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Every trading day, the price of CF Corp stock either goes up
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Updated on: 02 Feb 2015, 04:43
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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) 1/16 (B) 1/8 (C) 5/32 (D) 9/32 (E) 3/8
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Originally posted by anaik100 on 27 Jul 2010, 08:41.
Last edited by Bunuel on 02 Feb 2015, 04:43, edited 2 times in total.
Edited the question and added the OA




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Re: Every trading day, the price of CF Corp stock either goes up
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27 Jul 2010, 09:11
anaik100 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) 1/16 (B) 1/8 (C) 5/32 (D) 9/32 (E) 3/8 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: Every trading day, the price of CF Corp stock either goes up
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02 Feb 2015, 04:16
Hmmm, I have a very basic problem with this question...
Why is it that if we want the stock to go up exactly 3 points we need it to go up for 4 out of 5 days and not 3? Does it start by zero, so first day is zero, second 1, third 2 and fourth day is 3?
Would it be possible to also solve it using probability?
Thank you.



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Re: Every trading day, the price of CF Corp stock either goes up
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02 Feb 2015, 04:48
pacifist85 wrote: Hmmm, I have a very basic problem with this question...
Why is it that if we want the stock to go up exactly 3 points we need it to go up for 4 out of 5 days and not 3? Does it start by zero, so first day is zero, second 1, third 2 and fourth day is 3?
Would it be possible to also solve it using probability?
Thank you. It does not matter what is the initial price. We know that every trading day, the price, whatever it is, either goes up by $1 or goes down by $1 and the question asks about the probability that the price is up by exactly $3 from its initial price at the end of 5 trading days.
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Every trading day, the price of CF Corp stock either goes up
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Updated on: 19 Feb 2016, 13:25
Hi pacifist85, The specific details in this question factor in a great deal in how you must go about solving it. We're told that the price of the stock either increases by $1 OR decreases by $1 each day for 5 days. So EVERY DAY the price WILL change. To have a net increase of $3, 3 days of increases will NOT be enough. Here's why: 5 days total 3 days of increases = +$3 BUT the remaining 2 days will then be decreases... 2 days of decreases = $2 Net effect: +$3  $2 = +$1 This is NOT what we're looking for, so 3 days of increases would NOT be enough. With 4 days of increases though... 4 days of increase = +$4 1 day of decrease = $1 Net effect: +$4  $1 = +$3 And then there 5 days of increases, which = +$5 So, to get a net increase of $3, we need 4 days of increases (out of the 5 total days). GMAT assassins aren't born, they're made, Rich
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Re: Every trading day, the price of CF Corp stock either goes up
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19 Feb 2016, 10:45
anaik100 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) 1/16 (B) 1/8 (C) 5/32 (D) 9/32 (E) 3/8 any brute force method here plzzzzzz?



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Re: Every trading day, the price of CF Corp stock either goes up
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19 Feb 2016, 13:25
HI Nez, Since each day involves either an increase or a decrease, there are 2^5 = 32 possible arrangements. Bruteforcing EVERY one of those 32 options isn't really practical though. If you do a bit of work first to figure out what it would take for a 3 dollar increase to occur (4 increases and 1 decrease), THEN you can brute force the possible options from there. GMAT assassins aren't born, they're made, Rich
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Re: Every trading day, the price of CF Corp stock either goes up
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19 Feb 2018, 10:43
anaik100 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) 1/16 (B) 1/8 (C) 5/32 (D) 9/32 (E) 3/8 In order to have a net change of +$3, we must have 4 “up” days and 1 “down” day. Thus one of the sequences of 4 ups and 1 down is: up  up  up  up  down or UUUUD Let’s determine the probability of this sequence: P(UUUUD) = (1/2)^5 = 1/32 However, since a sequence of the 4 ups and 1 down (UUUUD) can be arranged in 5C4 = 5!/4! = 5 ways, the overall probability is 5 x 1/32 = 5/32. Answer: C
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