Problem Solving

Suppose that the probability is 0.5 that the value of the U.S.Dollar will rise against the Japanese yen over any given week, and that the outcome in one week is independent of that in any other week. What is the probability that the value of the U.S. Dollar will rise against the Japanese yen in 4 of the 7 weeks? What is the probability that it will rise against the yen in a majority of weeks over the period of 7 weeks?

There is a formula for this kind of probability:

P(success)^n * P(failure)^n * nCs
n=number of attempts = 7
nCs = number of combinations in which the requirement is met, in this case we choose 4 out of 7:

(0.5)^4 * (0.5)^3 * 7C4

1/16 * 1/8 * (7*6*5)/2*3

1/16 * 1/8 *7*5

1/128 * 35 = 35/128

This is the answer for the first part. The second part ask about the majority, so it asks about he probability that it will rise 4 or 5 or 6 or 7 times.
We can find each and sum them up. We can also, to save time, find the probability that it will not happen, and substruct this from 1.

So let's find the probability for 1,2,3 rises.

(0.5)^1 * (0.5)^6 * 7 = 1/2*1/64*7 = 1/128*7 = 7/128
(0.5)^2 * (0.5)^5 * 7C2 = 1/4*1/32* 7*3 = 21/128
(0.5)^3 * (0.5)^4 * 7C3 = 1/8*1/16* 7*5 = 35/128
(7+21+35)/128 = 63/128 This is minority!
Remember! 1-63/128 = 65/128

This formula was very helpful! But I noticed one thing--you forgot to account for the probability that it rose 0 times, which is 1/128, making the odds 64/128 or .5.