, so there are 5 possibilities for a $3 increase. Alternatively, we can use the combinations or anagrams method to calculate \(\frac{5!}{4!} = 5\).

Now, we need to count all the possible 5-day sequences of flips. Since each day can have 2 outcomes (H or T), we have \(2 \times 2 \times 2 \times 2 \times 2 = 32\) total possible outcomes over 5 days.

Finally, to compute the probability of a $3 increase over 5 days, we divide 5 successful outcomes by 32 total possibilities (of equal weight) to get \(\frac{5}{32}\).


Answer: C



Hi Bunuel,

Can you please tell me where I'm going wrong?
So I started with P(Up)=P(Down)=1/2
Exactly 3 out of 5 days= 5C3 * (1/2) * (1/2) * (1/2) * (1/2) * (1/2) = 5/16

Sorry if it's a silly doubt, just trying to understand. Also, thank you for sharing, helps a lot!
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Re: S96-15  [#permalink]

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New post 20 Jul 2016, 07:56
salonipatil wrote:
Bunuel wrote:
Official Solution:

Every trading day, the price of CF Corp stock either goes up by $1 or goes down by $1 with equal likelihood. At the end of 5 trading days, what is the probability that the price of CF Corp stock is up by exactly $3 from its initial price?

A. \(\frac{1}{16}\)
B. \(\frac{1}{8}\)
C. \(\frac{5}{32}\)
D. \(\frac{9}{32}\)
E. \(\frac{3}{8}\)


The daily change in CF Corp's stock price can be compared to a coin flip. Heads - the price goes up by $1. Tails - the price goes down by $1. Moreover, the coin is "fair": that is, each daily outcome is equally possible (meaning that the chance of heads is 50%, and the chance of tails is 50%). This also means that any particular sequence of flips is equally probable. As a result, our probability calculation is simplified. We just count the 5-day sequences that give us a $3 increase, then we compare that number to the total number of 5-day sequences.

To go up exactly $3, we need exactly 4 "up" days (heads) and 1 "down" day (tails). We don’t care about the order in which these days come. So we need to count the possible arrangements of 4 heads and 1 tails. We can simply list these out:

HHHHT

HHHTH

HHTHH

HTHHH

THHHH

The only question is what day of the week the "down" day falls on, so there are 5 possibilities for a $3 increase. Alternatively, we can use the combinations or anagrams method to calculate \(\frac{5!}{4!} = 5\).

Now, we need to count all the possible 5-day sequences of flips. Since each day can have 2 outcomes (H or T), we have \(2 \times 2 \times 2 \times 2 \times 2 = 32\) total possible outcomes over 5 days.

Finally, to compute the probability of a $3 increase over 5 days, we divide 5 successful outcomes by 32 total possibilities (of equal weight) to get \(\frac{5}{32}\).


Answer: C



Hi Bunuel,

Can you please tell me where I'm going wrong?
So I started with P(Up)=P(Down)=1/2
Exactly 3 out of 5 days= 5C3 * (1/2) * (1/2) * (1/2) * (1/2) * (1/2) = 5/16

Sorry if it's a silly doubt, just trying to understand. Also, thank you for sharing, helps a lot!


The price is up by exactly $3 in 5 days means that 4 days it went up (+4$) and one day it went down (-1$).

UUUUD --> \(P(U=4)=C^4_5*(\frac{1}{2})^4*\frac{1}{2}=\frac{5}{32}\).

Answer: C.

If the probability of a certain event is \(p\), then the probability of it occurring \(k\) times in \(n\)-time sequence is: \(P = C^k_n*p^k*(1-p)^{n-k}\)

In our case probability of having 4 days when price went up (4U) and one day when price went down (D):
\(n=5\) (5 days);
\(k=4\) (4 days when price wen t up);
\(p=\frac{1}{2}\) (probability of price going up is 1/2).

So, \(P = C^k_n*p^k*(1-p)^{n-k}=C^4_5*(\frac{1}{2})^4*(1-\frac{1}{2})^{(5-4)}=C^4_5*(\frac{1}{2})^5\)

OR: probability of scenario U-U-U-U-D is \((\frac{1}{2})^4*(\frac{1}{2})^4=(\frac{1}{2})^5\), but U-U-U-U-D can occur in different ways:

U-U-U-U-D;
U-U-U-D-U
U-U-D-U-U
U-D-U-D-U
D-U-U-U-U;

5 ways (# of permutations of five letters U-U-U-U-D, which is \(\frac{5!}{4!}=5=C^4_5\)).

Hence \(P=\frac{5!}{4!}*(\frac{1}{2})^5\).

Check this links for similar problems:
viewtopic.php?f=140&t=96468&p=742767&hilit=+google#p742767
viewtopic.php?f=140&t=56812&hilit=+probability+occurring+times
viewtopic.php?f=140&t=88069&hilit=+probability+occurring+times
viewtopic.php?f=140&t=87673&hilit=+probability+occurring+times

Also you can check Probability chapter of Math Book for more (link in my signature).

Hope it helps.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: S96-15  [#permalink]

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New post 01 Mar 2019, 00:08
Bunuel wrote:
Every trading day, the price of CF Corp stock either goes up by $1 or goes down by $1 with equal likelihood. At the end of 5 trading days, what is the probability that the price of CF Corp stock is up by exactly $3 from its initial price?

A. \(\frac{1}{16}\)
B. \(\frac{1}{8}\)
C. \(\frac{5}{32}\)
D. \(\frac{9}{32}\)
E. \(\frac{3}{8}\)
Hi Bunuel,
I know I have done something wrong, but still take a look please.
I have counted all the possibilities of stock price & since every outcome has same likelihood, just found the probability from there..
Zero day stock price-wise 5$(assuming)
1st day: 4/6
2nd day: 3/5/7
3rd day: 2/4/6/8
4th day: 1/3/5/7/9
5th day: 0/2/4/6/8/10

We need to find out about probability of 5+3 =8$
So P(8$)= 1/6
Can you please take a look.


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Re: S96-15   [#permalink] 01 Mar 2019, 00:08
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