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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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Question Stats:

70% (01:51) correct 30% (01:52) wrong based on 46 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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New post 16 Sep 2014, 01:51
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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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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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New post 20 Jul 2016, 07:56
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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.
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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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S96-15  [#permalink]

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New post 21 Jun 2019, 06:14
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


Sometimes I am a bit confused in deciding the number of possible outcomes.
May I know in this case why we cannot calculate the outcome from 10!/5!5!, instead we got the figure from 2^5 ?

I thought since there are 5U and 5D, then there are 10 items and we need to find the number of ways to pick 5 items out of those 10 items. Is my way of thinking wrong? Please help!
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Re: S96-15  [#permalink]

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New post 21 Jun 2019, 06:44
aidyn 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


Sometimes I am a bit confused in deciding the number of possible outcomes.
May I know in this case why we cannot calculate the outcome from 10!/5!5!, instead we got the figure from 2^5 ?

I thought since there are 5U and 5D, then there are 10 items and we need to find the number of ways to pick 5 items out of those 10 items. Is my way of thinking wrong? Please help!


I tried to elaborate this here: https://gmatclub.com/forum/s96-184686.html#p1712165 and also gave there links to similar topics.

Hope it helps.
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Re: S96-15   [#permalink] 21 Jun 2019, 06:44
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