Every trading day, the price of CF Corp stock either goes up

Question

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

(A) 1/16
(B) 1/8
(C) 5/32
(D) 9/32
(E) 3/8

Bunuel · Accepted Answer

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.

ScottTargetTestPrep · Answer

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

pacifist85 · Answer

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.

Bunuel · Answer

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  .

EMPOWERgmatRichC · Answer

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

Ekland · Answer

any brute force method here plzzzzzz?

EMPOWERgmatRichC · Answer

HI Nez,

Since each day involves either an increase or a decrease, there are 2^5 = 32 possible arrangements. Brute-forcing 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

MathRevolution · Answer

Apply Binomial distribution:  ^n{C_r} p^r q ^{n - r}

=> n = 5 and r = 4 (to be up by $3)

=> p =  \frac{1}{2}  (goes up by $1 - equal likelihood means 50%)

=> q =  1 - \frac{1}{2} = \frac{1}{2}  (p + q = 1 - where 'q' is failure of 'p')

=>  ^5{C_4} (\frac{1}{2})^3 (\frac{1}{2}) ^{2}

=>  5 * \frac{1}{8} * \frac{1}{4} = \frac{5}{32}

Answer C

RastogiSarthak99 · Answer

Believe everyone can arrive at 1/32 but to know that it has to be arranged through 5!/4! is the trick

+1 +1 +1 +1 -1 = +3

Four +1s and One -1 is required

5 spots and 4 are replicated

5!/4! ways * Probability of Four +1s and One -1

5/32

C.

bumpbot · Answer

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.

Every trading day, the price of CF Corp stock either goes up

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

Every trading day, the price of CF Corp stock either goes up

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards