Question

4. The stock market goes up more often that it goes down. Assume the probability that the stock market goes up in a randomly selected month is 60%. a. What is the probability the stock market goes up 12 months in a row? UUUUUUUUUUUU b. What is the probability it goes down 12 months in a row? DDDDDDDDDDDD c. What is the probability it goes up six months in a row and then goes down six months in a row? Note this is one specific pattern UUUUUUDDDDDD. d. How many different ways are there for the stock market to go up six times in 12 months? Hint: there is a function in R for “n choose k”; the syntax is “choose(n, k)”. e. Use your answers to the prior two questions to calculate the probability that the stock market goes up exactly six months out of 12 in any order. f. What is the probability the stock market goes up 6 or more months in a year? Hint: repeat the prior calculation for 7, 8, 9, 10, 11, and 12 times. Finally, calculate the probability for the union of these mutually exclusive events. g. What are the expected value and the standard deviation of the number of months the stock market rises in a year (12 months)?

Answer 1

Answer:

a: 0.0022

b: 0.00002

c: 0.0002

d: 924

e:  0.0025

f: 0.6652

g:  Expected value 7.8 months, standard deviation 1.77 months

Step-by-step explanation:

This is a binomial situation.  Either the market goes up, or it doesn't.  (It could stay the same, but it doesn't offer that as an option, so we don't include it).

When calculating probability of a binomial situation, use the formula

(nCr)(p^r)(q^(n-r))   where p is the probability of success, q is the probabilty of failure, n is the the number of times the situation occurs, and r is the desired number of successful outcomes.  The function nCr tells us how many ways you can choose r items from n total items.  In this case, we are choosing some number of months, r, from 12 months, n

For our situation we have:  n = 12, r = 12, p = 0.6, q = 0.4,

For a:   (12C12)(0.6^12)(0.4^0) = (1)(0.0022)(1) = 0.0022

We want all 12 months of success and no months of failure

For b:   (12C0)(0.6^0)(0.4^12) = (1)(1)(0.00002) = 0.00002

We want all months of failure, and no months of success

For c:   (0.6^6)(0.4^6) = 0.0002

We don't need the nCr function because it tells us that it we want 6 months of success followed by 6 months of failure.  There's only 1 way to do that, so there's nothing to count

For d:  12C6 = 924  

There are 924 different ways to the market can go up 6 times out of 12 months

For e:  (12C2)(0.6^2)(0.4^10) = (66)(0.36)(0.0001) = 0.0025

For f:  Add the following probabilities together...

(12C7)(0.6^7)(q^5) + (12C8)(0.6^8)(q^4) + (12C9)(0.6^9)(q^3)

+ (12C10)(0.6^10)(q^2) + (12C11)(0.6^11)(q^1) + (12C12)(0.6^12)(q^0)

= 0.6652

They are mutually exclusive, the probability for the union is zero because they can't happen at the same time

For g:  Find expected value with the formula E = np  , here n = 12, p = 0.6, so

E = 12(0.6) = 7.8 months

Find the standard deviation with the formula s = √(npq), or s = √(Eq), here q = 0.4, so s = √(7.8(0.4)) = 1.77 months