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Re: S96-15 [#permalink]
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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]
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.


Sent from my Redmi Note 5 using GMAT Club Forum mobile app
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Re: S96-15 [#permalink]
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]
Expert Reply
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]
Hi Bunuel,
Thanks for all your help.
What is the significance of the words "equal likelihood". How are these words been used in the official answer? Thanks in advance.
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Re: S96-15 [#permalink]
bedarkaryashas wrote:
Hi Bunuel,
Thanks for all your help.
What is the significance of the words "equal likelihood". How are these words been used in the official answer? Thanks in advance.



Equal likelihood means 50/50 up or down
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Re S96-15 [#permalink]
I think this is a high-quality question and I agree with explanation.
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