Every trading day

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Every trading day

by BTGmoderatorDC » Sun Feb 11, 2018 7:20 am
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

What is the easiest way to solve this problem?

OA C

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by EconomistGMATTutor » Sun Feb 11, 2018 9:16 am
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

What is the easiest way to solve this problem?

OA C
Hi lheiannie07,
Let's take a look at your question.

The price of CF Corp stock either goes up by $1 or goes down by $1 with equal likelihood, it means that,
$$p\left(Up\right)=\frac{1}{2}$$
$$p\left(Down\right)=\frac{1}{2}$$

We need to find the probability that the price of CF Corp stock is up by exactly $3 from its initial price at the end of 5 trading days.
The price can be $3 at the end of 5 trading days only if during the 4 trading days the price goes up and during on day the price goes down by $1.
For 4 days up the price will be $4 and for 1 day down the price will be $4 - $1 = $3
It means we need to find the probability P(UUUUD) where U represents up price and D represents down price.
Since we have only two outcomes "Up" and "Down", we can solve this using binomial theorem.
$$=\left(nCk\right)p^kq^{n-k}$$
$$=\left(5C4\right)p^4q^{5-4}$$
$$=5\left(\frac{1}{2}\right)^{^4}\left(\frac{1}{2}\right)$$
$$=5\left(\frac{1}{16}\right)\left(\frac{1}{2}\right)=\frac{5}{32}$$

The other ways to find the probability is to find the number of ways , we get 4 ups and 1 down during 5 days. It can be like
DUUUU or
UDUUU or
UUDUU or
UUUDU or
UUUUD or
So there can be 5 ways of getting 4 up days and 1 down day.
We will just multiply this 5 after finding the probability.
$$\Pr\left(Getting\ 4\ Up\ days\ and\ 1\ down\ day\right)=5\times\Pr\left(Getting\ 4\ Ups\right)\times\Pr\left(Getting\ 1\ down\right)$$
$$\Pr\left(Getting\ 4\ Up\ days\ and\ 1\ down\ day\right)=5\times\left(\frac{1}{2}\right)^{^4}\times\left(\frac{1}{2}\right)$$
$$\Pr\left(Getting\ 4\ Up\ days\ and\ 1\ down\ day\right)=\frac{5}{32}$$

Therefore, option C is correct.

Hope it helps.
I am available if you'd like any follow up.
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by [email protected] » Sun Feb 11, 2018 11:11 am
Hi lheiannie07,

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. Since each day involves either an increase or a decrease, there are 2^5 = 32 possible arrangements. 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 EXACTLY $3, we need 4 days of increases (out of the 5 total days). This can occur 5 different ways:

IIIID
IIIDI
IIDII
IDIII
DIIII

Thus, the answer to the question is 5/32

Final Answer: C

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by Scott@TargetTestPrep » Mon Feb 19, 2018 3:07 pm
lheiannie07 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

Scott Woodbury-Stewart
Founder and CEO
[email protected]

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