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Math Expert V
Joined: 02 Sep 2009
Posts: 58340

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Difficulty:   35% (medium)

Question Stats: 70% (01:51) correct 30% (01:52) wrong based on 46 sessions

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. $$\frac{3}{8}$$ _________________ Math Expert V Joined: 02 Sep 2009 Posts: 58340 Re S96-15 [#permalink] Show Tags 2 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 $$\frac{5}{32}$$.

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Intern  B
Joined: 20 Feb 2016
Posts: 6

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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! Math Expert V Joined: 02 Sep 2009 Posts: 58340 Re: S96-15 [#permalink] Show Tags 1 1 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 $$\frac{5}{32}$$.

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. _________________ Intern  B Joined: 22 Dec 2016 Posts: 20 Re: S96-15 [#permalink] Show Tags 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 Intern  B Joined: 19 Apr 2019 Posts: 35 Location: Indonesia GMAT 1: 690 Q49 V34 GPA: 3.8 WE: Project Management (Commercial Banking) S96-15 [#permalink] Show Tags 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 $$\frac{5}{32}$$.

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!
Math Expert V
Joined: 02 Sep 2009
Posts: 58340

Show Tags

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 $$\frac{5}{32}$$.

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