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

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

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