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 5-day sequences that give us a $3 increase, then we compare that number to the total number of 5-day 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 5-day 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: \(P = C^k_n*p^k*(1-p)^{n-k}\)

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#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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