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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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16 Sep 2014, 01:51
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 5day sequences that give us a $3 increase, then we compare that number to the total number of 5day 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 5day 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: S9615
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20 Jul 2016, 04:20
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 5day sequences that give us a $3 increase, then we compare that number to the total number of 5day 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 5day 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: S9615
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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 5day sequences that give us a $3 increase, then we compare that number to the total number of 5day 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 5day 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: \(P = C^k_n*p^k*(1p)^{nk}\) 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, \(P = C^k_n*p^k*(1p)^{nk}=C^4_5*(\frac{1}{2})^4*(1\frac{1}{2})^{(54)}=C^4_5*(\frac{1}{2})^5\) OR: probability of scenario UUUUD is \((\frac{1}{2})^4*(\frac{1}{2})^4=(\frac{1}{2})^5\), but UUUUD can occur in different ways: UUUUD; UUUDU UUDUU UDUDU DUUUU; 5 ways (# of permutations of five letters UUUUD, 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#p742767viewtopic.php?f=140&t=56812&hilit=+probability+occurring+timesviewtopic.php?f=140&t=88069&hilit=+probability+occurring+timesviewtopic.php?f=140&t=87673&hilit=+probability+occurring+timesAlso you can check Probability chapter of Math Book for more (link in my signature). Hope it helps.
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Re: S9615
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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 pricewise 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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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 5day sequences that give us a $3 increase, then we compare that number to the total number of 5day 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 5day 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: S9615
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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 5day sequences that give us a $3 increase, then we compare that number to the total number of 5day 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 5day 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/s96184686.html#p1712165 and also gave there links to similar topics. Hope it helps.
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